ALTERNATING CURRENT CHAPTER 7
Chapter Outline
7.1 AC Generation
7.2 Alternating Voltage Applied to a Resistor, Inductor, and Capacitor
7.3 Electrical Resonance Series L-C-R Circuit
7.4 L-C Oscillations
7.5 Transformer
Electricity and magnetism were considered as separate and unrelated. When a resistor is connected across the terminals of a battery, a current is established in the circuit. The current has a unique direction, it goes from the positive terminal to the negative terminal via the external resistor. The magnitude of the current also remains almost constant. This is called direct current (DC).
If the direction of the current in a resistor or in any other element changes alternately, the current is called an alternating current (AC). In this chapter, we shall study the alternating current that varies sinusoidally with time.
7.1 AC GENERATION
To get sinusoidally varying alternating current, we need a source which can generate sinusoidally varying emf. An AC generator, also called an AC dynamo, can be used as such a source. It converts mechanical energy into electrical energy, producing an alternating emf, by rotating its coil.
The total emf induced in the coil at any instant t is given by E= E0sin w t
We see that the emf varies sinusoidally with time with an angular frequency w and hence, with a time period π = ω 2 T . The maximum magnitude of the emf, known as peak emf, is E0.
We know,
When t = 0, then E = 0
When , 4 T t = then
When , 2 T t = then
When 3 , 4 T t = then
When t = T, then
If the instantaneous value of alternating emf is E=E 0 sin w t, then the corresponding instantaneous value of alternating current is given by I = I 0 sin wt, where I 0 and E 0 denote the peak values of current and emf, respectively.
Each cycle of AC consists of two half cycles. For one half cycle, the current is entirely positive whereas during the next half cycle, the current is entirely negative. One complete cycle of alternating current or alternating voltage is shown in figure (a) .
The instantaneous value of alternating current or alternating emf may also be represented by
I = I 0 cos w t and E = E0 cos w t
Figure (b) represents the alternating current (or) alternating emf as cosine function of time.
7.1.1 Alternating Current and Voltage
A lternating current is that current whose magnitude changes with time and direction reverses periodically.
Similarly, the voltage whose magnitude and direction change with time and attains the same magnitude and direction after definite time intervals is called alternating voltage.
The symbol used to represent the AC source in a given circuit is as shown in figure.
Other forms of AC: e or i t
Triangular saw-tooth form of AC t
Square form of AC
The maximum value of current or voltage in an AC circuit is called its peak value or crest value or amplitude.
The value of current or voltage in an AC circuit at any instant of time is called its instantaneous value. Instantaneous current,
=ω=ω+φ 00 sin()sin()IItorIIt
Instantaneous voltage,
=ω=ω+φ 00 sin()sin()EEtorEEt where ( w t + φ ) is called phase.
Advantages and Disadvantages of Alternating Current over Direct Current
Advantages:
i) The cost of generation of AC is less than that of DC.
ii) AC can be conveniently converted into DC with the help of rectifiers.
iii) ac is available in a wide range of voltages. These voltages can be easily stepped up or stepped down with the help of transformers.
iv) AC appliances are simple, robust, and require less care as compared to DC devices.
v) By supplying AC at high voltages, we can minimise transmission losses or line losses.
Disadvantages:
i) AC is more dangerous than DC.
ii) AC is transmitted more by the surface of the conductor. This is called skin effect. It is for this reason that several strands of thin insulated wire, instead of a single thick wire, need be used.
iii) For electroplating, electro-refining, electrotyping, only DC can be used but not AC.
1. Write the general equation for the instantaneous emf of a 50 Hz generator whose peak voltage is 250 V.
Sol: Here, E0 = 250V and ν = 50Hz
ω=πν=π×=π 2250100 and E = E0 sin w t = 250 sin 100 π t
Try yourself:
1. Equation of an alternating voltage is V = 100 sin 100 π t volt. Find the value of voltage at 1 s.
200 t =
Key Insights:
Ans: 100 volt
■ Since alternating current varies continuously with time, its effect is measured by defining either the mean value (average value ) of AC or by defining root mean square value or virtual value of A C.
■ Average value of a function f(t) from t 1 to t2 is defined as
Thus, we see that the average value of AC over one complete cycle is zero.
Similarly, we can prove that the average value of alternating voltage over one complete cycle is also zero.
The reason is that alternating current or voltage during first half cycle ( 0 to T/2) is positive and during other half cycle (T/2 to T) is negative. Therefore, we find the mean or average value of alternating current or voltage only over any half cycle.
Mean or Average Value of AC for Half Cycle
It is an arithmetic average of all the values of current in a sine wave for half cycle.
We can also find the value of ∫
fdt
graphically.
Average value is the area of f-t graph from t1 to t2
7.1.2 Mean (or) Average Value and RMS Value of AC
The value of current at any instant t is given by I = I 0 sin w t.
The average value of a sinusoidal wave over one complete cycle is given b y
The value of steady current which sends the same amount of charge through a circuit in a certain time interval as is sent by an alternating current through the same circuit in the same time interval (half the time period or half cycle) is known as mean or average value of alternating current over half cycle. (Either positive half cycle or negative half cycle)
The value of current at any instant t is given by I = I 0 sin w t . Hence, the mean or average value for half cycle is given by
Thus, the mean or average value of AC over positive half cycle is + 0.637 I 0 and that over negative half cycle is – 0.637 I0.
Similarly, for alternating voltage over half cycle also, the same relation holds good.
Key Insights:
■ I or E A t B Between A and B, Iavg = 0 Eavg = 0 (+) ()
■ Between A and B, Iavg = 0 Eavg = 0 B A () (+) I or E t
■ I or E A
B Between A and C or C and B (+)
2. An alternating current is represented by the equation I = I 0 cos w t. Find its average value over the time interval t = 0 to t = π /2 w
Try yourself:
2. An alternating current is represented by I = I0cos100 πt. If its average value over half cycle is 25/π A, find the amplitude of current.
Ans: 12.5 A
7.1.3 Root Mean Square (RMS) Value (or) Effective Value (or) Virtual Value of Alternating Current (irms)
The instantaneous current I may be positive or negative at a given instant but the quantity I2 is always positive. So, its average is always positive. Therefore, we calculate the average of I2 over a complete period and then take the square root of it. This gives the root mean square or rms current.
It is the square root of the average of squares of all the instantaneous values of current over one complete cycle.
Root mean square value of AC is defined as that steady current which produces the same amount of heat in a conductor in a certain time interval as is produced by the AC in the same conductor during the same time.
It is represented by I rms
Root mean square value of AC is also known as effective value ( Ieff) or virtual value (Iv).
Let us now calculate th e value of average of I
Clearly, rms value of an alternating current is 70.7% of its peak value.
Similarly, rms or virtual or effective value of alternating emf is given as
7.1.4 Form Factor and Peak Factor of AC
It is defined as the ratio of rms value to the average value during half cycle of alternating current or voltage.
i.e., Formfactor rmsrms avgavg IE IE ==
We know that = 0 2 rms I I and 0 2
General Expression for RMS Current
i) Suppose the current flowing through a circuit element is given by i = a+b cos wt, where a and b are constants, then its value is given by
ii) Suppose the current flowing through a circuit element is () cossin.iatbt =ω+ω
Then, its rms value is given by
iii) Suppose the current flowing through a circuit element is given by () cossin.iabtct =+ω+ω
Then, its rms value is given by
Peak factor: It is the ratio of peak value to the rms value of alternating current or voltage.
i.e., peak factor 00 rmsrms IE IE ==
3. A current is made of two components DC component i1 = 3 A and an AC component =ω 2 i42sint . Find the reading of hot wire ammeter.
Sol: ==+ω 12 ii+i342sint +ω ==
∫ 22 2 00 0 .(342sin) TT rmsT idttdt i T dt =+ω+ω ∫ 22 0 1 (9242sin32sin) T rms ittdt T ∴ i rms = 5 A
4. The electric current in a circuit is given by i = 3t. Find the rms current for the period t = 0 to t = 1 s.
Sol: i = 3t; i2 = 9t2 1 22 2 00 1 00 .9tdt 3
Try yourself:
3. An alternating current is given by =ω+ω i(3sint4cost)A . Find rms current.
Ans: 5 A 2
7.1.5 Power in AC Circuits
In DC circuits, power is given by P=VI. But in AC circuits, since there is some phase angle between voltage and current, therefore power is defined as the product of voltage and that component of the current which is in phase with the voltage.
Thus, P=EIcos φ , where E and I are rms values of voltage and current.
Power factor: The quantity cos φ is called power factor.
Instantaneous power: Suppose E = E 0sin w t and I = I0sin( w t + φ ) in a circuit then Pinstantaneous = EI = E 0 I0sin w t sin( w t + φ )
Average power (True power): The average of instantaneous power in an AC circuit over a full cycle is called average power. Its unit is watt.
Average power over complete cycle,
00 cos 2 ave WEI P T
EI EI
=×φ=φ 00 coscos 22 rmsrms
EI PEI
∴=φ=φ 00 cos.cos 2 avermsrms
PEI
=φ=φ 00 cos.cos 22 avermsrms EI
The component of current along emf is I cos φ, so the component I rms cos φ contributes to the power consumption. But the component I rms sin φ does not contribute to the power and it is called wattless component of current.
Apparent or virtual power: The product of apparent voltage and apparent current in an electric circuit is called apparent power. This is always positi ve
== 00 2 apprmsrms EI PEI
5. A vertical circular coil of radius 8 cm and 20 turns is rotated about its vertical diameter with angular velocity 50 rev/s in a uniform horizontal magnetic field of 3×10 –2 T. If the coil forms a closed loop of resistance, the average power loss due to joule’s heating effect is (R = 10 Ω )
Sol: emf induced in coil is E=NBA w . Sin( w t)
Max emf E 0 =NBA w=πω 2 ()NBr
=×××π×××π 24 0 2031064102(50) E
××π××× 224 60106410100 = 3.84 V
Max current, 0 0 0.384A E i R ==
avgrmsrms PEiEi =××Φ=×
Average power loss, 00 1 cos 2 cos1
= 2 0 1 2 E R = 0.737 W
Try yourself:
4. The rms value of voltage of an alternator is 200 V and the rms current delivered by it to a load is 3A. If the phase angle between the current and voltage is π /6, find the power.
Ans: 519.6 W
7.1.6 AC Instruments
I f ordinary DC measuring instruments (ammeter and voltmeter) are used for measuring alternating current and alternating voltage, they will show zero reading. This is because the average of positive and negative half cycles in AC is zero.
On the other hand, hot wire instruments depend on the heating effect caused by electric current for their functioning. Whether the current is flowing in one direction or in the reverse direction, heating effect will be there. This is because the heat produced is independent of the direction of flow of current. Moreover, heat produced is proportional to square of current and is always positive.
Hence, hot-wire instruments are largely employed in measuring AC voltages and currents. These instruments measure the virtual voltage or virtual current as the case may be. The alternating current is measured in ‘virtual ampere’ and the alternating voltage is measured in ‘virtual volt’.
AC ammeter and AC voltmeter reads the rms values of alternating current and voltage respectively. Their common name is ‘hot wire meters.’
One virtual ampere is that value of alternating current which when flowing through a resistance for a certain time generates the same quantity of heat as is done by a steady (direct) current of one ampere through the same resistance for the same time.
One virtual volt is that value of alternating emf which when applied across a resistance for a certain time generates the same amount of heat as is generated by a steady (direct) potential difference of one volt applied across the same resistance for same time.
Key Insights:
■ The given values of alternating current and voltage are virtual values unless otherwise specified. AC instruments can be used to measure both AC and DC. While the scales of DC instruments are linearly graduated, the scales of AC instruments are not evenly graduated.
7.1.7 Phasor and Phasor Diagrams
Let us consider a circuit in which emf and current vary sinusoidally with time and are mathematically expressed as
E=E 0 sin w t and I=I 0 sin( w t + φ )
Where φ is the phase angle between alternating emf and current, and E and I are the peak values of alternating emf and current.
The instantaneous values E and I may be regarded as projections of E 0 and I 0 respectively.
When a quantity varies sinusoidally with time and can be represented as projection of a rotating vector, then it is called as phasor.
A diagram, representing alternating emf and current (of same frequency) as rotating vectors (phasors) with phase angle between them is called as phasor diagram.
Fig
In figure (a), OA and OB represent two rotating vectors having magnitudes E 0 and I 0 in anti-clockwise direction with same angular velocity w . OM and ON are the projections of OA and OB on y -axis, respectively. So, 0 sin OMEt =ω and ON = () ω+φ 0 sin It , represent the instantaneous values of alternating emf and current. BOA ∠=φ represents the phase angle by which current I 0 leads the alternating emf E0. Figure (a) also represents the phasor diagram.
The phasor diagram, in a simple representation is shown in figure (b). Moreover, when we are interested only in phase relationship, the phasor diagram may also be represented as shown in figure (c).
4. The equation of the alternating voltage applied to our houses is e =(311sin100 πt ) volt. A hot wire instrument is used to measure the alternating voltage. The rms value of emf is
(1) 219.9 V (2)
(b)
TEST YOURSELF
1. The rms value of an alternating current, which when passed through a resistor produces heat three times of that produced by a direct current of 2 A in the same resistor, is
(1) 6 A (2) 3 A (3) 2 A (4) 23A
2. A 220 V main supply is connected to a resistance of 100 kΩ. The rms current is (1) 22 mA (2) 2.2 mA (3) 220 mA (4) 10 mA
3. The peak value of an alternating emf is given by E=E0cosωt is 10 volt and its frequency is 50 Hz. At a time t=(1/600) s, the instantaneous value of the emf is :
(1) 1 V (2) 5 V
(3) 53V (4) 10 V
V (3) 543.7 V (4)
Answer Key
V
(1) 4 (2) 2 (3) 3 (4) 1
7.2 ALTERNATING VOLTAGE APPLIED TO A RESISTOR, INDUCTOR, AND CAPACITOR
When an AC voltage is applied to a circuit containing a resistor, an inductor or a capacitor, each element responds differently due to its unique electrical properties.
7.2.1
Alternating Voltage Applied to a Resistor
An AC source applied to a resistor of resistance R is shown in figure. Such a circuit is known as a purely resistive circuit.
The alternating emf applied is given by 0 sin EEt =ω .....(i)
Let I be the current in the circuit at any instant t. The potential difference across resistor = IR.
So, or E EIRI R == Using equation (i), we get 0 0 sinorsin E ItIIt R =ω=ω ...(ii) where 0 0 E I R = is the peak value of alternating current.
Comparison of equations (i) and (ii) shows that the emf and current across the resistor are in phase.
The phasor diagram for pure resistance is shown in figure below
In a purely resistive circuit, there is no phase difference between emf and current i.e., emf and current will be in phase and have the same frequency but with peak emf more than peak current.
Resistance (R)–frequency (ν) of applied AC source (V) curve of a purely resistive circuit is shown in figure below. Clearly value of resistance R does not change with change in frequency.
Skin Effect
A direct current flows uniformly throughout the cross-section of the conductor. An alternating current, on the other hand, flows mainly along the surface of the conductor. This effect is known as skin effect. The reason is that when alternating current flows through a conductor, the flux changes in the inner part of the conductor are higher. Therefore, the inductance of the inner part is higher than that of the outer part. Higher the frequency of alternating current, more is the skin effect.
The depth up to which AC flows through a wire is called skin depth.
We know that the phase difference between voltage and current is 0o. The instantaneous power dissipated in the resistor is
(1)
Since AC of high frequency do not pass through the entire cross-section of the wire, the effective resistance of the wire for AC is much greater than that for DC. Hence, the conductor required to carry high frequency
AC consists of a number of strands of fine wire connected in parallel at their ends, and insulated throughout their length from each other. This increases the surface area and thus decreases the resistance .
Key Insights:
■ In te rms of rms values, the equation for power and relation between current and voltage in AC circuits are essentially the same as those for the DC case.
6. An alternating current is flowing through a resistor. The current measured on AC ammeter is 5 A and the voltage across the resistor measured by an AC voltmeter is 10 volt. Find the power dissipated in the resistor.
Sol: I rms = 5 A, V rms = 10 volt
∴ P avg = 10 × 5 watt = 50 watt.
Try yourself:
5. In the arrangement shown, reading of the AC voltmeter is 60 volt. Find the power dissipated in 5 Ω resistor.
The alternating emf applied is given by =ω 0 sin EEt .....(1)
The induced emf across the inductor ., L dI VL dt =− which opposes the growth of current in the circuit. As there is no potential drop across the circuit, so
.0 dI EL dt or, = dI LE dt or, = dIE dtL
Using equation (i), we get
00 sinor, sin dIEEtdItdt dtLL =ω=ω
Integrating both sides, we get =ω=ω∫∫∫ 00 sin sin EE dItdttdt LL () 00 cos or, cos EE t It LL −ω
Since, ()π −ω=ω− cossin 2 tt and peak
value of current, 0 0 = ω E I L = ω rms rms E I L
7.2.2 Alternating Voltage Applied to a Pure Inductor
An alternating voltage applied to an inductor of inductance L is shown in figure below. Such a circuit is known as purely inductive circuit.
∴ π =ω− 0 sin 2 IIt ....... (2)
Comparison of equations (1) and (2) shows that current lags behind the emf by an angle of π 2 in a purely inductive circuit. In other words, phase difference between alternating voltage and alternating current is π 2 in an AC circuit containing only an inductor as shown in the figure.
The phasor diagram for purely inductive circuit is shown in figure. O Y X O X Y E E (OR) I I
I nductive Reactance [X L ]
The opposition offered by an inductor to the flow of AC is called an inductive reactance.
Here, 0 0 E I L = ω .
Comparing it with 0 0 E I R = we conclude that, (wL) has the dimensions of resistance. The term ( w L) is known as inductive reactance.
∴ Inductive reactance is given by relation.
=ω=πν 2 L XLL where ν is frequency of AC.
Key Insights:
■ We have (2), L XLL=ω=πν∝ν L X
■ So, X L increases with the increase in the frequency of AC.
■ For high frequency of AC, →∞ L X , i.e., inductor behaves as good as an open circuit (i.e., I = 0)
■ Graphs between XL and ν and XL and L are as shown below
■ For dc, ν=0∴ XL = 0
For AC of high frequencies, ≈∞ L X
Hence, an inductor offers an easy path for DC and resistive path for AC.
i.e., DC can flow easily through the inductor. Inductor offers finite resistance to the flow of AC.
■ The SI unit of inductive reactance X L is ohm.
■ Variation of voltage and current with time in a purely inductive circuit can also be represented as shown in figure.
E
t I
■ Practically an inductor always possesses some resistance because of the wire used to make its coil. Therefore, a pure inductor is only a theoretical concept.
■ We can have resistance without inductance but we can not have inductance without resistance.
Power
The instantaneous power supplied to the inductor is, (knowing that E is the AC voltage across the inductor)
Power = P = π
=ωω−
00 sinsin 2 EIEItt
=−ωω=ω 00 00 sincossin2 2 EI EIttt
Average
Average power = 0
So, there is no power consumption in a purely inductive circuit.
Energy Cycle in an Inductor
The average power supplied to an inductor over one complete cycle is zero. Figure below explain the magnetisation and the demagnetisation of an inductor.
and the net flux becomes zero at the end of a half cycle. The voltage E is negative (since di/dt is negative). The product of voltage and current is negative, and energy is being returned to source
i) Current i through the coil entering at A increases from zero to a maximum value. Flux lines are set up, i.e., the core gets magnetised. With the polarity, shown voltage and current are both positive. So their product is positive. Energy is absorbed from the source.
iii) Current i becomes negative, i.e., it enters at B and comes out of A. Since the direction of current has changed the polarity of the magnet changes. The current and voltage are both negative. So, their product is positive and energy is absorbed.
ii) Current in the coil is still positive but is decreasing. The core gets demagnetised
iv) Current i decreases and reaches its zero value at 4 when core is demagnetised and flux is zero. The voltage is positive but the current is negative. The power is, therefore, negative. Energy absorbed during the 1/4 cycle is returned to the source in cycle 2-3.
7. The voltage applied to a purely inductive coil of self inductance 15.9 mH is given by the equation V = 100 sin 314t + 75sin942t + 50sin1570t. Find the equation for current wave.
Sol: The standard equation of the voltage can be written as 011022033 VVsintVsintVsint ωωω=++
On comparing this with the given equation, we have
ω==ω 1 L11 rad 314,XL s
3 31415.9105 =××=Ω
ω==ω 2 L22 942rad/s,XL 3 94215.91015 =××=Ω and XL3 3 157015.91025 =××=Ω
Hence, : 01 1 1 100 20 A, 5 o L V i X ===
02 2 2 75 5A, 15 o L V i X === and
03 3 3 50 2A. 25 o L V i X ===
Thus 011022 033 sin(-)sin(-) sin(-) =ωφ+ωφ +ωφ iitit it
20sin314t5sin942t 22
2sin1570t 2
Try yourself:
6. Find the maximum value of current when a coil of inductance 2 H is connected to 150 V, 50 cycles/s supply.
Ans: 0.337 A
7.2.3 Alternating Voltage Applied to a Pure Capacitor
Alternating voltage applied to a capacitor is shown in figure. Such a circuit is known as purely capacitive circuit.
0 sin EEt =ω
The alternating emf applied across the capacitor is given by =ω 0 sin EEt .....(1)
Let q be the charge on the capacitor at any instant.
∴ Potential difference across the capacitor, = C Vq C But =⇒==ω 0 sin C q VEEEt C (or) =ω 0 sin qECt
Now, ==() ω 0 sin dqd IECt dtdt ()() =ωω=ω ω 0 0 cos cos 1/ E ECtt C
Since, π ω=ω+ cossin 2 tt
The peak value of current =
∴=ω+ 0 sin 2 IIt .....(2)
Comparison of equations (1) and (2) shows that current leads the emf by an angle π 2 in a purely capacitive circuit.
The phasor diagram for purely capacitive circuit is as shown in figure.
900 O X Y X Y E E O (or) I I
Capacitive Reactance (XC )
The resistance offered by a capacitor to the flow of AC is called capacitive reactance.
Here, 0 0 1 E I C =
,
Comparing it with = 0 0 , E I R
We conclude that ω 1 C has the dimensions of resistance. The term ω 1 C is known as capacitive reactance.
∴ capacitive reactance, == ω×π 11 2 C X CCv
Key Insights:
■ We have ==⇒∝ ω×πνν 111 2 CC XX CC
So XC decreases with the increase in the frequency of AC.
For AC of high frequencies, → 0 C X i.e, capacitor acts like a high pass filter or a conducting wire.
■ Graphs of X C and ν and X C and C are as shown below
■ For DC , ν=0∴==∞ 1 0 C X
For AC of high frequencies, → 0 CX
Thus, capacitor offers infinite resistance to the flow of DC. So DC cannot pass through the capacitor, however small the capacitance of the capacitor is.
But capacitor offers small opposition to the flow of AC. So AC can be considered to pass through it easily.
Hence, a capacitor behaves as an open circuit for DC and it acts like a high pass filter or a conducting wire at high frequencies of AC.
■ The SI unit of capacitive reactance is ohm.
■ The variation of voltage and current with time in a purely capacitive circuit can also be represented as shown in figure.
t I E
■ As ω→→→∞ 0(),0,LCdcXX
■ As ω→∞→∞→ (),,0 LCacXX
Power
The instantaneous power supply to the capacitor P = π =ωω+ 00 sinsin 2 EIEItt =ω 00 sin2 2 EI t
Average power =
Average power = 0
So, there is no power consumption in a purely capacitive circuit.
Energy Cycle in a Capacitor
Average power supplied to a capacitor over one complete cycle is zero.
ii) The current i reverses its direction. The accumulated charge is depleted, i.e., the capacitor is discharged during this quarter cycle. The voltage gets reduced but is still positive. The current is negative. Their product, the power is negative. The energy absorbed during the first 1/4 cycle is returned during this quarter.
i) The current i flows as shown and from the maximum at 0, reaches a zero value at 1. The plate A is charged to positive polarity while negative charge q builds up in B reaching a maximum at 1 until the current becomes zero. The voltage = / c VqC is in phase with q and reaches maximum value at 1. Current and voltage are both positive.
One complete cycle of voltage/current. Note that thecurrent leads the voltage.
iii) As i continues to flow from
So P = Vci is positive. Energy is absorbed from the source during this quarter cycle as the capacitor is charged.
A to B, the capacitor is charged to reverse polarity, i.e., the plate B acquires positive and A acquires negative charge. Both the current and the voltage are negative. Their product p is positive. The capacitor absorbs energy during this 1/4 cycle.
iv) The current i reverse its direction at 3 and flows from B to A. The accumulated charge is depleted and the magnitude of the voltage VC is reduced. VC becomes zero at 4 when the capacitor is fully discharged. The power is negative. Energy absorbed during iii) is returned to the source. Net energy absorbed is zero.
8. A50 µF capacitor is connected to a 100 V, 50 Hz AC supply. Determine the rms value of the current in the circuit.
Sol: Here, 65 50 F5010F510F C =µ=×=×
E rms = 100 V, ν = 50 Hz
Capacitive reactance,
ω×πν 11 2 C X CC = ×××× 5 1 51023.1450 = 63.69 Ω
== 100 63.69 rms rms C I X = 1.57 A
Try yourself:
7. A 10 µF capacitor is connected to an AC supply given by V = 100 sin100 tV . Then write the expression for instantaneous current.
Ans: 0.1cos100 t A
7.2.4 Impedance and Admittance
Impedance (Z): The opposition offered by all the circuit elements to the flow of AC is called impedance (z). Impedance (Z) can be defined as the ratio of peak (or) rms value of applied emf to peak (or) rms value of current in an AC circuit.
i.e., == 0 0 rms rms EE Z II
The SI unit of impedance of the circuit is ohm.
Admittance (Y): Reciprocal of impedance of a circuit is called admittance of the circuit.
i.e., admittance (Y) = 1 Z
SI unit of admittance (Y) of the circuit is ohm-1, i.e., mho or siemens.
7.2.5 Alternating Voltage Applied to L-R Series Circuit
Consider an AC source of emf E = E 0 sin w t connected to a series combination of a pure inductor of inductance L and a resistor of resistance R as shown in figure.
fig (a)
Let I be the rms value of current flowing through the circuit. The potential difference across the inductor is given by,VL = IXL.....(1)
Current I lags the voltage VL by an angle of π 2 , when AC flows through the inductor.
In figure (b), VL is represented by OB along y-axis and current I along x-axis. The potential difference across the resistor,
VR = IR.....(2)
Since, the voltage and current are in phase when AC flows through the resistor, so VR is represented along x-axis (the direction along which I is represented).
(or) (or) R
fig (b)
Therefore, the resultant of V L and V R is represented by OC and is given by
22 22 or, RLOCOAOBEVV =+=+
Using equations ( 1 ) and (2), we get 222222 LL EIRIXIRX =+=+
where L XL =ω is the inductive reactance. 22 or, LLR EE I RXZ == + .....(3)
From the above equations of E and I we have =+=+ω 22222 LRL ZRXRL
where ZLR is effective resistance offered by the LR circuit to ac, and it is called impedance of LR circuit.
Let φ be the angle made by the resultant of VL and VR with the x-axis, then from figure, we get
φ==== tan LL R ACOBVIX OAOAVIR
or, φ==ω tan L XL RR
In series LR circuit, emf leads the current or in other words, the current is said to lag behind the emf by an angle φ given by the above equation.
∴ Current in L-R series circuit is given by ==ω−φ 0 sin() LRLR EE It ZZ (or) =ω−φ 0 sin()IIt power factor
φ==== +ω 2 2 cos R VIRRR EIZZRL
Average power of L-R circuit is =φ cos avermsrms PEI
Key Insights:
■ =+ω=+×π2222222 4 LR ZRLRLv
Variation of z with w is as shown in the figure. Z R O w
Thus, ZLR increases with the frequency of ac, so ZLR is low for lower frequency of AC and high for higher frequency of AC.
■ The phase angle between voltage and current
1 2 tan L R increases with the increase of frequency of AC.
Choke Coil
Choke coil (or ballast) is a device having high inductance and negligible resistance. It is used to control current in AC circuits and is used in fluorescent tubes. The power loss in a circuit containing choke coil is least.
If AC source is directly connected to a mercury tube, the tube will be damaged. To avoid this a choke coil is connected in series with the tube. This is a simple L-R circuit with impedance.
=+ω222 ZRL .
If the applied voltage is =ω 0 sin EEt , the peak current through the circuit is = +ω 0 0 222 E I RL
The rms current is given by
The rms voltage across the resistor is
rmsrmsR RE VRI RL
If choke coil were not used, the voltage across the resistor would be the same as the applied voltage. Thus, by using the choke coil, the voltage across the resistor is reduced by a factor
Rrmsrms RE VRI RL
The advantage of using a choke coil to reduce the voltage is that an inductor does not consume power. Hence, we do not lose electrical energy in the form of heat. If we connect an additional resistor in series with the tube, to reduce the voltage instead of a choke coil, power will be lost due to Joule’s heating (H = i2Rt).
7) Choke coil for different frequencies are made by using different substances in their core.
For low frequency, L should be large. Thus, iron core choke coil is used. For high frequency AC circuit, L should be small, so air cored choke coil is used.
9. A 50 V, 10 W lamp is run on 100 V, 50 Hz AC mains. Calculate the inductance of the choke coil required.
Sol: Voltage marked on lamp, V = 50 V Power, P = 10 W
∴ Resistance of lamp, × ===Ω 2 5050 250 10 V R P
Current rating of lamp , 101 A 505 rms P I V ===
The given circuit is equivalent to LR circuit.
Here E rms = 100 V, ν = 50 Hz
When the lamp is worked on a.c., the impedance of the circuit is
1) It consist of a Cu coil wound over a soft iron laminated core.
2) Thick Cu wire is used to reduce the resistance (R) of the circuit.
3) Soft iron is used to improve inductance (L) of the circuit.
4) The inductive reactance or effective opposition of the choke coil is given by XL = ω=π 2 LvL
5) For an ideal choke coil, r = 0, no electric energy is wasted, i.e. average power P = 0.
6) In actual practice, choke coil is equivalent to a R-L circuit.
187500 1.9 98596 or, L = 1.38 H
Try yourself:
8. A series combination of a coil of inductance L and a resistor of resistance 12 Ω is connected across a 12 V, 50 Hz supply. Calculate L if the circuit current is 0.5 A.
Ans: 0.066 H
7.2.6 C-R Series Circuit with Alternating Voltage
Let an alternating source of emf E =E0sinwt is connected to a series combination of a pure capacitor of capacitance (C) and a resistor of resistance (R) as shown in figure (a).
VC VR R C E fig (a)
Let I be the rms value of current flowing through the circuit.
The potential difference across the capacitor,
VC = IXC.....(i)
The current leads the emf by an angle of π/2 when AC flows through the capacitor. So in figure (b), VC is represented by OB along negative y-axis and the current I is represented along x-axis.
The potential difference across the resistor,
VR = IR.....(ii)
The emf and current are in phase when AC flows through the resistor. So, VR is represented by OA along x-axis.
Therefore, the resultant potential difference of VC and VR is represented by OC and is given by () 22 22 or RCOCOAOBEVV =+=+
Using equations (i) and (ii), we get =+=+222222 CC EIRIXIRX
where = ω 1 C X C is the capacitive reactance. or == + 22 CCR EE I RXZ ..........(3)
From the above equations of I and E, we have
2 222 1 CRC ZRXR C
Where Z C- R is the effective opposition offered by the C-R circuit to ac, which is the impedance of C-R circuit. Let φ be the angle made by E with x-axis, then from the above phase diagram, we have
φ=== tan CC R ACVIX OAVIR
or, tan φ== ω XC I RCR
In series C-R circuit, emf lags behind the current or in other words, the current is said to lead the emf by an angle φ given by the above equation.
∴ Current in C-R series circuit is given by ==ω+φ 0 sin() CRCR EE It ZZ
(or) =ω+φ 0 .sin()IIt Power Factor
2 2 cos 1 R VIRRR EIZZ R c average power of L-R circuit is =φ cos avermsrms PEI
+ ω
Key Insights:
■ Impedance of C-R circuit, =+=+ ω 222 22 1 CRC ZRXR C
=+ πν 2 222 1 4 R C
Thus, ∝ ν 1 CR Z .
For very high frequency (ν) of AC, → ZR and for very low frequency of AC, →∞ Z
■ Phase angle between voltage and current is given by φ== ωπν 11 tan 2 CRCR
As ν increases, phase angle φ decreases.
10. A series circuit contains a resistor of 20 Ω , a capacitor and an ammeter of negligible resistance. It is connected to a source of 200 V, 50 Hz. If the reading of ammeter is 2.5 A, calculate the reactance of the capacitor.
Sol: Here, R =20 Ω , E rms = 200 V, ν = 50 Hz, I rms = 2.5 A
The circuit is C-R circuit. Impedance of circuit, ===Ω 200 80 2.5 rms CR rms E Z I
But, =+22 CRC ZRX
or, =+ 222 CRC ZRX
or, =− 222 CCR XZR
or, ()()=−=− 22 22 8020 CCR XZR = 77.46 ohm
Try yourself:
9. A 10 µF capacitor is in series with a 50 Ω resistance and the combination is connected to a 220 V, 50 Hz line. Calculate (i) the capacitive reactance, (ii) the impedance of the circuit and (iii) the current in the circuit.
Ans: (i) 318.5 Ω (ii) 322.4 Ω (iii) 0.68 A
7.2.7 L -C Series Circuit with Alternating Voltage
let an alternating source of emf E = E0sinwt is connected to the series combination of a pure capacitor of capacitance (C) and an inductor of inductance (L) is shown in fig.
Let I be the rms value of current flowing the circuit
The potential difference across L is VL = I.XL
The current I lags VL by an angle π/2
The Potential difference across capacitance c is VC = I.XC
The current I leads VC by an angle π/2 .
The voltage VL and VC are represented by OB and OC respectively.
The resultant potential difference of VL and VC is
1 LC ILIZ C
From the above equations, impedance of L -C circuit is () =ω
1 LC ZL C
ω
a) If ω> ω 1 L C i.e, X L>X C then V L>V C
Potential difference V=VL–VC.
Now current lags behind voltage by π/2
b) If ω< ω 1 L C , then V L<V C Resultant potential difference ( V) = VC – VL Now current leads emf by π/2
c) If ω= ω 1 L C , then =ω−= ω 1 0 ZL C and 1 rLC
ω=ω=
Current, ==∞ E I Z .
Variation of |Z Le| with w is as shown in the figure.
|ZLC| w r w
In L-C circuit, the phase difference between voltage and current is always π/2 . Power factor, cos φ = cos π/2=0 .
So, power consumed in L-C circuit is =××φ=cos0 rmsrms PVI
∴ In L-C circuit no power is consumed
Key Insights:
■ In L-C circuit, the impedance 1 ZL C =ω− ω and current, E I Z = So, the impedance and current vary with frequency. At a particular angular frequency ( w 0 ), ω= ω 1 L C and current, = E I Z becomes maximum ( I 0 ) and resonance occurs.
At resonance, Z = 0 and 0 0 E I Z ==∞ and resonance angular frequency,
ω= 0 1 LC resonance frequency, = π 1 2 n LC
11. A 100 mH inductor and a 1000 µF capacitor are connected in series combination. For what value of supply frequency f, current in the circuit becomes infinity?
Sol: I → ∞ when z = 0 36 1 i.e., when 0 1 1111 Hz 22 1001010 = 16 Hz. L c LC fLC ω∝= ω
⇒ω= ∴== ππ××
Try yourself:
10. A 100 mH inductor and a 1000 µF capacitor is connected in series combination and the combination is connected across a 220 V, 50 Hz supply. Find the maximum value of current in the circuit.
Ans: 11 A
7.2.8 L-C-R Series Circuit with Alternating Voltage
A circuit containing pure inductor of inductance (L), pure capacitor of capacitance (C) and resistor of resistance (R), all joined in series, is shown in figure. Let E be the rms value of the applied alternating emf to the L-C-R circuit.
Let I be the rms value of current flowing through all the circuit elements.
The potential difference across resistance R is
VR = IR..........(i)
This potential difference V R is in phase with the current I.
In phase diagram,
The potential difference VR is represented by OA in the direction of I.
The potential difference across inductance L is
VL = IXL..........(ii)
The potential difference VL leads current I. by an angle of π/2
This value V L is represented by OB perpendicular to the direction of I, The potential difference across C is
VC = IXC .........(iii)
This potential difference VC lags current I by an angle of π/2
This the value V C is represented by OC perpendicular to the direction of I.
Y - Y O D E A X I VC VL VR L C (VL VC ) B or XL XC ( ( or R
Case-I:
If XL>XC, i.e., ω> ω 1 L C , then VL > VC
Since VL and VC are in opposite phase, so their resultant (VL –VC) is represented by OD
The resultant of VR and (VL –VC) is given by OL. The magnitude of OL is given by
()()=+ 22OLOAOD
()=+− 2 2 RLC VVV
or, ()=+− 2 2 RLC EVVV
()=+− 2 22 LCEIRIXIX
()=+− 2 2 LCIRXX
or, ()=+− 2 2 LC E RXX I
Now, = , E Z I the effective opposition of L-C-R circuit to AC, is called impedance (Z) of the circuit.
Impedance (Z) of L-C-R circuit is given by
()=+− 2 2 LCZRXX
Again, = E I Z () ∴= + 2 2LC E I RXX = +ω− ω 2 2 1 E RL C 1 and LC XLX C =ω= ω
Let φ be the phase angle between E and I, then from above figure,
φ=== tan LCLCLC R VVIXIXXX VIRR
Putting =ω= ω 1 and LC XLX C ω− ω φ= 1 tan L C R
∴ Current in L-C- R series circuit is given by
==ω−φ 0 sin() EE It ZZ (or)
=ω−φ 0 .sin()IIt
Since the emf leads the current by angle of φ , the L-C-R circuit is known as inductance dominated circuit.
Power factor:
Average power of L-C-R circuit is =φ cos avermsrms PEI
Case-II:
If X L< X C, i.e., ω< ω 1 L C , then V L < V C and their resultant ( V C – V L ) is represented by OB. The resultant of VR and (VC –VL) is given by OC.
Current in L-C- R series circuit is given by
==ω+φ 0 sin() EE It ZZ
(or) =ω+φ 0 .sin()IIt
Since the current leads the emf by angle of φ , the L-C-R circuit is known as capacitance dominated circuit.
Case-III:
When XL = XC or, 1 , L C ω= ω then VL = VC so, the resultant potential difference is equal to VR, i.e., E = VR = IR tan 00 CL R VV V
Thus, there is no phase difference between current and emf. Therefore, the given L-C-R circuit is equivalent to a pure resistive circuit.
The impedance of such L-C-R circuit is given by Z = R. It is minimum and is independent of the frequency of AC.
Now, power factor φ== 0 coscos01
Average power
The magnitude of OC is given by
The phase difference φ between voltage and current is given by
=φ= cos avermsrmsrmsrms PEIEI
Wattless Current or Idle Current
If the voltage and current differ in phase by π/2, then Power factor, φ== 0 coscos900 . In this case, as there is no resistance to the flow of current, the average power consumed by the current is zero. Such a current is, therefore, called wattless current. Since this current does not perform any work, therefore, this current may also be called idle current. Such a current flows only in purely inductive, purely capacitive or L–C circuits.
The average power consumed per cycle in a pure inductive or capacitive or L –C circuit is zero. Such current for zero power consumption is called Wattless current.
The average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle φ between them. The quantity cos φ is called the power factor. Let us discuss the following cases:
Case-I:
Resistive circuit: If the circuit contains only pure R. It is called resistive. In that case φ =0, cosφ =1. There is maximum power dissipation.
Case-II:
Purely inductive or capacitive circuit: If the circuit contains only an inductor or capacitor, we know that the phase difference between voltage and current is π/2. Therefore, cosφ=0, and no power is dissipated even though a current is flowing in the circuit. This current is sometimes referred to as wattles current.
Case-III:
L-C-R series circuit: In an L-C-R series circuit, power dissipated is given by P = VI cosφ where φ = tan-1(XC – XL)/R. So, φ may be non zero in a RL or RC or RCL circuit. Even in such cases, power is dissipated only in the resistor.
Case-IV:
Power dissipated at resonance in L-C-R circuit: At resonance X C – X L=0 and φ = 0. Therefore, cos φ =1 and P = I 2Z = I 2 R . That is maximum power is dissipated in a circuit (through R) at resonance.
Relation between applied pd and pd’s across the components in L-C-R circuit:
For DC,
V=VR+VL+VC (only before steady state) For ac
C R
V= IZ ()=+− 2 2 LCIRXX ()()=+− 22 LCIRIXIX ==ωLL VIXIL where ==ωLL VIXIL and
IR
Key Insights:
■ Rules to be followed for various combinations of AC circuits
a) Compute effective resistance of the circuit as R
b) Calculate the net reactance of the circuit as X = XL – XC where =ω= ω 1 , LC XLX C
■ Resistance offered by all the circuited elements to the flow of AC is impedance (Z) ()=+=+− 2 222 LCZRXRXX
■ Calculate the peak value of current as = 0 0 E I Z
■ The phase difference between emf & current can be known by constructing an AC triangle as
Variation of Z with w is shown in figure, z zmin = R ww r
Analytical Solution
The voltage equation for the circuit is 0 =sin diq LRiVEt dtC ++=ω
we know that i=dq/dt. Therefore, di/dt=d2q/dt2 .
Thus, in terms of q, the voltage equation becomes ++=ω 2 0 2 sin dqdqq LREt dtdtC
This is like the equation for a forced, damped oscillator.
Let us assume a solution =ω+θ 0 sin()qqt
So that, =−ωω+θ 0 cos()dqqt dt and =−ωω+θ 2 2 0 2 sin()dqqt dt
Substituting these values in above equation, we get [] ωω+θ+−ω+θ 0 cos()()sin() cL qRtXXt =ω 0 sin Et
where we have used the relation =ω1/, C XC . L XL =ω
Multiplying and dividing equation by =+−22 ()ZCL RXX , we have,
ωω+θ+ω+θ
0 () cos() sin() qZCL RXX tt ZZ
=ω 0 sin Et
Now, let =φ cos R Z and =φ () sin CLXX Z
So that, φ= 1 tan CLXX R substituting this in equation and simplifying, we get,
ωω+θ−ω φ= 00cos()sin qZtt E
Comparing the two sides of this equation, we see that, ω == 000 EqZIZ
where =ω00Iq and ππ θ−φ=−θ=−+φ or 22
Therefore, the current in the circuit is ==ωω+θ 0 cos() dq iqt dt
=ω+θ 0 cos()IIt or, =ω+φ 0 sin()IIt
where == +− 00 0 22 () CL I EE ZRXX and φ= 1 tan CLXX R .
12. A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series L-C-R circuit in which R=3 Ω , L = 25.48 mH, and C=796 µ F. Find
(a) the impedance of the circuit; (b) the phase difference between the voltage across the source and the circuit; (c) the power dissipated in the circuit; and (d) the power factor.
Sol: a) To find the impedance of the circuit, we first calculate XL and XC.
=π 2 L XvL 3 23.145025.48108 =××××=Ω = π 1 2 C X vC
6 1 4
23.145079610 ==Ω ××××
Therefore, ()()=+−=+−2222 384 LCZRXX = 5 Ω
b) Phase difference, φ= 1 tan CLXX R 1 48 tan 53.1 3 ==−°
Since φ is negative, the current in the circuit lags the voltage across the source.
c) The power dissipated in the circuit is P = I2R
Now, 0 1283 40 A 5 22 i I ===
Therefore, ()2 40 A3 4800 W P =×Ω=
d) Power factor = φ== 0 coscos53.10.6
Try yourself:
11. An inductor, a capacitor and a resistor are in series with an alternator of frequency 50 Hz. The potential differences across them are 50 V, 80 V and 40 V respectively. Find the voltage of the alternator?
Ans: 50 V
TEST YOURSELF
1. The frequency for which a 5 μF capacitor has a reactance of 1 1000 Ω is given by
(1) 106 Hz π (2) 108 Hz π
(3) 1 Hz 1000 (4) 1000 Hz
2. The magnetic field energy in an inductor changes from maximum value to minimum value in 5.0 ms when connected to an AC source. The frequency of the source is
(1) 20 Hz (2) 50 Hz
(3) 200 Hz (4) 500 Hz
3. An alternating emf of e = 200sin100 πt is applied across a capacitor of capacitance 2 μF. The capacitive reactance and the peak current is
(1) 3 2 510 , 410A × Ω×π π
(2) 3 2 510 410, A × ×πΩ π
(3) 2 3 410 , 510A × Ω× π
(4) 4 Ω, 5 × 10–3 A
4. An alternating voltage 2002sin(100)VEt = is connected to a 1 μF capacitor through an a.c. ammeter. What will be the reading of the ammeter?
(1) 20 mA (2) 40 A
(3) 30 mA (4) 102mA
5. At a frequency ω o the reactance of a certain capacitor equals that of a certain inductor. If frequency is changed to 2 ω o .The ratio of reactance of the inductor to that of the capacitor is
(1) 4 : 1 (2) 2:1
(3) 1:22 (4) 1 : 2
6. An inductance of 0.2 H and resistance 100 Ω are connected to an AC 180 V− 50 Hz. The current in the circuit is
(1) 0.52 A (2) 5.2 A
(3) 1.52 A (4) 15.2 A
7. A circuit when connected to an AC source of 12 V gives a current of 0.2 A. The same circuit when connected to a DC source of 12 V, gives a current of 0.4 A. The circuit is
(1) series LR
(2) series RC
(3) series LC
(4) series LCR
8. A series circuit has a resistance of 60 Ω and impedance100 Ω. If potential difference applied across the circuit is 120 V. The power consumed in the circuit will be
(1) 86.4 W (2) 864 W
(3) 240 W (4) 480 W
9. In a series circuit, R = 300 Ω, L = 0.9 H, C = 2.0 μF and w=1000 rad/s. The impedance of the circuit is
(1) 1300 Ω (2) 900 Ω
(3) 500 Ω (4) 400 Ω
10. The power in AC circuit is given by cos. rmsrms PEI=φ The value of cosϕ in series L-C-R circuit at resonance is :
(1) Zero (2) 1
(3) 1 2 (4) 1 2
11. In a series L-C-R circuit, R = 10 Ω and the impedance Z = 20 Ω. Then the phase difference between the current and the voltage is
(1) 60° (2) 30°
(3) 45° (4) 90°
12. In an AC circuit the applied potential difference and the current flowing are given by :
V=200 sin 100t volt, 5sin100A 2 It
The power of consumption is equal to:
(1) 1000 W (2) 40 W
(3) 20 W (4) zero
13. A coil of inductive reactance 31 Ω as a resistance of 8 Ω. It is placed in series with a condenser of capacitive reactance 25 Ω. The combination is connected to an AC source of 110 V. The power factor of the circuit is
(1) 0.56
(2) 0.64
(3) 0.80
(4) 0.33
Answer Key
(1) 2 (2) 2 (3) 1 (4) 1
(5) 1 (6) 3 (7) 1 (8) 1
(9) 3 (10) 2 (11) 1 (12) 4
(13) 3
7.3 ELECTRICAL RESONANCE SERIES L-C-R CIRCUIT
Electrical resonance is said to take place in a series L-C-R circuit, when the circuit allows maximum current for a given frequency of alternating supply, at which capacitive reactance becomes equal to the inductive reactance.
The current ( I ) in a series L-C-R circuit is given by
From the above equation (i), it is clear that current I will be maximum if the impedance (Z) of the circuit is minimum.
At low frequencies, ω=×π 2 LLv is very small and = ω×π 11 2 CCv is very large.
At high frequencies, L w is very large and 1 C ω is very small.
For a particular frequency ( ν 0), ω= ω 1 L C
i.e., XL = XC and the impedance (Z) of L-C-R circuit is minimum and is given by Z = R
Therefore, at the particular frequency ( ν 0), the current in L-C-R circuit becomes maximum. The frequency ( ν 0) is known as the resonant frequency and the phenomenon is called electrical resonance.
Again, for electrical resonance (XL –Xc) = 0.
i.e., XL = XC or, () 2 11 or L CLC ω=ω= ω
(or) ()() 0 11 or2 v LCLC ω=π=
(or) = π 0 1 2 v LC .....(ii)
This is the value of resonant frequency.
The resonant frequency is independent of the resistance R in the circuit. However, the sharpness of resonance decreases with the increase in R.
Series L-C-R circuit is more selective when resistance of this circuit is small.
a) Net reactance, X = 0
b) XL = XC
c) Impedance, Z = R ( minimum )
d) peak value of current == 0 0 0 EE ZR I (maximum but not infinity)
e) Resonant frequency, 0 1 2 LC ν= π
f) Voltage and current will be in phase
g) power factor φ= cos1
h) Re so nant frequency is independent of value of R
i) A series L-C-R circuit behaves like a pure resistive circuit at resonance.
Key Insights:
Figure (A)
Applications: Series L-C-R circuit at resonance admit maximum current at particular frequencies, so they can be used to tune the desired frequency or filter unwanted frequencies. That is why, it is called acceptor circuit. They are used in transmitters and receivers of radio, television and telephone carrier equipment etc.
At resonance in L-C circuit:
a) Net reactance, X = 0
b) XL=XC
c) Impedance, Z = 0
d) peak value of current, ==∞ 0 0 E Z I
e) Resonant frequency, 0 1 2 LC ν= π
f) Voltage and current differ in phase by π 2
g) Power factor φ= cos0
At resonance in L-C-R circuit:
■ R esonant circuits have a variety of applications, for example, in the tuning mechanism of a radio or a TV set. The antenna of a radio accepts signals from many broadcasting stations. The signals picked up in the antenna acts as a source in the tuning circuit of the radio, so the circuit can be driven at many frequencies. But to hear one particular radio station, we tune the radio. In tuning, we vary the capacitance of a capacitor in the tuning circuit such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal received. When this happens, the amplitude of the current with the frequency of the signal of the particular radio station in the circuit is maximum.
■ It is important to note that resonance phenomenon is exhibited by a circuit only if both L and C are present in the circuit. Only then do the voltages across L and C cancel each other (both being out of phase) and the current amplitude is E0/R. The total source voltage appearing across R . This means that we cannot have resonance in a RL or RC circuit.
The Working Principle of Metal Detector
Metal detector works on the principle of resonance in AC circuits. When a person walks through a metal detector, in fact he walks through a coil of many turns connected to a capacitor tuned suitably.
If he carries some metal, impedance of the circuit changes, bringing significant changes in the current. This change in current is detected and the electronic circuit causes the alarm.
7.3.1 Sharpness of Resonance
The amplitude of the current in the series L-C-R circuit is given by
and is maximum when ==ωω0 1/. LC The maximum value is 0,max0 / IER =
For values of w other than w0 ,the amplitude of the current is less than the maximum value. Suppose we choose a value of w for which the current amplitude is 1/2 times its maximum value. At this value, the power dissipated by the circuit becomes half. From the curve in figure we see that there are two such values of w, say, w1 and w2, one greater and the other smaller than w 0 and symmetrical about w 0. We may write
ω=ω+∆ωω=ω−∆ω 1020 ;
The difference ω−ω=∆ω 12 2 is often called the bandwidth of the circuit. The quantity ω∆ω 0 (/2) is regarded as a measure of the sharpness of resonance. The smaller th e ∆ w , we note that the current amplitude i 0 is ω=ω+∆ω 0max 10 (1/2) I for . Therefore, at w 1,at
+ω−= ω 2 22 1 1 1 2 RLR C ; ω−=
1 1 1 LR C which may be written as, ω+∆ω−= ω+∆ω 0 0 1 () () LR C
0 0 0 0 1 1 1 LR C
Using ω= 2 0 1 LC in the second term on the left hand side, we get
0 0 0 0 1 1 L LR we can approximate, 000 1 1as1since,1.
Therefore,
∆ω∆ω ω+−ω−= ωω 0 00 11 0 LLR ∆ω ω= ω
0 0 2 LR ; band width, ∆ω= 2 R L The sharpness of resonance is given by. ωω = ∆ω 00 2 L R
The ratio ω0 L R is also called the quality factor, Q of the circuit.
ω = 0 L Q R
From the above equation we see that ω ∆ω= 0 2 Q
So, larger the value of Q, the smaller is the value of 2 ∆ w or the bandwidth and sharper is the resonance. Using ω= 2 0 1/ LC we can write =ω0 1/ QCR . We see from figure. that if the resonance is less sharp, not only is the maximum current less, the circuit is close to resonance for a larger range ∆ w of frequencies and the tuning of the circuit will not be good. So, less sharp the resonance, less is the selectivity of the circuit or vice versa. From equation. we see that if quality factor is large. ie., R is low or L is large, the circuit is more selective.
Half Power Frequencies and Band Width
The frequencies at which the power in the circuit is half of the maximum power (The power at resonance), are called half power frequencies.
i) The current in the circuit at half power frequencies (HPF) is 1 2 or 0.707 or, 70.7% of maximum current (current at resonance).
(b) ω→ 2 called upper half power frequency. It is greater than w 0 . At this frequency, the circuit is inductive.
iii) Band width ( ∆ w ) : The difference of half power frequencies,w1 and w2 is called band width. ∴∆ w=w 2−w 1.
Quality Factor (Q) of Series Resonant Circuit
i) The characteristic of a series resonant circuit is determined by the quality factor (Q - factor) of the circuit.
ii) It defines sharpness of I - V curve at resonance. When Q-factor is large, the sharpness of resonance curve is more and vice versa.
Resonantfrequency
iii) Q
Bandwidth
For series resonant circuit, it can be proved that,
at resonance, ω=
iv) Q - Factor tells the relation between voltage across the inductor or capacitor and peak value of voltage. The sharpness or selectivity of a resonance circuit is measured by Q-factor, called quality factor.
The Q-factor of series resonance circuit is defined as the ratio of voltage developed across the inductance or capacitance at resonance to the applied voltage (which is the voltage across R).
ii) There are two half power frequencies
(a) ω→ 1 called lower half power frequency. At this frequency, the circuit is capacitive.
voltageacrossLorC AppliedVoltage(= voltageacross R)
factor
v) The quality factor (Q) is also defined as2π times the ratio of the energy stored in L (or C) to the average energy loss per period.
maximumenergystoredincircuit 2 energylossperperiod Q
The maximum energy stored in inductor, = 2 0 1 2 ULI
The energy dissipated per second, == 2 2 0 . 2 Rrms IR UIR
Energy dissipated per time period, = 2 0 . 2 R IR UT
Substituting these values, we get
Different forms of Q-factor
a) ω == ∆ω 0 1 L Q RC
b) = maximumP.DacrossC maximumP.DacrossR Q
e)
lossofenergyinonetimeperiod Q π
2×maximumenergystored Q= T×averageenergyexpended
Key Insights:
■ Q is just a number
■ Q-factor is also called voltage amplification factor or efficiency of the circuit.
■ Q-factor will be large, i.e., the circuit will have more sharpness if R is low or L is large or C is low
■ Q-factor of L-C circuit at resonance is infinity.
13. In series L-C-R circuit, L = 2.0 H, C = 32 µF and R =10 Ω find Q-factor of resonance circuit
Sol: Q-factor
c) = maximumP.DacrossL maximumP.DacrossR Q
Try yourself:
12. In a series L-C-R circuit sharpness is Q. If the resistance of the circuit is doubled, inductance is quadrupled and capacitance remains unchanged, find the new sharpness. Ans: Q
TEST YOURSELF
1. In an L-C-R series circuit, the voltages across R, L and C at resonance are 40 V, 60 V and 60 V respectively. The applied voltage is (1) 60 V (2) 40 V (3) 160 V (4) 22 (40)(120)V +
2. A resistance of 10 Ω, a capacitance of 0.1 μF and an inductance of 2 mH are connected in series across a source of alternating emf of variable frequency. At what frequency does maximum current flow?
(1) 11.25 kHz (2) 23.76 kHz
(3) 35.46 kHz (4) 46.72 kHz
3. An inductor of inductance 100 mH is connected in series with a resistance, a variable capacitance and an AC source of frequency 2.0 kHz. The value of the capacitance so that maximum current may be drawn into the circuit.
(1) 50 nF (2) 60 nF (3) 63 nF (4) 79 nF
4. The following series L - C - R circuit, when driven by an emf source of angular frequency 70 kilo-radians per second, the circuit effectively behaves like 10 Ω
(1) purely resistive circuit
(2) series R-L circuit
(3) series R-C circuit
(4) series L-C circuit with R=0
5. An inductance of 1 mH and a charged capacitor capacitance of 10 μF are connected across each other. The resonating angular frequency of circuit is
(1) 105 rad/s (2) 104 rad/s (3) 103 rad/s (4) 102 rad/s
6. A series L-C -R circuit is connected to a source of alternating emf 50 V and if the potential differences across inductor and capacitor are 90 V and 60 V respectively, the potential difference across resistor is (1) 400 V (2) 40 V (3) 80 V (4) 160 V
7. In a circuit, L , C and R are connected in series with an alternating voltage source of frequency f. The current leads the voltage by 45°. The value of C is
(1) () 1 22ffLRππ+ (2) () 1 2 ffLRππ+ (3) () 1 22ffLRππ− (4) () 1 2 ffLRππ−
Answer Key
(1) 2 (2) 1 (3) 3 (4) 3 (5) 2 (6) 2 (7) 1
7.4 L-C OSCILLATIONS
A capacitor ( C ) and an inductor ( L ) are connected as shown in the figure. Initially the charge on the capacitor is Q ++ C L i
∴ Energy stored in the capacitor = 2 E Q U 2C
The energy stored in the inductor, UB = 0.
The capacitor now begins to discharge through the inductor and current begins to flow in the circuit. As the charge on the capacitor decreases, U E decreases but the energy = 2 B 1 ULI 2 in the magnetic field of the inductor increases. Energy is thus transferred from capacitor to inductor. When the whole of the charge on the capacitor disappears, the total energy stored in the electric field in the capacitor gets converted into magnetic field energy in the inductor. At this stage, there is maximum current in the inductor.
Energy now flows from inductor to the capacitor except that the capacitor is charged
oppositely. This process of energy transfer continues at a definite frequency (ν). Energy is continuously shuttled back and forth between the electric field in the capacitor and the magnetic field in the inductor.
+q0 -q0 t
If no resistance is present in the L-C circuit, the L-C oscillation will continue infinitely as shown.
However, in an actual L-C circuit, some resistance is always present due to which energy is dissipated in the form of heat. So, L-C oscillation will not continue infinitely with same amplitude, as shown.
+q0 -q0 t
Let q be the charge on the capacitor at any time t and di dt be the rate of change of current. Since no battery is connected in the circuit, 0 qdi L cdt −= but, idq dt =− from the above equations, we get 2 2 0 qdq L Cdt += 2 2 1 0 dq dtLCq+=
The above equation is analogues to 2 2 2 0 dr r dt +ω=
(differential equation of SHM)
Hence, on comparing, we get
The charge therefore oscillates with a frequency 1 2 LC ν= π and varies sinusoidally with time.
7.4.1 Comparison of L-C Oscillations with SHM
The L-C oscillations can be compared to SHM of a block attached to a spring
In L-C oscillations, ω= 0 1 LC
While mechanical oscillations ω= 0 K m where K is the spring constant
In L-C oscillations, = 1 V Cq tells us the potential difference required to store a unit charge
In a mechanical oscillation, = F K x tells us the external force required to produce a unit displacement of mass
In L-C oscillations current is the analogous quantity for velocity of the mass in mechanical oscillations
In L-C oscillations energy stored in capacitor is analogous to potential energy in mechanical oscillations
In L-C oscillations energy stored in inductor is analogous to kinetic energy of the mass in mechanical oscillations
In L-C oscillations maximum charge on capacitor q 0 is analogous to amplitude in mechanical oscillations
V max = A w in mechanical oscillations, =ω 000Iq in L-C oscillations
Analogies between Mechanical and Electrical Quantities
Mechanical System Electrical System
Mass m Inductance L
Force constant k Reciprocal capacitance 1/C
Displacement x Charge q
Velocity v = dx/dt Current I = dq/dt
Mechanical energy Electromagnetic energy
7.4.2 Energy of L-C Oscillations
Let q 0 be the initial charge on a capacitor. Let the charged capacitor be connected to an inductor of inductance L . The angular frequency of oscillation of L-C circuit is
At an instant t, charge q on the capacitor and the current i are given by; =ω 0 ()cos qtqt =−ωω 0 ()sin itqt
Energy stored in the capacitor at time t is
2 2 22 0 11 cos() 222 E
Energy stored in the inductor at time t is
2222 0 11 sin() 22 M ULiLqt
=ωω= 2 22 0 sin()1/ 2 qtLC C
Sum of energies 22 22 00 (cossin) 22 TEM qqUUUtt CC =+=ω+ω=
As q0 and C, both are time-independent, this sum of energies stored in capacitor and inductor is constant in time. Note that, it is equal to the initial energy of the capacitor.
Key Insights:
■ Note that the above discussion of L-C oscillations is not realistic for two reasons :
(i) Every inductor has some resistance. The effect of this resistance is to introduce a damping effe ct on the charge and current in the circuit and the oscillations finally die away.
(ii) Even if the resistance were zero, the total energy of the system would
not remain constant. It is radiated away from the system in the form of electromagnetic waves (discussed in the next chapter). In fact, radio and TV transmitters depend on this radiation.
■ You may like to compare the treatment of a forced oscillator discussed in with that of an L-C-R circuit when an AC voltage is applied in it. Let us therefore list the equivalence between different quantities in the situations.
Forced oscillations
Driven L-C-R circuit
Displacement, x charge on capacitor
Time, t Time, t
Mass, m Self inductance, L
Damping constant, b Resistance, R
Spring constant, k Inverse capacitance, 1/C
Driving frequency, w d Driving frequency, w
Natural frequency of oscillations, w Natural frequency of L-C-R circuit, w 0
Amplitude of forced oscillations, A Maximum charge stored, qm
Amplitude of driving force, Fo
Amplitude of applied voltage E0
You must note that since x corresponds to q, the amplitude A (maximum displacement) will correspond to the maximum charge stored, q m . Amplitude of oscillations in terms of other parameters, which we reproduce here for convenience:
Replace each parameter in the above equation by the corresponding electrical quantity, and see what happens. Eliminate L, C, w , and w 0, using =ω=ω ,1/LC XLXC and ω= 2 0 1/ LC
TEST YOURSELF
1. A condenser of capacity C is charged to a potential difference of V 1. The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to V2 is (1)
2. A capacitor of capacitance C has initial charge Q0 and connected to an inductor of inductance L as shown. At t = 0 switch S is closed. The current through the inductor when energy in the capacitor is three times the energy of inductor is
3. A capacitor of capacitance 2 μF is connected in the tank circuit of an oscillator oscillating with a frequency of 1 kHz. If the current flowing in the circuit is 2 mA, the voltage across the capacitor will be (1) 0.16 V (2) 0.32 V (3) 79.5 V (4) 15 V
4. A coil of inductance 8 μH is connected to a capacitor of capacitance 0.02 μF. To what wavelength is this circuit tuned?
(1) 7.54 × 103 m (2) 4.12 ×
m (3) 15.90 × 103 m (4) 7.54 × 102 m
7.5 TRANSFORMER
The transformer is a device which is used to change the voltage of an alternating current without change in frequency and power. As such they are of two types.
When it changes the low alternating current at high voltage into high alternating current at low voltage, then it is known as STEP-DOWN TRANSFORMER.
On the other hand, if it changes high alternating current at low voltage into low alternating current at high voltage then it is called STEP-UP TRANSFORMER.
A transformer is called step-up or stepdown type depending upon whether it increases or decreases the voltage respectively.
Principle: It works on the principle of mutual induction i.e. when current flowing through a coil or magnetic flux linked with a coil changes, an induced emf is produced in the other coil.
Construction: It consists of two coils of copper wire wound separately over a rectangular and laminated soft iron core [The core is made by placing soft-iron strips one above the other. These strips are insulated from each other to reduce the eddy currents and Hence, to reduce the loss of energy in the core]. These coils are kept insulated from each other as well as from the iron core. The two coils are known as primary coil P and secondary coil S. The AC source is connected across the primary coil while transformed voltage or transformer output is obtained across the secondary coil S.
In case of a step-up transformer, the primary coil consists of smaller number of turns of thick copper wire, while the secondary coil consists of larger number of turns of thin copper wire. In step down transformer, the order is just reversed.
Theory: When alternating current flows through the primary coil, a magnetic field is produced inside the coil which also varies both in magnitude and direction as per nature of input AC source.
Due to this, core is first magnetised in one direction and then in opposite direction in each cycle of current. Hence, the electrical energy of the primary coil is transformed into the magnetic energy of the core. Since the secondary coil is also wrapped on the same core and the core is made of soft iron, the magnetic flux passing through it also changes continuously due to repeated magnetisation and demagnetisation of the core. As no leakage of magnetic flux occurs in the process of transformation, an alternating emf of the same frequency is induced in the secondary coil. Thus, magnetic energy of iron-core is now transformed into the electrical energy of the secondary coil. The magnitude of induced emf produced in the secondary coil depends upon the ratio of number of turns in the two coils.
Let φ be the magnetic flux linked with each turn and NP and NS be the number of turns in the primary and secondary coils respectively.
Because there is no leakage of magnetic flux in the process of transformation, same magnetic flux φ will be linked with each turn of primary and secondary coils at every instant.
Therefore, according to Faraday’s law of electromagnetic induction, induced emf produced in primary and secondary coils are given by
Hence, ss pp eN eN = ....(1)
If there is no loss of energy in the primary coil and its resistance is assumed to be negligible then induced emf produced in the primary coil will be nearly equal to the applied potential difference VP between its ends.
Similarly, because the secondary coil is open, Hence, potential difference across its ends will be equal to emf induced in it. i.e., under these ideal conditions,
........(2)
Here, r is called Transformer ratio. For step-up transformer, r > 1 and for step down transformer, r < 1.
If we assume that there are no energy losses in the process of transformation, then instantaneous output power = instantaneous input power
or, .or, pss sspp spp IVN VIVIr IVN ====
Thus, a step-up transformer increases the voltage by decreasing the current and a stepdown transformer decreases the voltage by increasing the current. This happens according to law of conservation of energy. In other words, a transformer simply transforms the voltages and currents but does not generate electricity.
In real practice, the energy obtained from the secondary coil is always a little bit less than the energy given to primary coil. This is because of the fact that some energy of the primary coil is spent in heating its wire and some energy is spent in the process of magnetisation and demagnetisation of core.
Efficiency of a transformer : The ratio of output power to the input power is called efficiency ( η ) of transformer.
η==outputss inputpp PEI PEI
∴ Percentage of efficiency =× 100 SS PP EI EI
For an ideal transformer, η = 1 (100%). Due to transformer losses, the efficiency is less than one (less than 100%)
In actual transformers, small energy losses do occur due to the following reasons:
(i) Flux leakage: There is always some flux leakage; that is, not all of the flux due to primary passes through the secondary due to poor design of the core or the air gaps in the core. It can be reduced by winding the primary and secondary coils one over the other.
(ii) Resistance of the windings: The wire used for the windings has some resistance and so, energy is lost due to heat produced in the wire (I2R). In high current, low voltage windings, these are minimised by using thick wire.
(iii) Eddy currents: The alternating magnetic flux induces eddy currents in the iron core and causes heating. The effect is reduced by having a laminated core.
(iv) Hysteresis: The magnetisation of the core is repeatedly reversed by the alternating magnetic field. The resulting expenditure of energy in the core appears as heat and is kept to a minimum by using a magnetic material which has a low hysteresis loss. It is important to mention that transformer is an AC device and it cannot work on DC.
Key Insights:
■ The large scale transmission and distribution of electrical energy over long distance is done with the use of transformers. The voltage output of the generator is stepping-up (so that current is reduced and consequently, the I 2R loss is cut down).
It is then transmitted over long distances to an area substation near the consumers. There the voltage is stepped down. It is further stepped down at distributing substations and utility poles before a power supply of 240 V reaches our homes.
14. A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two line wires carrying power is 0.5 Ω /km. The town gets power from the lines through a 4000-220 V step down transformer at a substation in the town.
a) Estimate the line power loss in the form of heat
b) How much power must the plant supply, assuming there is negligible power loss due to leakage?
c) Characterise the step up transformer at the plant
Sol: 440V 4000/220 220V Town R
Power requirement of the town is given by EI, where E is the voltage at the receiving end of the line and I is the line current.
Here, P = 800 × 1000 W; E = 4000 V, then 8001000 200 A 4000 P I E × ===
(i) Line power loss ()()==××× 2 2 2002150.5 IR
= 600,000 W = 600 kW
(ii) Total power delivered by power plant
= 800 + 600 = 1400 kW
(iii) Voltage at sending end of line
= Receiving end line voltage + voltage drop in line
= 4000 + (IR) = 4000 + 200 (2×15× 0.5)
= 4000 + 3000 = 7000 V
∴ Step up transformer of 7000 440 V V is required
15. A transformer having efficiency 90% is working on 100 V and at 2.0 kW power. If the current in the secondary coil is 5A, calculate (i) the current in the primary coil and (ii) voltage across the secondary coil.
Sol: Here, 9 90%,5 A 10 s I η=== , E p = 100 V, (i) E p I p = 2 kW = 2000 W = 2000 p p I E or, == 2000 20 100 p IA (ii) η== Outputpower Inputpower ss pp EI EI 9 or, 20001800 W 10 sspp EIEI=η×=×= 18001800 360volt 5 s s E I ∴===
Try yourself:
13. A step down transformer converts a voltage of 2200 V into 220 V in the transmission line. Number of turns in primary coil is 5000. Efficiency of transformer is 90% and its output power is 8 kW. Calculate (i) number of turns in secondary coil and (ii) input power.
Ans: (i) 500 (ii) 8.9 kW
Parallel L-C-R Circuits:
Current and Phase Difference
From phasor diagram current
()=+− 2 2 RCL iiii and phase difference
φ=()() = 11 tan tan CLCL R iiSS iG iC iR iL i V φ
Admittance (Y) of the Circuit : From Equation of Current
2 1111 LC Y ZRXX
()=+− 2 2 LCGSS
Resonance
At resonance:
i)
ii)
=⇒= min CLn iiii
=⇒=⇒Σ= 0 CL CL VV SSS XX
iii) == max R V ZR i
iv) φ=⇒ 0 p.f = φ==cos1 maximum
v) Resonant frequency
⇒= π 1 2 v LC
Parallel L-C Circuits
If inductor has resistance (R) and is connected in parallel with capacitor as shown, then
i) At resonance ~ R L i V=V0sinωt C
a) == max min 1 L Z YCR
b) current through the circuit is minimum and = 0 min VCR i L
c) =⇒=⇒=∞ 11 LC LC SSX XX
d) Resonant frequency
ω=− 2 0 2 1 /sec Rrad LCL =− π 2 0 2 11 2 R VHz LCL
(Condition for parallel resonance is < L R C )
e) Quality factor of the circuit
=− 2 2 11 1 CRR LCL in the state of resonance the quality factor of the circuit is equivalent to the current amplification of the circuit.
ii) If inductance has no resistance : If R = 0 then circuit becomes parallel L-C circuit.
iL iC i φ
condition of resonance =⇒=cL CL VV ii XX
⇒=CLXX . At resonance current i in the circuit is zero and impedance is infinite.
Resonant frequency: = π 0 1 2 VHz LC
In an AC circuit R = 0 ⇒ cos φ = 0, so, P avg = 0 i.e., In resistance less circuit the power consumed is zero. Such a circuit is called the wattless circuit and the current flowing is called the wattles current.
The component of current which does not contribute to the average power dissipation is called wattless current.
i) The average of wattless component over one cycle is zero.
ii) Amplitude of wattless current is i 0sin φ and rms value of wattless curr ent
φ=φ 0 sinsin 2 rms i i i φ V i cosφ i sinφ
It is quadrature (90°) with voltage.
TEST YOURSELF
1. In a step up transformer, the ratio of turns of primary to secondary is 1 : 10 and primary voltage is 230 V. If the load current is 2 A,
CHAPTER REVIEW
AC Generation
■ Given AC source voltage is RMS voltage.
■ AC ammeter and voltmeter read rms value, i.e., effective value of alternating current and voltage respectively.
then current in the primary is (1) 20 A (2) 10 A (3) 2 A (4) 1 A
2. In a transformer, number of turns in the primary are 140 and that in the secondary are 280. If current in primary is 4 A then that in the secondary is (1) 4 A (2) 2 A (3) 6 A (4) 10 A
3. The number of turns in primary and secondary windings of a transformer are 1000 and 100 respectively. If 200 V DC voltage is impressed across the primary terminals, the voltage across secondary terminals is
(1) 22 V (2) 2200 V
(3) 220 V (4) zero
4. The number of turns in primary and secondary coils of a transformer is 50 and 200. If the current in the primary coil is 4 A, then the current in the secondary coil is (1) 1 A (2) 2 A (3) 4 A (4) 5 A
5. A transformer has 1500 turns in the primary coil and 1125 turns in the secondary coil. If the voltage in the primary coil is 200 V, then the voltage in the secondary coil is (1) 100 V (2) 150 V (3) 200 V (4) 250 V
Answer Key
(1) 1 (2) 2 (3) 4 (4) 1 (5) 2
■ AC can be measured by using hot wire ammeters or hot wire voltmeters because the heat generated is independent of the direction of current.
■ AC produces the same heating effects as that of DC of magnitude i = i rms
■ V volt AC means its maximum value is 2V volt, But V volt DC means its maximum value is also V volt. Hence, AC is more dangerous than DC.
■ If V = V0cos w t is the voltage supplied to a
circuit and I = I 0 cos (wt + φ) is the current flowing through the circuit, then average power consumed by the circuit is
P avg = 1/2 V0 I0cos φ = V rms . I rms .cos φ .
■ If the current through a circuit element is i = (a+b coswt + c sin wt), the rms value of current is, i rms = 22 2 2 bc a + +
If a = 0, I rms 22 2 bc + =
If b = 0, I rms 2 2 2 c a =+
Alternating Voltage Applied to a Resistor, Inductor, and Capacitor
■ When voltage V = V 0 cos w t is supplied to a pure inductor L , its reactant is x L = w L =2 π fL and the current flowing through it is, i 0 .cos() 2 =− L V t x π ω . Current lags the voltage by π /2.
■ For supply voltage V = V0 cos w t to pure inductor, power consumed by the inductor is zero.
■ When voltage V = V 0 cos t is supplied to a capacitor of capacitance C , its reactance 11 2 == xC CfCωπ and the instantaneous current flowing throu gh it is 0 .cos. 2 c V it x π =ω+ current leads the voltage by π /2.
■ Impedance of a circuit with an AC supply is given by max max rms rms VV Z II ==
■ For L-R series circuit impedance,
222 ZRL =+ω
■ For L-R series circuit power factor 222 cos R RL φ= +ω
■ For C-R series circuit, 2 22 1 ZR C =+ ω
■ For C-R series circuit, 2 22 cos 1 R R C φ= + ω
■ I n L-C -R series circu it impedance 2 2 1 ZRL C =+ω− ω
■ In L-C -R series circuit when 1 L C ω> ω ,the circuit is inductive and when 1 L C ω< ω , the circuit is capacitive.
■ In L-C-R series circuit power factor, 2 2 cos 1 R RL C φ= +ω− ω , and 1 tan L C R ω− φ=ω . Electrical Resonance Series L-C-R Circuit
■ In L-C -R series circuit current is maximum when supply frequency 1 Hz. 2 r ff LC == π
■ In L-C-R series circuit, sharpness or quality factor is given by 11 r r LL Q RCRRC ω === ω .
■ In L-C-R series circuit band width 2 r Q ∆ω=ω .
■ When average power consumed by a circuit
with AC supply is zero, current flowing in the circuit is called wattless current. If cos φ is the power factor of a circuit, then wattless current is I rms .sin φ .
■ Wattless current flows only in purly inductive, purely capacitive or L-C circuits.
■ An alternating current flows mainly on the surface of the conductor. This effect is known as skin effect.
■ Higher the frequency of alternating current, more is the skin effect.
■ For high frequency AC,the effective resistance of the wire is much greater then that for DC.
■ Choke coil is a device having high inductance and negligible resistance. It is used to control current in AC circuits.
■ Power dissipated in choke coil is least.
L-C Oscillations
■ An L-C oscillator is analogous to a springmass system.
■ In L-C oscillator, the charge oscillates with a frequency 1 2 f LC π = .
■ Inductance in electrical system is analogous to mass in mechanical system.
■ Capacitance in electrical system is analogous to the reciprocal (1/K) of spring constant K.
■ Charge in electrical system (Q) is analogous to displacement (x) of mechanical system.
■ Current( i = dQ / dt ) in electrical system is analogous to velocity ( V = dx / dt ) of mechanical system.
■ Magnetic flux (φB = Li) in electrical system is analogous to linear momentum (p = mv) of mechanical system.
■ Transformer works on the principle of mutual induction.
■ In an ideal transformer the primary winding has negligible resistance and all the flux in the core links both primary and secondary windings.
■ If φ be the flux in each turn in the core and if a voltage V p is applied in primary, then emf induced in secondary with N s turns is
d eN dt φ =− and emf induced in primary with N p turns is pp d eN dt φ =−
Here, e P = V P (since it is assumed that primary has zero resistance.
■ If the secondary is on open circuit or if current taken from secondary is small, then e s = Vs, where V s is the voltage across secondary ss Pp VN VN ∴=
■ If the tra nsformer is assumed to be 100% efficient (No loss of energy) then Pss
■ In a step-up transformer, primary is made of thick insulated copper wire and secondary is made of thin wire.
■ In a step down transformer primary is made of thin insulated copper wire and secondary is made of thick wire.
■ Efficiency of transformer () Outputpower 100%100% Inputpower ss pp VI VI η=×=×
■ Eddy current loss in the core of a transformer is kept to a minimum by having laminated core.
■ Hysteresis loss in a transformer is kept to a minimum by using soft iron core.
JEE MAIN LEVEL
Level-I
AC Generation
Single Option Correct MCQs
1. A 220 V main supply is connected to a resistance of 100 kΩ. The rms current is:
(1) 22 mA (2) 2.2 mA
(3) 220 mA (4) 10 mA
2. The peak value of an alternating emf given by E = E0sin w t is 10 volt and its frequency is 50 Hz. At a time
t second, the instantaneous value of the emf is
(1) 1 V (2) 5 V
(3) 53 V (4) 10 V
3. The equation of an alternating emf is ()V120sin100t =π volt. Find the peak value of its emf and its frequency.
(1) 120 V, 100 Hz (2) 120 V,100Hz 2
(3) 60 V, 200 Hz (4) 120 V, 50 Hz
4. An alternating current is given by i = i1 coswt + i2 sin w t. The rms current is given by (1) 12 2 ii + (2) 12 2 ii + (3) 22 12 2 ii + (4) 22 12 2 ii +
5. Write the equation of an alternating emf of 120 V and frequency of 60Hz.
(1) e = (169.7sin120πt) volt
(2) e = (152.7sin110πt) volt
(3) e = (143.7sin150πt) volt
(4) e = (156.7sin130πt) volt
Alternating Voltage Applied to a Resistor, Inductor and Capacitor
Single Option Correct MCQs
6. When a coil is connected to a DC source of emf 12 V, a current of 4 A flows in it. If same coil is connected to 12 V, 50 Hz AC source, the current is 2.4 A. The self inductance of the coil is (in henry)
(1) 1 20 π (2) 1 10 π (3) 1 25π (4) 1 5π
7. An electric bulb is designed to operate at 12 volts DC. If this bulb is connected to an AC source and gives normal brightness, what would be the peak voltage of the source?
(1) 37 V (2) 17 V (3) 18 V (4) 10 V
8. A capacitance of 0.4 μF is connected to an alternating emf of 100 cycles/second. What is the capacitive reactance?
(1) 3981 Ω (2) 4687 Ω
(3) 5768 Ω (4) 6843 Ω
9. In a pure inductive circuit, the current (1) lags behind the applied emf by a phase of π. (2) lags behind the applied emf by a phase of 2 π . (3) leads the applied emf by a phase of 2 π . (4) and applied emf are in same phase.
10. A bulb and a capacitor are connected in series to a source of alternating current. If its frequency is increased, while keeping the voltage of the source constant, then (1) bulb will give more intense light. (2) bulb will give less intense light. (3) bulb will give light of same intensity as before. (4) bulb will stop radiating light.
11. An emf E = 4cos(1000t ) volt is applied to an LR - circuit of inductance 3 mH and resistance 4 ohms. The amplitude of current in the circuit is
(1) 4 A 7 (2) 1.0 A (3) 4 7 A (4) 0.8 A
12. An inductance of 0.2 H and resistance 100 Ω are connected to an AC 180 V - 50Hz. The current in the circuit is ______.
(1) 0.52 A (2) 5.2 A
(3) 1.52 A (4) 15.2 A
13. An AC voltage source V = V 0 sin w t is connected across resistance R and capacitance C as shown in figure. It is given that 1 R C = ω . The peak current is I0. If the angular frequency of the voltage source is changed to 3 ω keeping R and C fixed, then the new rms current in the circuit is
R
14. A choke coil has negligible resistance. The alternating potential drop across it is 220 volt and the current is 5 mA. The power consumed is
(1) 5 220 W 1000 × (2) 220 W 5 (3) zero (4) 2.20 × 5 W
15. A coil of self – inductance 1 H
is connected in series with a 300 Ω resistance. A voltage of 200 V at frequency 200 Hz is applied to this combination. The phase difference between the voltage and the current will be (1) 1 4 tan 3
Numerical Value Questions
16. A coil has inductance 0.7 H and is joined in series with a resistance of 220 Ω. When AC emf of 220 V at 50 Hz is applied, wattless component of current is 1/ n, then find the value of n
= 2
17. A current of 6 A flows in a coil when connected to a DC source of 18 V. If the same coil is connected to an AC source of 20 V, 50 rad/s, a current of 4 A flows in the circuit. If the inductance of the coil is 10x mH, then find out the value of x.
18. An alternating voltage E (in volt) 2002sin(100) t is connected to one microfarad capacitor through an AC ammeter. The reading of the ammeter (in mA) shall be____.
19. A 100 Hz AC is flowing in a coil of inductance 10 mH. What is the reactance of coil? (in Ω) [Take π = 3.14]
Electrical Resonance Series L-C-R Circuit
Single Option Correct MCQs
20. A capacitor of capacitance 100 μF and a coil of resistance 50 Ω and inductance 0.5 H are connected in series with a 110 V, 50 Hz AC source. Find the rms value of the current.
(1) 0.12 A (2) 0.82 A (3) 0.53 A (4) 0.22 A
21. A series AC circuit contains an inductor (20 mH), a capacitor (100 μF), a resistor (50 Ω) and an AC source of 12 V, 50 Hz. Find the energy dissipated in the circuit in 1000 s.
(1) 2.286 kJ (2) 22.86 kJ (3) 0.2286 kJ (4) 228.6 kJ
22. In a series L-C-R circuit, R = 10 Ω and the impedance Z = 20 Ω. Then the phase difference between the current and the voltage is
(1) 60° (2) 30° (3) 45° (4) 90°
23. In a series circuit R = 300 Ω, L = 0.9 H, C = 2.0 μF and w = 1000 rad/s. The impedance of the circuit is
(1) 1300 Ω (2) 900 Ω
(3) 500 Ω (4) 400 Ω
24. The resonance point in XL− f and XC− f curves is
(1) P (2) Q (3) R (4) S
25. An electric lamp designed for operation on 110 V AC is connected to a 220 V AC supply, through a choke coil of inductance 2 H, for proper operation. The angular frequency of the AC is 10010rad/s. If a capacitor is to be used in the place of the choke coil, its capacitance must be
(1) 1 μF (2) 2 μF
(3) 5 μF (4) 10 μF
26. A choke coil has (1) high inductance and low resistance (2) low inductance and high resistance (3) high inductance and high resistance (4) low inductance and low resistance
Numerical Value Questions
27. In an AC circuit, the instantaneous values of emf and current are e = 200sin(314t) volt and sin314 3 it π =+
ampere. The average power consumed in watt is .
28. In an L-C-R series circuit with resistance R = 100 Ω and AC source (200 V, 50 Hz), if the
capacitor is removed, the current lags behind voltage by 60° and when only inductor is removed, the current leads the voltage by 60°. Power (in W) dissipated in L-C-R circuit is
29. A lamp with a resistance of 8 Ω is connected to an ideal choke coil. This arrangement is connected to an alternating source of 110 V, 60 Hz.The current in the circuit is 11 A. What is the value of inductive reactance of the choke coil (in ohm)?
L-C Oscillations
Single Option Correct MCQs
30. A 16 μF capacitor is charged to a 20 volt potential. Battery is then disconnected and inductor of inductance 40 mH is connected across the capacitor, so that L-C oscillations are step-up. Maximum current in the coil is (1) 0.4 A (2) 2 A (3) 0.8 A (4) 0.2 A
31. A capacitor of 10 μF and an inductor of 1 H are joined in series. An AC of 50 Hz is applied to this combination. What is the impedance of the combination?
(1) () 2 5 50 π− Ω π (2) () 2 10 100 −π Ω π (3) () 2 5 10 π− Ω π (4) () 2 10 500 −π Ω π
32. What is the mechanical equivalent of spring constant k in L-C oscillating circuit? (1) 1 L (2) 1 C (3) L C (4) 1 LC
33. An L-C circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed at t = 0. After what minimum time, the energy stored completely magnetic. (Assume that at t = 0, the capacitor is fully charged)
(1) t = 0 (2) t = 1.57 ms
(3) t = 3.14 ms (4) t = 6.28 ms
Numerical Value Questions
34. A telegraph line of length 100 km has a capacity of 0.01 μF/km and it carries an alternating current at 0.5 kilo cycle per second. If minimum impedance is required, then the value of the inductance that needs to be introduce d in series is mH (take 10 π= )
Transformer
Single Option Correct MCQs
35. In a transformer, the number of turns in the primary and secondary coils are 1000 and 3000 respectively. If the primary is connected across 80 V AC, the potential difference across each turn of the secondary will be (1) 240 V (2) 0.24 V
(3) 0.8 V (4) 0.08 V
36. A transformer is based on the principle of (1) mutual induction (2) self induction
(3) Ampere’s law
(4) Coulomb’s law
37. A step down transformer has primary voltage 1100 V. The transformer ratio is 1 : 5. If the primary current is 10 A, find the secondary voltage, secondary current assuming the transformer to be an ideal transformer.
(1) 1100 V, 10 A (2) 2200 V, 20 A (3) 3300 V, 30 A (4) 220 V, 50 A
38. In an ideal step up transformer, if the input voltage and input power are V 1 and P1 respectively and the output voltage and output power are V2 and P2 respectively, then (1) V1 = V1 ; P1 = P2 (2) V1 > V1 ; P1 > P2 (3) V1 < V1 ; P1 < P2 (4) V1 < V1 ; P1 = P2
39. A 220 V input is supplied to a transformer. The output circuit draws a current of 2 A at 440 V. If the efficiency of the transformer is 90%, the current drawn by the primary windings of the transformer is (1) 3.6 A (2) 2.8 A
(3) 3.4 A (4) 4.4 A
40. The transformer ratio of a transformer is 5. If the primary voltage of the transformer is 400 V, 50 Hz, the secondary voltage will be (1) 2000 V, 250 Hz (2) 80 V, 50 Hz
(3) 80 V, 10 Hz (4) 2000 V, 50 Hz
41. The efficiency of a transformer is 98%. The primary voltage and current are 200 V and 6 A. If the secondary voltage is 100 V, the secondary current is (1) 11.76 A (2) 12.25 A (3) 3.06 A (4) 2.94 A
42. The figure shows an ideal transformer that has 100 turns in the primary and 500 turns in the secondary. If the current in the primary is () 22sin200tA. π Then the potential difference in the secondary is (in Volts) 10 Ω
(1) ()22sin200t π
(2) ()23sin200t π
(3) ()32sin200t π
(4) ()42sin200t π
Numerical Value Questions
43. A transformer is used to light a 140 W, 24 V bulb from a 240 V AC mains. The current in the main cable is 0.7 A. The efficiency of the transformer is %
Level-II
AC Generation
Numerical Value Questions
1. The rms value of the waveform shown is . Y t +10
2. Calculate the average value of the voltage wave (in volts) shown in figure.
passes through the circuit, the AC ammeter reads 4 A. Then the reading of this ammeter (in ampere), if a DC and an AC flow through the circuit simultaneously is
Alternating Voltage Applied to a Resistor, Inductor and Capacitor
Single Option Correct MCQs
5. When resistance is connected with AC source, the emf
(1) leads current by 2 π degrees.
(2) leads current by 2 π radians.
(3) is in phase with current.
(4) leads current.
6. Voltage and current for a circuit with two elements in series are expressed as ()170sin6280V 3 Vtt π
() 8.5sin6280A 2 itt π
. What are the values of the elements?
(1) R = 27.32 Ω, C = 25.92 µ F
(2) R = 17.32 Ω, C = 15.92 µ F
(3) R = 7.32 Ω, C = 5.92 µ F
(4) R = 10.32 Ω, C = 5.92 µ F
3. The average value of current shown graphically in figure, from t = 0 to t = 2s is
4. An AC ammeter is used to measure current in a circuit. When a given direct current passes through the circuit, the AC ammeter reads 3 A. When another alternating current
7. When an AC source of emf e = E0sin(100t) is connected across a circuit, the phase difference between the emf e and the current i in the circuit is observed to be 4 π , as shown in the diagram. If the circuit consists possibly only of RC or L-C or LR in series, find the two elements. I or e e i t ф
(1) R = 1 kΩ, C = 10 µ F
(2) R = 1 kΩ, C = 1 µ F
(3) R = 1 kΩ, L = 10 H
(4) R = 1 kΩ, L = 1 H
8. An AC source of angular frequency w is fed across a resistor R and a capacitor C in series. The current registered is I. If now the frequency of source is changed to 3 ω (but maintaining the same voltage), the current in then circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency w
(1) 3 5 (2) 2 5 (3) 1 5 (4) 4 5
9. If the power factor is 1 2 in a series RL circuit with R = 100 Ω. If AC mains, 50 Hz is used, then L is
(1) 3 henry π (2) π henry (3) 3henry (4) 3henry π
10. The virtual current of 4 A and 50 Hz flows in an AC circuit containing coil. The power consumed in the coil is 240 W. If the virtual voltage across the coil is 100 V, then its resistance will be (1) 1 3 Ω π (2) 1 5 Ω π (3) 15 Ω (4) 1 Ω π
Numerical Value Questions
11. Two resistors are connected in a series across 5 V rms source of alternating potential. The potential difference across 6 Ω resistor is 3 V. If R is replaced by a pure inductor L of such magnitude that current remains same, then the potential difference across L is . (in volts)
12. An ideal choke takes a current of 10 A when connected to an AC supply of 125 V and 50 Hz. A pure resistor under the same conditions take a current of 12.5 A. If the two are connected to an AC supply of 100 V and 40 Hz, then the peak current in series combination of above resistor and inductor is 10x ampere. What is the value of x?
13. A capacitor of capacitance 1 μF is charged to potential difference 2 V and is connected to an inductor of 1 mH. At an instant when potential difference across the capacitor is 1 V, the current in the circuit is 2 10 10 3 x ampere. Find out the value of x
14. Three alternating voltage sources V 1 = 3 sint volt, V2 = 5 sin(t + φ1) volt and V3 = 5 sin(t – φ2) volt connected across a resistance R as shown in the figure (where ϕ1 and ϕ2 corresponds to 30° and 127° respectively). Find the peak current (in A) through the resistor.
Electrical Resonance Series L-C-R Circuit
Single Option Correct MCQs
15. A 60 V, 10 W lamp is to be run on 100 V, 60 Hz AC mains. The inductance of the choke coil required is (1) 7 H π (2) 12 H π (3) 32 H
(4) 4 H
L-C Oscillations
Single Option Correct MCQs
16. In an L-C circuit, the capacitor has maximum charge q0.The value of max
is
(3) 0 1 q
(4) 0 1 q
17. In oscillating L-C circuit, the total stored energy is U and maximum charge upon capacitor is Q. When the charge upon the capacitor is 2 Q , the energy stored in the inductor is (1) 2 U (2) 4 U (3) 4 3 U (4) 3 4 U
18. A fully charged capacitor C with initial charge q0 is connected to a coil of self inductance L at t = 0. The time at which the energy is stored equally between the electric and the magnetic field is (1) 4 LC π (2) 2 LC
19. The frequency of oscillation of current in the inductor is
Numerical Value Questions
20. In an oscillating L-C circuit in which C = 4.00 μF, the maximum potential difference across the capacitor during the oscillation is 1.50 V and the maximum current through the inductor is 50.0 mA. If the frequency of the oscillations is x × 10 Hz, then find the value of x.
21. The natural frequency of an L-C circuit is 125 kHz. When the capacitor is totally filled with a dielectric material, the natural frequency decreases by 25 kHz. Dielectric constant of the material is .(Round off to nearest integer)
22. In the circuit as shown C = 1 F, L = 11 H. Charge in the capacitor is 4 C at time t = 0 and it is decreasing at the rate of 3C/s . Find the maximum charge (in C) in the capacitor. L C
23. An ideal transformer with purely resistive load operates at 12 kV on the primary side. It supplies electrical energy to a number of nearby houses at 120 V. The average rate of energy consumption in the houses served by the transformer is 60 kW. The value of resistive load (Rs) required in the secondary circuit will be mΩ.
Level-III
Multiple Concept Questions
Single Option Correct MCQs
1. In the shown AC circuit, phase difference between currents I1 and I2 is
(1) 1 tan 2 L X R π
(2) 1 tan LCXX R
(3) 1 tan 2 L X R π + (4) 1 tan 2 LCXX R +π
2. In figure below, if Z L = Z C and reading of ammeter is 1 A, find the value of source voltage V.
3. In series L-C-R circuit at resonance, current versus frequency graph is shown below for different resistances. Choose correct alternative. l f
(1) 80 volt
(2) 60 volt
(3) 100 volt
(4) None of these
(1) R1 > R2 > R3
(2) R1 < R2 < R3 (3) R1 = R2 = R3
(4) impedance of circuit will be same for any value of R.
4. Find the rms and average value of the waveform shown in figure.
(1) 8.5, 10
(2) 10.3, 20 (3) 15.3, 15 (4) 2.6, 5
5. Determine the rms value of a semi-circular current wave which has a maximum value of a
(1) 2.515a (2) 1.815a (3) 0.615a (4) 0.818a
6. The current in a discharging LR circuit is given by i = i 0 e t /τ , where τ is the time constant of the circuit. Calculate the rms current for the period t = 0 to t = τ.
(1)
7. A resistor of resistance 100 Ω is connected to an AC source ε = 12sin(250 π t) V. Find the energy dissipated as heat during t = 0 to t = 1.0 ms.
(1) 0.60 × 104 J (2) 0.61 × 10–4 J
(3) 2.61 × 10–4 J (4) 2.61 × 10–6 J
8. A lamp consumes only 50% of peak power in an AC circuit. What is the phase difference between the applied voltage and the circuit current?
(1) 6 π (2) 3 π (3) 4 π (4) 2 π
9. A constant voltage at different frequencies is applied across a capacitance C as shown in the figure. Which of the following graphs correctly depicts the variation of current with frequency?
Numerical Value Questions
11. A circuit consisting of a capacitor and a coil in series is connected to the mains. Varying the capacitance of the capacitor, the horse power generated in the coil was increased η = 1.7 times. How much (in per cent) was the value of cosϕ changed in the process?
12. In an AC circuit, V and I are given by V = 100 sin(100 t ) V; 100sin100 mA. 3 π =+
It
The power dissipated in the circuit is watt.
13. As shown in the figure, the value of inductive reactance X L (in ohms) will be (if source voltage is 100 volt)
R1=20Ω l=2A R2=20Ω X=30Ω X=40Ω XL
14. Power factor of the A.C. circuit as shown in the figure is t.
VL=80V R=40Ω
i=1A
VC=50V
V.50Hz
10. In the given L-C-R series circuit, the readings are VR = 220 V, VL = 110 V, then the value of capacitance is p μF. What is the value of p?
C
10 (2)
15. A series LRC circuit is connected to an AC source (1 V, f). The voltage drops across R, L and C respectively, are 1 V, 4 V and 4 V. Find the current reading (in milliampere). 1V 4V
0.4mH 500Ω pF 1 V, f mA 4V
16. An L-C-R series circuit with 100 Ω resistance is connected to an AC source of 200 V and angular frequency 300 rad/s. When only the capacitance is removed, the current lags behind the voltage by 60°, when only the inductance is removed the current leads the voltage by 60°. Calculate the current in L-C-R circuit in amperes.
17. A circuit draws a power 550 watt from a source of 220 volt, 50 Hz. The power factor of the circuit is 0.8 and the current lags in phase with the potential difference. To make the power factor of the circuit as 1.0, if a capacitance of 1.5n μF will have be connected with it, then find the value of n. (to nearest integer)
18. A current of 4 A flows in a coil when connected to a 12 V DC source. If the same coil is connected to a 12V, 50 Hz π AC source, a current 2.4 A flows in the circuit. If the inductance of the coil is y × 10 −2 H, then find the value of y
19. An L-C-R series circuit with 10 Ω resistance is connected to an AC source of 200 V and angular frequency 300 rad/s. When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60°. Calculate the power (in kW) dissipated in the L-C-R circuit.
20. The current in a circuit is given as I = I1|sin w t| and its mean value is I0. If the rms value of the current is 0 I x π , then find the value of x.
21. It takes time, t0, for a direct current, I0, to charge a storage battery. Now an AC source whose effective current is I0, is connected along with half wave rectifier. If it takes 0 t x π time to charge the same battery, then find the value of x
22. The average and rms values for the shape shown are V1 and V2. Then 2 1 22 2 Vk V = π , where k =
23. A 20 Ω resistor is connected to an AC power supply of frequency 50 Hz with a voltage output that varies from 4 V to –2 V at equal time intervals as shown in the figure. If the average heating power dissipated in the resistor is W 2 x
, then find the value of x.
24. In the given AC circuit, the rms voltage of the source is 60 V and the rms current passing through the source is (10 + x) A. The value of x is XL=3Ω XC=3Ω
C1 L2 L1 60V(rms) XL=7Ω XC=19Ω R=5Ω
25. A coil having a resistance of 5 Ω and an inductance of 0.02 H is arranged in parallel with another coil having a resistance of 1 Ω and an inductance of 0.08 H. Calculate the power absorbed (in watt) to nearest integer when a voltage of 100 V at 50 Hz is applied. 1Ω 5Ω 0.02H 0.08H 100V 50Hz
26. An AC series circuit contains 4 Ω resistance and 3 Ω inductive reactance. What is the impedance (in Ω)of the circuit?
27. At resonance, the value of the power factor in an L-C-R series circuit is .
28. In the adjoining figure the impedance (in ohm) of the circuit will be .
= 30 Ω
= 20 Ω
29. An inductive circuit contains a resistance of 10 ohm and an inductance of 2.0 henry. If an AC voltage of 120 volt and frequency of 60 Hz is applied to this circuit, the current (in A) in the circuit would be .
THEORY-BASED QUESTIONS
Single Correct MCQs
1. In a pure inductive AC circuit, the current
(1) Lags behind the applied emf by an angle π
(2) Lags behind the applied emf by an angle π/2
(3) Leads the applied emf by an angle π/2
(4) Applied emf are in same phase
2. The resistance of a coil for DC is 5 Ω. In case of AC, the resistance : (1) will remain 5 Ω (2) will decrease (3) will increase (4) will be zero
3. An AC source is connected to a capacitor. The current in the circuit is I . Now, a dielectric slab is inserted into the capacitor, then the new current is : (1) equal to I (2) more than I (3) less than I (4) sometimes more and sometimes less than I
4. Alternating current cannot be measured by DC ammeter because:
30. As given in the figure, a series circuit connected across a 200 V, 60 Hz line consists of a capacitor of capacitive reactance 30 ohm , a non-inductive resistor of 44 ohm, and a coil of inductive reactance 90 ohm and resistance 36 ohm. The power dissipated in the coil is (watt) . XC = 30 Ω
(1) AC cannot pass through DC ammeter
(2) AC changes direction
(3) Average value of current for complete cycle is zero
(4) DC ammeter will get damped
5. The phase difference between the alternating current and emf is π/2. Which of the following combination is not used
(1) C alone (2) R,L
(3) L, C (4) L alone
6. An inductor used to reduce the high frequency components of an alternating signal by presenting a higher impedance, is called :
(1) inverter (2) converter
(3) cell (4) choke coil
7. An inductance L , a capacitance C and a resistance R may be connected to an AC source of angular frequency ω, in three different combinations of RC, RL and L-C
in series. Assume that 1 L C ω= ω The power drawn by the three combinations are P1, P2, P 3 respectively. Then
(1)
(3)
8. If LCR–circuit if resistance increases, quality factor
(1) increases finitely
(2) decreases finitely (3) remains constant
(4) none of these
9. To reduce the resonant frequency in an L-C-R series circuit with a generator
(1) the generator frequency should be reduced
(2) another capacitor should be added in parallel to the first (3) the iron core of the inductor should be removed
(4) dielectric in the capacitor should be removed
10. In series L-C-R circuit, at resonance.
a) The circuit behaves as a pure resistor.
b) Potential difference across inductor and capacitor are equal and opposite.
c) Current in the circuit is maximum.
(1) Only a, b are correct
(2) Only b, c are correct
(3) All a, b, c are wrong
(4) All a, b, c are correct
11. A lamp is connected in series with a capacitor
A) When a DC source is connected to the combination, the lamp will not glow in steady state
B) When AC source is connected to the combination, the brightness of glow of lamp increases with increases of frequency of AC.
(1) A is true B is false
(2) A is false B is true
(3) A and B are true
(4) A and B are false
12. A step up transformer is used to
(1) increase the current and increase the voltage
(2) decrease the current and increase the voltage
(3) increase the current and decrease the voltage
(4) decrease the current and decrease the voltage
13. If I1, I2 and I 3 represent the average value, rms value and peak value of an alternating current, which of the following is the correct ascending order of them?
(1) I 3 < I2 < I1 (2) I2 < I1 < I 3
(3) I1 < I 3 < I2 (4) I1 < I2 < I 3
14. Consider the following statements A and B and select the appropriate choice from those given below
A: Transformers work on the principle of mutual induction
B: In a transformer, the ratio of primary turns to secondary turns is called the transformation ratio
(1) A is true, B is false
(2) A is false, B is true
(3) Both A and B are false
(4) Both A and B are true
15. Given below are two Statements:
Statement I : The main reason for preferring use of AC voltage over DC voltage is that AC voltages can be easily and efficiently converted from one voltage to the other by means of transformers.
Statement II : When bulk pieces of conductors are subjected to changing magnetic flux, induced currents are produced in them, their flow patterns resemble swirling eddies in water, these currents are called eddy currents.
In light of the given statements, choose the correct answer from the options given below
(1) Both statement I and statement II are true
(2) Both statement I and statement II are false
(3) Statement I is true but statement II is false
(4) Statement I is false but statement II is true
16. A small signal voltage V ( t )= V o sin w t is applied across an ideal capacitor C
(1) Current I(t) is in phase with voltage V(t)
(2) Current I(t) leads voltage V(t) by 180°
(3) Current I(t) lags voltage V(t) by 90°
(4) Over a full cycle the capacitor C does not consume any energy from the voltage source.
17. In series L-C -R resonance circuit , if we change the resistance only, from a lower to higher value:
(1) The quality factor and the resonance frequency will remain constant
(2) The quality factor will increase
(3) The bandwidth of resonance circuit will increase
(4) The resonance frequency will increase
18. Match List-I with List-II
List I List II
(A) Phase difference between current and voltage in a purely resistive AC circuit (I) 2 π ;Current leads voltage
(B) Phase difference between current and voltage in a pure inductive AC circuit (II) zero
(C) Phase difference between current and voltage in a pure capacitive AC circuit (III) 2 π ;current lags voltage
(D) Phase difference between current and voltage in an L-C-R series circuit (IV)
Choose the most appropriate answer from the options given below:
(A) (B) (C) (D)
(1) III IV II I
(2) II III IV I
(3) I III IV II
(4) II III I IV
19. Match List‐I with List‐II : List I List II
(A) 1 L C ω> ω (I) Current is in phase with emf
(B) 1 L C ω= ω (II) Current lags behind the applied emf
(C) 1 L C ω= ω (III) Maximum current occurs
(D) Resonant frequency (IV) Current leads the emf
Choose the most appropriate answer from the options given below:
(A) (B) (C) (D)
(1) II I IV III
(2) II I III IV
(3) III I IV II
(4) IV III II I
20. A series L-C-R circuit is tuned to resonance. The impedance of the circuit now is
(1)
(2)
(3)
(4) R
21. Given below are two Statements:
Statement I : The reactance of an AC circuit is zero. It is possible that the circuit contains a capacitor and an inductor. Statement II : In AC circuit, the average poser delivered by the source never becomes zero. In the light of the above statements, choose the correct answer from the options given below: In light of the given statements, choose the correct answer from the options given below
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Assertion and Reason Type Questions
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A)
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A)
(3) if (A) is true but (R) is false
(4) if both (A) and (R) are false
In light of the given statements, choose the correct answer from the options given below
22. (A) : A choke coil is used instead of a resistor in fluorescent tubes with AC mains.
(R) : A choke coil reduces voltage across tube without wasting power, whereas a resistor would waste power as heat.
23. (A) : Capacitor serves as a block for DC and offers an easy path to AC.
(R) : Capacitive reactance is inversely proportional to frequency.
24. (A) : When capacitive reactance is smaller than the inductive reactance in L-C-R series circuit, emf leads the current.
(R) : The phase angle is the angle between the alternating emf and alternating current of the circuit.
25. (A) : An inductance and a resistance are connected in series with an A.C. circuit. In this circuit the current and the potential difference across the resistance lag behind potential difference across the inductance by an angle π/2.
(R) : In L-R circuit voltage lags the current by phase angle which depends on the value of inductance and resistance both.
26. (A) : The DC and AC both can be measured by a hot wire instrument.
(R) : The hot wire instrument is based on the principle of magnetic effect of current.
27. (A) : AC is more dangerous than DC for human body.
(R) : Frequency of AC is dangerous for human body.
28. (A) : An AC can be transmitted over long distances without much power loss.
(R) : An AC can be stepped up or down with the help of a transformer.
29. (A) : An inductor and a capacitor are called low pass filter and high pass filter respectively.
(R) : Reactance of an inductor is low for low frequency signals and that of a capacitor is high for high frequency signals.
JEE ADVANCED LEVEL
Multiple Option Correct MCQs
1. A current of 4 A flows in a coil when connected to 12 V DC source. If the same coil is connected to a 12 V, 50 rad/s source, a current of 2.4 A flows in the circuit. Then
(1) R = 4 Ω
(2) R = 3 Ω
(3) L=4 H
(4) L=0.08 H
2. Choose correct statement if capacitance increases from zero (0) to infinity (∞).
30. (A) : Inductance coils are usually made of thick copper wire.
(R) : Induced current is more in wire having less resistance.
(1) Current increases from 0 (zero) to maximum then decreases to zero.
(2) Reading of voltmeter first increases and it will be maximum when XL = XC and then decreases.
(3) Power factor of circuit first increases then decreases.
(4) V1 may be greater than V, V1 may be equal V, V1 may be less than V, where V1 is reading of voltmeter and V is source voltage.
3. If 21 LCXX > and R1 = R2 for the circuit shown, then choose the correct alternative.
Hz
(1) Source voltage V will lead the current.
(2) Power factor of box 1 will be greater than power factor of box 2.
(3) Power factor of box 1 is less than power factor of circuit.
(4) Voltage across box-1 will lag current.
4. Choose the correct alternative:
(1) Inductive reactance of choke coil is much lesser.
(2) Power factor of ideal choke coil is nearly zero.
(3) Choke coil is connected in parallel across tube light.
(4) In tube light initial induced voltage across choke coil is greater than supply voltage.
5. In the series L-C-R circuit shown, choose correct option.
(1) Reading of voltmeter V2 may be greater than V (source voltage).
(2) Reading of voltmeter V4 may be equal to V1.
(3) Reading of V4 may be equal to V (source voltage).
(4) Reading of V1, V2 and V3 may be equal.
6. In the shown circuit, if the readings of ideal AC voltmeters V1 and V3 are 300 volt each, then choose correct option/options if reading of ammeter is 10 A:
400 volts
(1) V2 = 300 V (2) V2 = 400 V
(3) R = 10 Ω (4) R = 20 Ω
7. In the given circuit, the AC source has w = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is (are)
(1) the current through the circuit, I, is 0.3 A.
(2) the current through the circuit, I is 0.32 A .
(3) the voltage across 100 Ω resistor = 102 V
(4) the voltage across 50 Ω resistor = 10 V.
8. The voltage of an AC source varies with time according to the relation E = 100sin(100πt). cos(100πt) volt.
(1) The peak voltage of the source is 100 volt.
(2) The peak voltage = 50 volt.
(3) The peak voltage = 1002volt
(4) The frequency of source voltage is 100 Hz.
9. Current in an AC circuit is given by i = (3sinωt + 4cosωt) A, then
(1) rms value of current is 5 A.
(2) mean value of this current in positive one-half period will be 6 A π .
(3) If voltage applied is V = V m sinωt, then the circuit may contain resistance and capacitance only.
(4) If voltage applied is V = V m cosωt, then the circuit may contain resistance and inductance only.
10. In an AC circuit, the power factor
(1) is unity when the circuit contains an ideal resistance only.
(2) is unity when the circuit contains an ideal inductance only.
(3) is zero when the circuit contains an ideal resistance only.
(4) is zero when the circuit contains an ideal inductance only.
11. The current in a certain circuit varies with time as shown. The peak value of current is I0. If I v and I m represent the virtual (rms) and mean value of current for a complete cycle respectively, then
12. Which of the following plots may represent the reactance of a series L-C combination: d Frequency Reactance b c a
(1) a (2) b (3) c (4) d
13. The graphs given below depict the dependence of two reactive impedances X1 and X2 on the frequency of the alternating emf applied individually to them. We can then say that
(1) X1 is an inductor and X2 is a capacitor. (2) X1 is an resistor and X2 is a capacitor. (3) X1 is a capacitor and X2 is an inductor. (4) X1 is an inductor and X2 is a resistor.
14. An inductor-coil, a capacitor and an AC source of rms voltage 24 V are connected in series. When the frequency of the source is varied, a maximum rms current of 6.0 A is observed. If this inductor coil and capacitor is connected to a battery of emf 12 V and internal resistance 4.0 Ω, what will be the current?
(1) 2 A (2) 1.5 A (3) 0.5 A (4) 2.5 A
15. Switch is in position A for a long time. At time t = 0, it is shifted to position B. Find the maximum charge that will accumulate on capacitor.
(1) () E LC R (2) () E LCR (3) LE CR
(4)
16. A charged capacitor and an inductor are connected in series. At time t = 0, the current is zero but the capacitor is charged. If T is the period of resulting oscillations, then the time after which current in the circuit becomes maximum, is
(1) T (2) 4 T (3) 2 T (4) 3 4 T
17. In the L-C circuit, at this time +Q –Q C I L
(1) I is increasing and Q is increasing.
(2) I is increasing and Q is decreasing. (3) I is decreasing and Q is increasing. (4) I is decreasing and Q is decreasing.
18. An L-C series circuit has an oscillation frequency f . Two isolated inductors, each with inductance L and two capacitors each with capacitance C, all are connected in series and circuit is completed. The oscillation frequency is
(1) 4 f (2) 2 f
(3) f (4) 4 f
19. In an AC circuit, the instantaneous voltage is zero when the instantaneous current is maximum. Then AC circuit connected to the source may be a
(1) pure inductor.
(2) pure capacitor.
(3) pure resistor.
(4) combination of inductor and a capacitor.
20. In which of the following AC circuits, there occurs a phase difference of 2 π between voltage and current?
(1) circuit having pure inductance.
(2) circuit having pure capacitance.
(3) circuit having pure inductance and pure capacitance.
(4) none of the above
21. In the circuit shown in the figure, the AC source gives a voltage V = 20cos(2000t) volt. Neglecting source resistance, the voltmeter and ammeter reading will be
(1) 0 V, 0.47 A
(2) 1.68 V, 0.47 A
(3) 0 V, 1.4 A
(4) 5.6 V, 1.4 A
22. Two impedances Z1 and Z2 when connected separately across a 230 V, 50 Hz supply consumed 100 W and 60 W at power factors of 0.5 lagging and 0.6 leading respectively. If these impedances are now connected in series across the same supply, find
A) total power absorbed.
B) the value of the impedance to be added in series so as to raise the overall power factor to unity.
(1) 19 W, 295 Ω
(2) 19 W, 95 Ω
(3) 43.2 W, 195 Ω
(4) 75 W, 195 Ω
23. A step up transformer works on 220 V and gives 2 A to an external resistor. The turns ratio between the primary and secondary coils is 2 : 25. Assuming 100% efficiency, find the secondary voltage, primary current and power delivered.
(1) 2750 V, 25 A, 5500 W
(2) 5500 V, 0.16 A, 2750 W
(3) 1600 V, 1 A, 5000 W
(4) 4000 V, 1 A, 5500 W
24. An AC source rated 100 volts (rms) supplies a current of 10 A (rms) to a circuit. The average power delivered by the source
(1) must be 1000 W.
(2) may be 1000 W.
(3) may be greater than 1000 W.
(4) may be less than 1000 W.
25. In which of the following AC circuits, consumed power will be zero?
(1) Circuit having pure inductance.
(2) Circuit having pure capacitance.
(3) Circuit having a combination of pure inductance and pure capacitance.
(4) none of the above
26. A resistance R = 12 Ω, inductance L = 2 H and capacitance C = 5 mF are connected in series to an AC generator of frequency 50 Hz.
(1) At resonance, the circuit impedance is zero
(2) At resonance, the circuit impedance is 12 ohm
(3) The resonance frequency of the circuit is 1 Hz 0.2 π .
(4) The inductive reactance is less than capacitive reactance.
27. In the L-C-R circuit shown in figure.
XC =20Ω XL =10Ω R =10Ω
V = 200 √ 2 sin w t
(1) current will lead the voltage.
(2) rms value of current is 20 A.
(3) power factor of the circuit is 1 2 .
(4) voltage drop across resistance is 100 V.
28. In an LRC series circuit, the current is given by i = Icos w t. If the voltage amplitudes for the resistor, inductor and capacitor are VR, VL and VC respectively, then
(1) the instantaneous power into the resistor is pR = VRIcos2 w t
(2) the value of instantaneous power into the inductor is pL = –VLIsin w tcos w t
(3) the instantaneous power into the capacitor is pC = –VCIsin w tcos w t
(4) pR + pL + pC equals total power p supplied by the source at each instant of tim e.
29. An alternating emf of frequency 1 2 v LC
is applied to a series L-C-R circuit. For this frequency of the applied emf,
(1) the circuit is at resonance and its impedance is made up only of a reactive part.
(2) the current in the circuit is in phase with the applied emf and the voltage across R equals this applied emf.
(3) the sum of the p.d across the inductance and capacitance equals the applied emf which is 180° ahead of phase of the current in the circuit.
(4) the quality factor of the circuit is w L/R or 1/ωC-R and this is a measure of the voltage magnification (produced by the circuit at resonance) as well as the sharpness of resonance of the circuit.
30. A choke coil of resistance 5 Ω and inductance 0.6 H is in series with a capacitance of 10 μF. If a voltage of 200 V is applied and the frequency is adjusted to resonance, the current and voltage across the inductance and capacitor are I0, V0, I1 and V1 respectively. We have
(1) I 0 = 40 A
(2) V0 = 9.8 kV
(3) V1 = 9.8 kV
(4) V1 = 19.6 kV
Numerical Value Questions
31. The RMS value of an alternating current (in ampere), which when passed through a resistor produces heat three times of that produced by a direct current of 2 A in the same resistor, is____.
32. In a certain circuit, current changes with time according to 2 it = . The rms value of current between t = 2 to t = 4 will be_____.
33. A 120 volt AC source is connected across a pure inductor of inductance 0.70 henry. If the frequency of the source is 60 Hz, the current passing through the inductor is ____A.
34. An emf E = 4cos(1000t) volt is applied to an LR circuit of inductance 3 mH and resistance 4 ohms. The amplitude of current in the circuit is _____ A.
35. One 10 V, 60 W bulb is to be connected to 100 V line. The required induction coil has self inductance of value ____ mH. (f = 50 Hz)
36. In the circuit shown in the figure, the AC source gives a voltage V = 20cos(2000t) V. Neglecting source resistance, the ammeter reading (in ampere) will be______.
37. The impedance of a circuit consists of 3 ohm resistance and 4 ohm reactance. The power factor of the circuit is ________.
38. For the series L-C-R circuit shown in the figure, what is the amplitude of the current at the resonating frequency (in A)?
39. In a series L-C-R circuit, C = 2 μF, L = 1 mH and R = 10 Ω. When the current in the circuit is maximum, at that time the ratio of the energies stored in the capacitor and the inductor will be_____.
40. In an L-C-R series circuit, an inductor 30 mH and a resistor 1 Ω are connected to an AC source of angular frequency 300 rad/s. If the value of capacitance for which the current leads the voltage by 45° is 3 1 10 F x × , then the value of x is_____.
41. The reading of ammeter (in ampere) in the circuit shown will be____. XC = 5 Ω
V XL = 5 Ω R = 55 Ω
42. In a series resonant circuit, the AC voltage across resistance R , inductance L and capacitance C are 5 V, 10 V and 10 V respectively. The AC voltage applied to the circuit will be __ V.
Passage-based Questions
Passage-I:
The currents i1 and i2 in AC circuit are given as (in A): 124sinand4sin33 itit
43. The current i3 can be given as
(1) () 43sin2/3 A t ω−π
(2) () 23cos/3 A t ω+π
(3) 4 sinωt A
(4) 8 A
44. The rms value of i3 is
(1) 26A (2) 6A
(3) 32A (4) 22A
Passage-II:
In the circuit shown in the figure, R = 50 Ω, 12 253 volt and 256sin volt EEt ==ω where w = 100π rads−1. The switch S is closed at time t = 0, and remains closed for 14 minutes, then it is opened.
45. Find the amount of heat produced in the resistor is
(1) 64000 J (2) 56000 J
(3) 63000 J (4) 75000 J
46. If total heat produced is used to raise the temperature of 3 kg of water at 20 °C, what would be the final temperature of water?
(1) 15 °C (2) 25 °C (3) 45 °C (4) 75 °C
47. Find the value of the direct current that will produce same amount of heat in the resistor in same time as combination of DC source and AC source has produced. Specific heat of water = 4200 J/kg°C.
(1) 0.23 A (2) 1.22 A
(3) 2.24 A (4) 3.25 A
Passage-III:
A resistor and a capacitor are connected to an AC supply of 200 V, 50 Hz in series. The current in the circuit is 2 A. If the power consumed in the circuit is 100 watt, then
48. The resistance in the circuit is (1) 100 Ω (2) 25 Ω (3) 12575×Ω (4) 400 Ω
49. In the above problem, the capacitive reactance in the circuit is (1) 100 Ω (2) 25 Ω (3) 12575×Ω (4) 400 Ω
50. In the above problem, the capacitance in the circuit is
(1) 100 F 100 π (2) 25 F 100 π (3) 12575 F 100 × π (4) 1 F 10012575 π×
Passage-IV:
In an L-C circuit shown in figure: L C q + –
C = 1 F, L = 4 H. At time t = 0, charge in the capacitor is 4 C and it is decreasing at a rate of 5 C/s .
51. Maximum charge in the capacitor can be: (1) 6 C (2) 8 C
(3) 10 C (4) 1 d 5 Cs d q t =
52. Charge in the capacitor will be maximum after time t =.......... second
(1) 1 2 2sin 3
(3) 1 2 2n 3 ta
(2) 1 2 2sin 3
(4) None of these
53. Choose the correct option.
(1) Maximum current in the circuit is 4 A.
(2) When current is half its maximum value, charge in capacitor is less than half its maximum value.
(3) Both (1) and (2) are correct.
(4) Both (1) and (2) are wrong.
Passage-V:
In the circuit shown in figure, R 1 = 10 Ω , 2 33 H, 20 and mF 10 2 LRC ==Ω= . Current in L-R1 circuit is I1, in C–R2 circuit is I2 and the main current is I. C R2 R1 L V – 200√2 sin(100t)V
54. Phase different between I1 and I2 is
(1) 0° (2) 90°
(3) 180° (4) 60°
55. At some instant, current in L–R1 circuit is 10A. At the same instant, current in C–R2 branch will be
(1) 5 A (2) 52 A
(3) 56 A (4) 53 A
56. At some instant, I1 in the circuit is 102 A , then at this instant current I will be
(1) 26.5 A (2) 102 A
(3) 202 A (4) 25 A
Passage-VI:
An inductor 20 × 10−3 henry, a capacitor 100 μF and a resistor 50 Ω are connected in series across a source of emf V = 10sin(314t)volt
57. Find the impedence in the circuit.
(1) 56 ohm (2) 40 ohm
(3) 36 ohm (4) 60 ohm
58. Find the energy dissipated in the circuit in 20 minutes.
(1) 615 J (2) 750 J
(3) 953 J (4) 250 J
59. If resistance is removed from the circuit and the value of inductance is doubled, then find the variation of current with time in the new circuit.
(1) 0.43cos(314t) A (2) 0.52cos(214t) A
(3) 0.25cos(114t) A (4) 0.52cos(314t) A
Matrix Matching Questions
60. In Column-I, variation of current I with time t is given in the figure. In Column-II, root mean square current Irms and average current is given. Match the Column-I with corresponding quantities given in ColumnII:
Column I
(A) i
Column II
(I)
(B) i
(II) average current for positive half cycle is i0. (C) i
T/2 T t i0
(III) average current for positive half cycle is 0 . 2 i
(D) i O T/2 T t i0 (IV) For full cycle, average current is zero.
(A) (B) (C) (D)
(1) IV I,IV,III IV,II II
(2) II I,IV,III IV,II IV
(3) IV IV,II IV,II IV
(4) IV IV,II I,IV,III II
61. Four different circuit components are given in each situation of column I and all the components are connected across an AC source of same angular frequency ω = 200 rad/s. The information of phase difference between the current and source voltage in each situation of Column-I is given in Column-II. Match the circuit components in Column-I with corresponding results in Column-II.
Column II
Column I
(A) 500µF 10Ω (I) The magnitude of required phase difference is 2 π .
(B) 5 H (II) The magnitude of required phase difference is 4 π
(C) 3µF 4 H (III) The current leads in phase to source voltage.
(D) 1 kΩ 5 H (IV) The current lags in phase to source voltage.
(A) (B) (C) (D)
(1) II,IV IIV I,III II,III
(2) II,III I,II,III,IV I,IV II,IV
(3) II,III I,IV I,III II,IV
(4) I,III I,IV II,III II,IV
62. Referring to the given circuit, match Column-I with Column-II. 1µF 100 Ω 10 mH
63. In Column-I, some AC circuits are given and in Column-II some circuit quantities are given. Match the entries of Column-I with the entries of Column-II.
Column I Column II
V = 10 sin wt (Volt)
Column I
Column II
(A) For ω = 8000 rad/s (I) Peak current in the circuit is less than 0.1 A.
(B) For ω = 10000 rad/s (II) Voltage across the combination and the current are in same phase.
(C) For ω = 10500 rad/s (III) Voltage across the combination leads the current.
(D) For ω = 12500 rad/s (IV) Current through the circuit leads the voltage across it.
(A) (B) (C) (D)
(1) I,IV II I,III I,III
(2) I II III IV
(3) I,III II I,III I,IV
(4) I,III I,III II I,IV
(D)
(IV) Power factor of series L-C-R circuit is 3 5
(A) (B) (C) (D)
(1) I,II,III,IV I,III I,II,III III
(2) I I,II,III,IV I,II,III,IV III,IV
(3) II I,IV I,IV IV
(4) I,II I,II,III,IV III I,II,III,IV
BRAIN TEASERS
1. A 10 Ω resistor, 10 mH inductor and 100 μF capacitor are connected in series to a 50 V (rms) source having variable frequency. The energy that is delivered to the circuit during one period if the operating frequency is twice the resonance frequency, is J x π . The value of x is_____.
2. In the given circuit, the AC source has ω = 100rads−1. Considering the inductor and capacitor to be ideal, the correct choice (s) is /are
(2) The phase difference between current supplied by source and current in the capacitor B is zero.
(3) The current in the capacitor A is 1 A 510 .
(4) Power dissipated through resistor is 4 watt.
4. In the circuit shown in figure, find the maximum value of (VA – VB)(in volts) after closing the switch in steady state.
(1) The current through the circuit, I is 0.3 A.
(2) The current through the circuit, I is 0.32 A .
(3) The voltage across 100 Ω resistor is 102 A .
(4) The voltage across 50 Ω resistor is 10 V.
3. A resistor R of resistance 2000 Ω and two similar capacitor A and B of capacitance 1 μC each are connected with an AC source of 200 V and 500 rad/s. Now switch is closed at t = 0, then choose the correct option(s).
5. Three alternating voltage sources V1 = 3sinωt volt, V2 = 5sin(ωt + ϕ) volt and V3 = 5sin(ωt − ϕ2) volt are connected across a resistance as shown in the figure (where ϕ 1 and ϕ 2 corresponds to 30° and 127°respectively). Find the rms current (in A) through the resistor. [ Take 2 = 1.4] (Round off to nearest integer)
(1) The phase difference between applied voltage and voltage across the capacitor
6. Figure shows a system of inductor and parallel plate capacitor made of 2 parallel circular plates of area A and filled with dielectric liquid of dielectric constant K as shown. A small leak develops in capacitor and liquid starts to fill the inductor of same dimensions having n turns per unit length. Find the ratio of magnitude of initial to final reactance of circuit after liquid fills the inductor completely.
l l A A K
Given : w 2A2n2 = c2 w→ angular frequency of AC c → speed of light and µ r → relative permeability of liquid
(1) () () 1 1 r K K µ+ (2) () () 1 1 1 r K K −µ (3) () () 1 1 r K K +µ + (4) () () 1 1 1 r K K +
FLASHBACK (Previous JEE Questions)
JEE Main
1. An alternating voltage V(t) = 220 sin100πt volt is applied to a purely resistive load of 50Ω. The time taken for the current to rise from half of the peak value to the peak value is: (2024)
(1) 5 ms (2) 3.3 ms
(3) 7.2 ms (4) 2.2 ms
2. In series LCR circuit, the capacitance is changed from C to 4C. To keep the resonance frequency unchanged, the new inductance should be: (2024)
(1) reduced by 1 4 L
(2) increased by 2L
(3) reduced by 3 4 L
(4) increased by 4L
3. An AC voltage V = 20sin 200πt is applied to a series LCR circuit which drives a current 10sin200 3 It π =π+ . The average power dissipated is: (2024)
7. In the given circuit the AC source has w = 100 rad.s–1. Considering the inductor and capacitor to be ideal, what will be the current flowing through the circuit?
Ω
V f (1) 5.9 A (2) 4.24 A (3) 0.94 A (4) 6 A
(1) 21.6 W (2) 200 W (3) 173.2 W (4) 50 W
4. A capacitor of capacitance 100 μF is charged to a potential of 12 V and connected to a 6.4 mH inductor to produce oscillations. The maximum current in the circuit would be: (2024)
(1) 3.2 A (2) 1.5 A (3) 2.0 A (4) 1.2 A
5. A series LCR circuit with 3 10010 L,CF mH == ππ ,is connecte d across an ac source of 220 V, 50 Hz supply. The power factor of the circuit would be ________. (2024)
6. In an a.c. circuit, voltage and current are give n by : V =100sin(100t)V and 100sin100 3 It π =+ mA respectively. The average power dissipated in one cycle is : (2024)
(1) 5 W (2) 10 W (3) 2.5 W (4) 25W
7. A series L,R circuit connected with an ac source E = (25sin1000t)V has a power factor of 1 2 .If the source of emf is changed to E = (20sin2000t)V, the new power factor of the circuit will be: (2024) (1) 1 2 (2) 1 3 (3) 1 5 (4) 1 7
8. A transformer has an efficiency of 80% and works at 10 V and 4 kW. If the secondary voltage is 240 V, then the current in the secondary coil is : (2024) (1) 1.59 A (2) 13.33 A (3) 1.33 A (4) 15.1 A
9. Primary side of a transformer is connected to 230 V, 50 Hz supply. Turns ratio of primary to secondary winding is 10 : 1. Load resistance connected to secondary side is 46 Ω. The power consumed in it is : (2024)
(1) 12.5 W (2) 10.0 W (3) 11.5 W (4) 12.0 W
10. Primary coil of a transformer is connected to 220 V ac. Primary and secondary turns of the transforms are 100 and 10 respectively. Secondary coil of transformer is connected to two series resistance shown in shown in figure. The output voltage (V0) is: (2024)
(1) 7 V (2) 15 V (3) 44 V (4) 22 V
11. A power transmission line feeds input power at 2.3 kV to a step down transformer with its primary winding having 3000 turns. The output power is delivered at 230 V by the transformer. The current in the primary of the transformer is 5A and its efficiency is 90%. The winding of transformer is made of copper. The output current of transformer is_____A. (2024)
12. A capacitor of capacitance 150.0 μF is connected to an alternating source of emf given by E = 36sin(120π t ) V. The
maximum value of current in the circuit is approximately equal to (2023)
(1) 22 A (2) 2 A
(3) 2 A (4) 1 A 2
13. Given below are two statements:
Statement I : Maximum power is dissipated in a circuit containing an inductor, a capacitor and a resistor connected in series with an AC source, when resonance occurs
Statement II : Maximum power is dissipated in a circuit containing pure resistor due to zero phase difference between current and voltage
In the light of the above statements, choose the correct answer from options given below (2023)
(1) Statement I is true but Statement II is false
(2) Both Statement I and Statement II are false
(3) Statement I is false but Statement II is true
(4) Both Statement I and Statement II are true
14. As per the given graph, choose the correct representation for curve A and curve B. {Where, XC = reactance of pure capacitive circuit connected with AC source, X L = reactance of pure inductive circuit connected with AC Source, R = impedance of pure resistive circuit connected with AC source, Z = Impedance of L-C-R series circuit} (2023)
Curve A Impedance
Curve B
f(HZ)
(1) A = XL, B = R (2) A = XC, B = XL
(3) A = XC, B = R (4) A = XL, B = Z
15. Given below are two statements:
Statement I : When the frequency of an AC source in a series L-C-R circuit increases, the current in the circuit first increases, attains a maximum value and then decreases.
Statement II : In a series L-C-R circuit, the value of power factor at resonance is one.
In the light of given statements, choose the most appropriate answer from the options given below. (2023)
(1) Statements I is correct but Statement II is false.
(2) Statement I is incorrect but Statement II is true.
(3) Both Statement I and Statement II are false.
(4) Both Statement I and Statement II are true.
16. Given below are two statements:
Statement I : An AC circuit undergoes electrical resonance if it contains either a capacitor or an inductor.
Statement II : An AC circuit containing a pure capacitor or a pure inductor consumes high power due to its non-zero power factor.
In the light of given statements, choose the most appropriate answer from the options given below. (2023)
(1) Both statement I and Statement II are true
(2) Statement I is false but statement II is true
(3) Both statement I and Statement II are false
(4) Statement I is true but statement II is false
17. In an L-C oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes x times its initial resonant frequency ω 0. The value of x is (2023)
(1) 1 4 (2) 4 (3) 16 (4) 1 16
18. For the given figures, choose the correct options. (2023)
40
(1) The rms current in circuit (b) can never be larger than that in (a).
(2) The rms current in circuit (b) can be larger than that in (a).
(3) The rms current in figure (a) is always equal to that in figure (b).
(4) At resonance, current in (b) is less than that in (a).
19. In a series LR circuit with X L = R , power factor is P 1. If a capacitor of capacitance C with X C = X L is added to the circuit the power factor becomes P 2. The ratio of P 1 to P 2 will be (2023)
(1) 1 : 2 (2) 1:2 (3) 1 : 3 (4) 1 : 1
20. In the given circuit, rms value of current (Irms) through the resistor R is (2023) XL = 200 Ω XC =100 Ω R
(1) 1 A 2 (2) 20 A
(3) 22 A (4) 2 A
21. If R, XL and XC represent resistance, inductive reactance and capacitive reactance. Then which of the following is dimensionless. (2023)
(1) L C X R X (2) LC R XX
(3) RXLXC (4) LC R XX
22. An oscillating L-C circuit consists of a 75 mH inductor and a 1.2μF capacitor. If the maximum charge to the capacitor is 2.7 μC, then the maximum current in the circuit will be ______mA. (2023)
23. A series combination of resistor of resistance 100 Ω, inductor of inductance 1 H and capacitor of capacitance 6.25 μF is connected to a source. The quality factor of the circuit will be____. (2023)
24. In the circuit shown in the figure, the ratio of the quality factor and the bandwidth is _______s. (2023)
25. An L-C-R series circuit of capacitance 62.5 nF and resistance of 50 Ω , is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is ____mH. (Take π2 = 10) (2023)
26. A series L-C-R circuit is connected to an AC source of 220 V, 50 Hz. The circuit contains a resistance, R = 80 Ω, an inductor of inductive reactance, X L = 70 Ω, and a capacitor of capacitive reactance, XC = 130 Ω. If the power factor of circuit is 10 x , then the value of x
is___. (2023)
27. To increase the resonant frequency in series L-C-R circuit, (2022)
(1) source frequency should be increased.
(2) another resistance should be added in series with the first resistance.
(3) another capacitor should be added in series with the first capacitor.
(4) the source frequency should be decreased.
28. The rms value of conduction current in a parallel plate capacitor is 6.9 μA. The capacitance of this capacitor, if it is connected to 230 V AC supply with an angular frequency of 600 rad/s, will be (2022)
(1) 5 pF (2) 50 pF
(3) 100 pF (4) 200 pF
29. Match List I with List II.
List I List II
(A) AC generator (I) Detects the presence of current in the circuit
(B) Galvanometer (II) Converts mechanical energy into electrical energy
(C) Transformer (III) Works on the principle of resonance in AC circuit
(D) Metal detector (IV) Changes an alternating voltage for smaller or greater value
Choose the correct answer from the options given below (2022)
(A) (B) (C) (D)
(1) II I IV III
(2) II I III IV
(3) III IV II I
(4) III I II IV
30. A series L-C-R circuit has L = 0.01 H, R = 10 Ω and C = 1 μF and it is connected to AC voltage of amplitude( V m ) 50 V. At frequency 60% lower than resonant frequency, the amplitude of current will be approximately (2022)
(1) 466 mA (2) 312 mA
(3) 238 mA (4) 196 mA
31. An alternating emf E = 440 sin(100π t ), is applied to a circuit containing an inductance of 2 H π . If an AC ammeter is connected in the circuit, its reading will be (2022)
(1) 4.4 A (2) 1.55 A
(3) 2.2 A (4) 3.11 A
32. A transformer operating at primary voltage 8 kV and secondary voltage 160 V serves a load of 80 kW. Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be (2022)
(1) 800 Ω and 1.06 Ω
(2) 10 Ω and 500 Ω
(3) 800 Ω and 0.32 Ω
(4) 1.06 Ω and 500 Ω
33. The equation of current in a purely inductive circuit is 5sin(49πt – 30°). If the inductance is 30 mH then the equation for the voltage across the inductor, will be: 22 Let 7
(2022)
35. For a series L-C-R circuit, I vs w curve is shown.
(a) To the left of ω r, the circuit is mainly capacitive.
(b) To the left of ω r, the circuit is mainly inductive.
(c) At ω r, impedance of the circuit is equal to the resistance of the circuit.
(d) At ω r, impedance of the circuit is 0. w r w I max I
Choose the most appropriate answer from the options given below. (2022)
(1) (a) and (d) only (2) (b) and (d) only
(3) (a) and (c) only (4) (b) and (c) only
36. A 110 V , 50 Hz, AC source is connected in the circuit (as shown in figure). The current through the resistance 55 Ω, at resonance in the circuit, will be............... A. (2022)
55 Ω L C
(1) 1.47sin(49πt – 30°)
(2) 1.47sin(49πt + 60º)
(3) 23.1sin(49πt – 30º)
(4) 23.1sin(49πt + 60º)
34. A direct current of 4 A and an alternating current of peak value 4 A flow through resistance of 3 Ω and 2 Ω respectively. The ratio of heat produced in the two resistances in same interval of time will be (2022)
(1) 3 : 2 (2) 3 : 1 (3) 3 : 4 (4) 4 : 3
AC source
37. A capacitor of capacitance 500 μF is charged completely using a DC supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an L-C circuit. The maximum current in L-C circuit will be ___A . (2022)
38. The frequencies at which the current amplitude in an L-C-R series circuit becomes 1 2 times its maximum value, are 212 rad
s–1 and 232 rad s–1. The value of resistance in the circuit is R = 5 Ω. The self-inductance in the circuit is _____ mH. (2022)
39. To light a 50 W, 100 V lamp it is connected, in series with a capacitor of capacitance 50 F x µ π , with 200 V, 50 Hz AC source. The value of x will be ____. (2022)
40. An inductor of 0.5 mH, a capacitor of 200 μF and a resistor of 2 Ω are connected in series with a 220 V AC source. If the current is in phase with the emf, the frequency of AC source will be _________ × 10 2 Hz. (2022)
JEE Advanced
41. A series L-C-R circuit is connected to a 45sinω t V source. The resonant angular frequency of the circuit is 105 rads −1 and current amplitude at resonance is I0. When the angular frequency of the source is ω = 8 × 104 rads−1, the current amplitude in the circuit is 0.05I0. If L = 50 mH, match each entry in list-I with an appropriate value from list-II and choose the correct option. (2023)
List-I
List-II
(A) I 0 in mA (I) 44.4
(B) The quality factor of the circuit (II) 18
(C) The bandwidth of the circuit in rad s−1 (III) 400
(D) The peak power dissipated at resonance in watt (IV) 2250 (V) 500
(A) (B) (C) (D)
(1) II III V I
(2) III I IV II
(3) IV V III I
(4) IV II I V
Passage-I:
A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported
to a place 20 km away from the power plant for consumers' usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers' end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the power cable is purely resistive and the transformers are ideal with power factor unity. All the currents and voltages mentioned are rms values.
(2013)
42. If the direct transmission method with a cable of resistance 0.4 Ω km–1 is used, the power dissipation(in %) during transmission is___.
43. In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1 : 10. If the power to the consumers has to be supplied at 200 V, the ratio of the number of turns in the primary to that in the secondary in the stepdown transformer is____.
44. An AC voltage source of variable angular frequency w and fixed amplitude V 0 is connected in series with a capacitance C and an electric bulb of resistance R (inductance zero). When ω is increased (2010)
(1) the bulb glows dimmer.
(2) the bulb glows brighter.
(3) total impedance of the circuit is unchanged.
(4) total impedance of the circuit increases.
CHAPTER TEST – JEE MAIN
Section – A
1. In pure inductive circuit, the curves between frequency f and reciprocal of inductive reactance L 1 X is (1) f XL 1
(2) f XL 1 (3) f XL 1 (4) f XL 1
2. A sinusoidal voltage V(t) = 210sin(3000 t) volt is applied to a series L-C-R circuit in which L = 10 mH, C = 25 μF and R = 100 Ω. The phase difference (Φ) between the applied voltage and resultant current will be (1) tan–1(0.17) (2) tan–1(9.46) (3) tan–1(0.30) (4) tan–1(13.33)
3. In a series circuit R = 300 Ω, L = 0.4H, C = 8 μF and w = 1250 rad/s. The impedance of the circuit is (1) 1300 Ω (2) 900 Ω (3) 500 Ω (4) 400 Ω
4. The phase difference between the voltage and the current in an AC circuit is 4 π . If the frequency is 50 Hz, then this phase difference will be equivalent to a time of (1) 0.02 s (2) 0.25 s (3) 2.5 ms (4) 25 ms
5. In the figure, which of the phasor diagram represents RL-C circuit driven at resonance?
6. A series L-C-R circuit with inductance 10 H, capacitance 10 μF, resistance 50 Ω is connected to an AC source of voltage, V = 200 sin(100 t) volt. If the resonant frequency of the L-C-R circuit is ν 0 and the frequency of the AC source is ν, then
(1) 0 50 Hz vv== π
(2) 0 50 Hz,50Hzvv== π
(3) 0 100 100Hz, Hz vv== π
(4) v0 = v = 50Hz
7. A circuit containing capacitors C1 and C2 as shown in the figure are in steady state with key K1 closed and K2 is opened. At the instant t = 0, if K1 is opened and K2 is closed, then the maximum current in the circuit will be
10. A 300 Ω resistor is connected in series with a parallel-plate capacitor of capacitance across the terminals of a 50 Hz AC generator. The ratio of current when the gap between the capacitor has vacuum to that when ruby mica of dielectric constant k = 2 is inserted between the plates, is equal to
(1) 5 13 (2) 13 5
(3) 22 7 (4) 7 22
11. The current passing through a choke coil of 5 henry is decreasing at the rate of 2 A/s. The emf developing across the coil is (1) 10 V (2) –10 V
(3) 2.5 V (4) –2.5 V
(1) 1 A (2) 1 A 2 (3) 2 A (4) 3 A 2
8. In the circuit, 1 and 2 are joined for long time. After disconnecting 1 and 2, 2 and 3 are connected. The charge on the capacitor when the current in the inductor (ideal) decreases by 3 A is [nearly]
12. A choke coil and capacitor are connected in series and the current through the combination is maximum for AC of frequency n. If they are connected in parallel, at what frequency is the current through the combination minimum?
(1) n (2) 2 n
(3) 2n (4) None of these
13. An alternating emf of angular frequency w is applied across an inductance. The instantaneous power developed in the circuit has an angular frequency
(1) 4 ω (2) 2 ω (3) w (4) 2 w
9. The number of turns in the coil of an AC generator is 5000 and the area of the coil is 0.25 m2. The coil is rotated at the rate of 100 cycles/s in a magnetic field of 0.2 T. The peak value of the emf generated is nearly (1) 786 kV (2) 440 kV (3) 220 kV (4) 157.1 kV
14. A virtual current of 4 A and 50 Hz flows in an AC circuit containing a coil. The power consumed in the coil is 240 W. If the virtual voltage across the coil is 100 V, its inductance will be (1) 1 H 3
(2) 1 H
(3) 1
15. In the series L-C-R circuit as shown in figure, the heat developed in 20 seconds and amplitude of wattless current is XC = 4 Ω XL = 7 Ω R = 4 Ω E = 25 sin(100π t +π /2)
(1) 1500 J, 3 A (2) 1000 J, 4 A
(3) 8000 J, 3 A (4) 1000 J, 3 A
16. In an AC circuit, the instantaneous emf and current are given by ε = 100sin30 t , 20sin30 4 it π =−
. In one cycle of AC the average power consumed by the circuit and the wattless current are, respectively:
(1) 50W, 10 A (2) 1000 W,10A 2
(3) 50 W,0 2 (4) 50W, 0
17. For the given circuit, comment on the type of transformer used.
(1) Step down transformer
(2) Step up transformer
(3) Auto transformer
(4) Auxiliary transformer
18. A 50Hz, A.C. current of crest value 1 A 3 , flows through the primary of transformer. If mutual inductance between the primary and secondary is 0.75 H, then crest voltage induced in the secondary is
(1) 157 V (2) 80 V
(3) 78.5 V (4) 75 V
19. The equation of the alternating voltage applied to our houses is e = 311sin(100π t) volt. A hot wire instrument is used to
measure the alternating voltage. The rms value of emf is (1) 219.9 V (2) 329.1 V (3) 543.7 V (4) 432.7 V
20. The figure shows variation of R, XL and XC with frequency f in a series L-C-R circuit. Then for what frequency point, the circuit is inductive.
XL R
B C f (1) A (2) B (3) C (4) All points
Section – B
21. A series circuit contains a resistor of 20 Ω, a capacitor and an ammeter of negligible resistance. It is connected to a source of 200 V, 50 Hz. If reading of ammeter is 2.5 A, calculate the reactance of the capacitor (in Ω).
22. A 50 μF capacitor is connected to a 100 V, 50 Hz AC supply. Determine the rms value of the current in the circuit. (in A)
23. The rms value of current in an AC circuit is 20 A. What is its peak current in A’? [Take 2 = 1.414]
24. In an oscillating L-C circuit, the maximum charge on the capacitor is Q . The charge on the capacitor when the energy is stored equally between the electric and magnetic field is Q N . The value of N is____.
25. A circuit has an inductance of 100 mH and carries a current of 3 A. To prevent sparking when the circuit is switched off, a capacitor is connected across the switch. If the capacitor can withstand a maximum voltage of 300 V, the minimum capacitance in μF, must be equal to _____.
CHAPTER TEST – JEE ADVANCED
2022
P2 Model
Section – A
[Integer Value Questions]
1. In the circuit shown, reactance of each capacitor is 4R, and that of each inductor is 3 R . If R = 5Ω, then find reading of ammeter. (in amperes)
2R/3 R/3
C/4
C/4
C/4
C/4 L L L
V =(10√ 2cosw t) volt
2. A series RC combination is connected to an AC voltage of angular frequency w = 500 rads−1. If the impedance of the RC circuit is 1.25 R Ω , the time constant (in millisecond) of the circuit is____.
3. The power factor of the circuit shown in the figure is P × 10−1. What is the value of P?
R = 20 Ω XL = 100 Ω XC = 20 Ω 40 Ω
220 V. 50Hz
4. The value of L , C and R in an L-C-R series circuit are 4 mH, 40 pF and 100 Ω respectively. The quality factor of the circuit is 10n. What is the value of n?
5. In the series L-C-R circuit as shown in figure, if the value of V is 5 × 10P V, the value of P is_____. R = 20 Ω
6. In a series L-C-R circuit, the angular frequencies ω1 and ω2 at which the current amplitude falls to 1 2 of the current amplitude at resonance are separated by frequency interval R xL . The value of x is_____.
7. Three sources of emfs, V0sinωt, V0cosωt and V0 are connected in series as shown. If the rms value of current in the circuit is VP R , what is the value of P? R
8. In the circuit shown, XC = 100 Ω, XL = 200 Ω and R = 100 Ω. If the maximum current through the source is 2 A P , then what is the value of P? C R L
(200 sin w t ) V –
Section – B
[Multiple Option Correct MCQs]
9. A resistance of 103 Ω, a capacitance of 10−6 F and an inductance of 2 henry are connected in series with a source of 200sin(1000t) volt. Which of the following are correct?
(1) RMS current = 0.1 A.
(2) Impedance = 1414 Ω.
(3) The voltage leads the current by 45°.
(4) Power factor = 0.707.
10. A current of 4 A flows in a coil when connected to a 12 V DC supply. If the same coil is connected to a 12 V AC source of angular frequency 50 rad/s, a current of 2.4 A flows in the circuit.
(1) Inductance of the coil is 0.08 H
(2) Resistance of the coil is 3 ohm
(3) Reactance of the coil is 5 ohm
(4) Inductance of the coil is 0.16 H
11. In an L-R circuit, the value of L is 0.4 π henry and the value of R is 30 ohm. If in the circuit, an alternating emf of 200 volt at 50 cycles per second is connected, the impedance (Z) of the circuit and current ( I) will be
(1) Z = 50 ohm
(2) Z = 60 ohm
(3) I = 2 ampere
(4) I = 4 ampere
12. In an AC series circuit, it is given that R = 10 Ω, XL = 20 Ω and XC = 10 Ω. Then choose the correct options.
(1) Voltage function will lead the current function
(2) Total impedance of the circuit is 102 Ω
(3) Phase angle between voltage function and current function is 45°.
(4) Power factor of circuit is 1 2
13. In the L-C-R series AC circuit shown, the rms readings of voltmeters are V1 = 150 V, V2 = 50 V and the rms voltage of the source is 130 V. Then,
(1) rms voltage across inductor must be 90 V.
(2) rms voltage across capacitor must be 40 V.
(3) rms voltage across resistor must be 120 V.
(4) The power factor of the circuit (5/13).
14. In a certain series L-C-R circuit, it is found that XL = 2XC and phase difference between current and voltage of source is 4 π . If now, capacitance is made one-fourth times initial value and the inductance and resistance are doubled, then
(1) the new circuit is in resonance.
(2) average power in the circuit is maximum.
(3) the impedance in the new circuit versus angular frequency of the source is not a straight line.
(4) the rms voltage across the resistance is equal to the rms voltage across the source.
Section – C
[Single Option Correct MCQs]
15. An alternating emf of e = 200sin(100πt) volt is applied across a capacitor of capacitance 2μF. (Time t is in seconds). The capacitive reactance and the peak current is
(1) 3 2 510 ,410A × Ω×π π
(2) 3 2 510 410, A × ×πΩ π
(3) 2 3 410 ,510A × Ω× π
(4) 4 Ω, 5 × 10−3A
16. A pure resistive circuit element X when connected to an AC supply of peak voltage 200 V gives a peak current of 5 A which is in phase with the voltage. A second circuit element Y, when connected to the same AC supply also gives the same value of peak current but the current lags behind by 900. If the series combination of X and Y is connected to the same supply, the rms value of current is:
(1) 10 2 A (2) 5 2 A
(3) 5 2 A (4) 5 A
17. The following series L-C-R circuit, when driven by an emf source of angular frequency 70 kilo-radians per second, the circuit effectively behaves like 10 Ω 1µF 100µH
(1) purely resistive circuit
(2) series RL circuit
(3) series RC circuit
(4) series L-C circuit with R = 0
18. In a series L-C-R circuit, the frequency of a 10 V, AC voltage source is adjusted in such a fashion that the reactance of the inductor measures 15 Ω and that of the capacitor
ANSWER KEY
JEE Main Levels
Theory-based Questions (1) 2 (2) 3 (3) 2 (4) 3 (5) 2 (6)
11 Ω. If R = 3 Ω, the potential difference across the series combination of L and C will be
JEE Advanced Level
(11) 2,3 (12) 4 (13) 3 (14) 2 (15) 1 (16) 2,4 (17) 2 (18) 3 (19) 1,2,4 (20)1,2,3 (21) 4 (22) 3 (23) 1 (24) 2,4 (25) 1,2,3 (26)
Brain Teasers
Flashback
Chapter Test – JEE Main
Chapter Test – JEE Advanced
ELECTROMAGNETIC WAVES CHAPTER 8
Chapter Outline
8.1 Displacement Current
8.2 Electromagnetic Waves
8.3 Momentum and Radiation Pressure
8.4 Electromagnetic Spectrum
Maxwell formulated a set of equations involving electric and magnetic fields, and their sources, the charge, and current densities. These equations are known as Maxwell's equations. Together with the Lorentz force formula, they mathematically express all the basic laws of electromagnetism.
The most important prediction to emerge from Maxwell's equations is the existence of electromagnetic waves, which are (coupled) time-varying electric and magnetic fields that propagate in space. The speed of the waves, according to these equations, turned out to be very close to the speed of light (3×108 m/s), obtained from optical measurements. This led to the remarkable conclusion that light is an electromagnetic wave.
Hertz, in, 1885, experimentally demonstrated the existence of electromagnetic waves. Its technological use by Marconi and others, in due course, led to the revolution in communication that we are witnessing today.
In this chapter, we discuss the need for displacement current and its consequences. Then, we present a descriptive account of electromagnetic waves, starting from γ-rays (wavelength ~10 –12 m) to long radio waves (wavelength ~106 m)
8.1 DISPLACEMENT CURRENT
The magnetic field due to a conductor carrying current (ic) is determined by using the Ampere's circuital law,
0 c Bdli
When a time varying current is applied to a condenser, there is a flow of electricity. So, there must be a magnetic field created between the plates of the condenser.
However, Maxwell observed that Ampere's circuital law cannot be applied between the plates of the condenser, as in the case of a conductor.
Maxwell observed the inconsistency of Ampere's law in a circuit carrying time varying current with a gap.
To remove the inconsistency, he suggested the existence of an additional current, called displacement current ( id).
Conduction current i c is produced by the time varying magnetic field ∝ c dB i dt and = c idq dt
Displacement current id is produced by the time varying electric field, ∝ d dE i dt .
i) A changing electric field gives rise to a magnetic field. Consider the process of charging of a capacitor and apply Ampere's circuital law.
0 .() Bdlit ...... (1)
ii) To find magnetic field at a point P outside the parallel plate capacitor C, which is
a part of circuit through which a timedependent current i(t) flows.
(a) A loop of radius r, to determine magnetic field at a point P on the loop
iii) Consider a plane circular loop of radius r whose plane is perpendicular to the direction of the current-carrying wire and which is centered symmetrically with respect to the wire. From symmetry, the magnetic field is directed along the circumference of the circular loop and is the same in magnitude at all points on the loop so that if B is the magnitude of the field, then the equation (1) becomes π=µ 0 (2)() Brit ....... (2)
(c) A tiffin-shaped surface with the circular loop as its rim and a flat circular bottom S between the capacitor plates. The arrows show uniform electric field between the capacitor plates.
v) On applying Ampere's circuital law to such surfaces with the same perimeter, the left hand side of equation (1) does not change but the right hand side is zero and not µ 0i , since no current passes through the surface.
vi) So, we have a contradiction. Calculated one way, there is a magnetic field at a point P, whereas calculated another way, the magnetic field at P is zero.
vii) Since the contradiction arises from the use of Ampere's circuital law. This law must be missing something. The missing term must be such that one gets the same magnetic field at point P, no matter what surface is used.
(b) A pot-shaped surface passing through the interior between the capacitor plates with loop shown in (a) as its rim.
iv) Consider a different surface, which has the same boundary, like a pot surface which does not touch the current but has its bottom between the capacitor plates. Its mouth is the circular loop mentioned in the above diagram. Another such surface is shaped like a tiffin box (without lid), as shown in the fig. below.
viii)If the plates of the capacitor have an area A , and a total charge Q , the magnitude of the electric field E between the plates is (Q/A)/ ε 0. The field is perpendicular to the surface S. It has the same magnitude over the area A of the capacitor plates and vanishes outside it. The electric flux Φ E through the surface S, using Gauss's law, is
(3)
If the charge Q on the capacitor plates changes with time, there is a current i = (dQ/dt), so that using equation (3),
This implies that for consistency,
0 dEi dt
This is the missing term in Ampere's circuital law.
ix) The current carried by conductors due to flow of charges is called conduction current. The current due to changing electric field is called displacement current or Maxwell's displacement current.
x) The source of a magnetic field is not just the conduction electric current due to flowing charges but also the time rate of change of electric field. The total current i is the sum of the conduction current denoted by i c and the displacement current denoted by id = 0 dE dt
This means that outside the capacitor plates, we have only conduction current i c = i, and no displacement current, i.e., id = 0. On the other hand, inside the capacitor, there is no conduction current, i.e., i c = 0, and there is only displacement current, so that id = i
(d) The electric and magnetic fields E and B between the capacitor plates, at the point M
xi) The generalised Ampere's circuital law states that the total current passing through any surface of which the closed loop is the perimeter is the sum of the conduction current and the displacement current. The generalised law is
and it is known as Ampere−Maxwell law.
(e) A cross sectional view of (d)
xii) The displacement current has the same physical effects as the conduction current. In some cases, for example, steady electric fields in a conducting wire, the displacement current may be zero, since the electric field E does not change with time. In other cases, like the charging capacitor, both conduction and displacement currents may be present in different regions of space. In most of the cases, they both may be present in the same region of space, as there exists no perfectly conducting or perfectly insulating medium. There may be large regions of space where there is no conduction current, but there is only a displacement current due to time-varying electric fields. In such a region, we expect a magnetic field, though there is no (conduction) current source nearby.
1. A parallel plate capacitor has circular plate of radius R . During charging at any time t , charge on the plate is Q = at. Find the magnetic field at a radial distance R/2 from the common axis of the plates.
Try yourself:
1. A parallel plate capacitor has plate radius R. During charging, charge density on plate is s = s 0t . Find the magnetic induction at a radial distance 3R/2 from the common axis of the plates.
Ans: 00 3 R µσ
8.1.1 Maxwell's Equations
i) Gauss’ law for electricity, with the usual notation
0 ../ netEdAq ...... (1)
a) It establishes a relation between the net charge enclosed in a closed volume to the total electric flux associated with its surface area.
b) It shows that isolated charge does exist and electric lines of force start from positive charge and terminate at negative charge.
ii) Gauss’ law for magnetism,
..0BdA ...... (2)
a) The right hand side of eq. (2) confirms that are equal number of magnetic lines of force enters and leaves a certain volume, as they are closed curves.
b) It shows that isolated magnetic monopoles do not exist.
iii) Faraday’s law in EMI,
EdldB dt .......(3)
This law shows that a time varying magnetic field induces an electrical field.
iv) Ampere−Maxwell law relates the magnetic field induced to the changing electric flux as well as to the current due to flow of charges, as
The above four equations, viz., eq. (1) to (4), are known as Maxwell’s equations.
Lorentz force: The force acting on a charge q moving in a region where electric and magnetic fields, similar to EM wave, are existing simultaneously, is
The Maxwell's equations, along with the Lorentz force, gives the complete description of all magnetic interactions.
TEST YOURSELF
1. A parallel plate condenser consists of two circular plates each of radius 2 cm separated by a distance of 0.1 mm. A time varying potential difference of 5 × 1013 V/s is applied across the plates of the condenser. The displacement current is
(1) 5.50 A (2) 5.56 × 102 A
(3) 5.56 × 103 A (4) 2.28 × 104 A
2. A parallel plate condenser has conducting plates of radius 12 cm separated by a distance of 5 mm. It is charged with a constant charging current of 0.16 A. The rate at which the potential difference between the plates changes is
(1) 1 × 109 Vs-1
(2) 2 × 1010 Vs-1
(3) 3 × 1012 Vs-1
(4) 2 × 109 Vs-1
3. A parallel plate condenser of capacity 100 pF is connected to 230 V of AC supply of 300 rad/s frequency. The rms value of displacement current is
(1) 6.9 μA
(2) 2.3 μA
(3) 9.2 μA
(4) 4.6 μA
4. An AC rms voltage of 2V, having a frequency of 50 kHz, is applied to a condenser of capacity of 10 μF. The maximum value of the magnetic field between the plates of the condenser, if the radius of plate is 10 cm, is (1) 0.4 μT (2) 4π μT (3) 42 Tπµ (4) 40π μT
Answer Key (1) 3 (2) 4 (3) 1 (4) 3
8.2 ELECTROMAGNETIC WAVES
Electromagnetic waves are a type of energy propagation that doesn't require a medium to travel through, unlike mechanical waves, such as sound waves. Electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of wave propagation.
8.2.1 Source of Electromagnetic Waves
i) Neither stationary charges nor charges in uniform motion (steady currents) can be sources of electromagnetic waves. The former produces only electrostatic fields, while the latter produces magnetic fields that, however, do not vary with time.
ii) Only accelerated charges radiate energy in the form of em waves. So, they are the source of an em wave.
iii) An oscillating (accelerating) charge produces an oscillating electric field in space, which produces an oscillating magnetic field, which in turn is a source of oscillating electric field, and so on.
iv) The oscillating electric and magnetic fields regenerate each other and propagate through space as waves, called electromagnetic waves.
v) An electric charge oscillating harmonically with frequency ν produces electromagnetic waves of the same frequency ν
vi) The energy associated with the propagating wave comes at the expense of the energy of the source— the accelerated charge.
8.2.2 History of Electromagnetic Waves
i) In 1865, Maxwell predicted the electromagnetic waves theoretically. According to him, an accelerated charge sets up a magnetic field in its neighbourhood.
ii) In 1887, Hertz produced and detected electromagnetic waves experimentally at a wavelength of about 6 m, (low frequency region).
iii) Seven years later, J.C. Bose became successful in producing electromagnetic waves of wavelength in the range 5 mm to 25 mm. His experiments, like that of Hertz's, were confined to the laboratory.
iv) In 1896, Marconi discovered that if one of the spark gap terminals is connected to an antenna and the other terminal is earthed, the electromagnetic waves radiated could go up to several kilometres.
v) The antenna and the earth wires form the two plates of a capacitor, which radiates radio frequency waves. These waves could be received at a large distance by making use of an antenna earth system as a detector.
vi) Using these arrangements in 1899, Marconi first established wireless communication across the English channel, i.e., across a distance of about 50 km.
8.2.3 Nature of Electromagnetic Waves
Electric and magnetic fields oscillate sinusoidally in space and time in an electromagnetic wave, giving rise to and sustaining each other. The oscillating electric and magnetic fields E and B are perpendicular to each other and to the direction of propagation of the electromagnetic wave. If E is along the x-axis and B is along the y-axis, the direction of propagation is along ×
EB , i.e., along the z-axis.
A linearly polarised electromagnetic wave, propagating in the z-direction with the oscillating electric field E along the x-direction and the oscillating magnetic field B along the y-direction
The electric and magnetic fields shown in the above figure are mathematically represented by
0 ˆˆ sin() x EEiEkzti ==−ω
Electromagnetic waves are self-sustaining oscillations of electric and magnetic fields in free space or vacuum. Electromagnetic waves travel through vacuum with the speed of light c, where 8 00 1 310m/s c ==× µ∈
0 ˆˆ sin() y BBjBkztj ,
where E 0 and B 0 are the amplitudes of the electric field and magnetic field, respectively.
E and B are the instantaneous values.
The planes in which E vector and B vector exist are perpendicular to each other at a given instant. These two planes are perpendicular at all instants but the orientation of planes are variable with respect to time. This is the reason we get unpolarised light.
ω is angular frequency. k = π λ 2 is the magnitude o f the wave vector k and its direction describes the direction of propagation of the wave. The speed of propagation of the wave is ( ω / k).
ω = ck, where =µε00 1/ c
The relation ω = ck is the standard one for waves. In terms of frequency, ν(=ω/2π) and wavelength, λ=π(2/) k ,
The magnitudes of
E and B are related by
where µ 0 = permeability of free space
∈ o= permittivity of free space
The speed of em waves or light in any other medium of permittivity ∈ and permeability µ is Cmed, where 0 med 00 11 C =
therefractiveindex rr med C n C
The velocity of light depends on electrical and magnetic properties of the medium.
Hertz not only showed the existence of electromagnetic waves but also demonstrated that the waves, which had wavelength ten million times that of the light waves, could be diffracted, refracted, and polarised. Thus, he conclusively established the wave nature of the radiation.
Key Insights:
■ The velocity of electromagnetic waves in free space or vacuum is an important fundamental constant. Electromagnetic waves are of different wavelengths but velocity is same.
■ The Poynting vector, × =×= µ
represents the direction of energy flow per unit area per second along the direction of wave propagation. Its unit is Wm–2. It gives the instantaneous value of intensity; average value of intensity of the EM wave is 2 1 2 avgom ICE =∈
■ The medium offers opposition to the propagation of wave. Such a hindrance is called wave impedance (Z) and its value in a medium is given by
{} µ=µµε=εε 00rr and where µ r and ε r are relative permeability and relative permittivity of medium.
a) For vacuum or free space, µ ==Ω ε 0 0 376.6 Z
b) For a conducting medium, the impedance
µω = σ e Z
µ – permeability of medium
ω – angular frequency of incident EM wave
s – conductivity of medium
c) For higher frequencies, Z e is larger. Hence, the EM wave is reflected. So, good conductors are good reflectors.
d) For a perfect conductor, s is very high and hence, Ze is zero.
e) For a transparent medium, refractive index
== Impedanceinfreespace Impedanceinmedium freespace medium Z n Z
2. In an electromagnetic wave in free space, the root mean square value of electric field
is 6 V/m. Find the peak value of magnetic field.
Sol. E m = V 262 m rms E = \ B m = 8 8 62 T2210 T. 310 m E c ==× ×
Try yourself:
2. The amplitude of the magnetic field part of a electromagnetic wave in vacuum is Bo=500 nT. Find the amplitude of electric field part of the wave.
Ans:150 V/m
8.2.4 Average Energy Density of Electromagnetic Waves
Electromagnetic waves carry energy and they travel through the space. This energy is shared equally between the electric and magnetic fields.
In free space, the average energy density uE of static electric field and average energy density uB of static magnetic field are given by =ε= µ 2 2 0 11 and 22EoB B uEu Total energy density due to both static fields is =+=ε+
In an electromagnetic wave, both electric and magnetic fields vary sinusoidally in space and time. Therefore, average density u of an electromagnetic wave can be obtained by replacing E and B by their rms values in the above expression.
∴=ε+ µ 22 0 0 11 22rmsrms uEB Now, == 00 and 22 rmsrms EB EB
∴=ε+ µ 2 2 0 00 0 11 224 E uB
=ε+ µ 22 000 0 11 44 uEB
Since E0 = cB0 and = µε 2 00 1 c
Average electric energy density = () 2 2 22 0 00000 00 111 444 E B uEcB=ε=ε=ε µε
== µ 2 0 1 4 BBu = Average magnetic energy density
Hence, in an electromagnetic wave, half the energy of the wave is contained in the electric field and half in the magnetic field.
\ Average energy density of electromagnetic wave is =+==22 EBEB uuuuu or =ε= µ 2 2 0 00 0 11 22 B uE
Intensity of Electromagnetic Waves
Intensity of electromagnetic wave is defined as the energy crossing per second per unit area of a surface perpendicular to the direction of propagation of the wave. It is denoted by I.
Consider a plane electromagnetic wave propagating along x-axis with speed c. Let us imagine that the wave crosses an area A, normally. In a small time ∆ t, the wave moves a distance = c ∆ t. The energy that has crossed A in the time ∆ t is the energy that occupies the volume ∆ V = A × c ∆ t = Ac ∆ t. If u av is the average energy density of the electromagnetic wave, then, total energy crossing area A in time ∆ t is =×∆=×∆ avav uuVuAcT
\ Intensity of electromagnetic wave is energy
In terms of electric field, =ε 2
In terms of magnetic field, =
Either eq. (i) or eq. (ii) may be used to find the intensity of electromagnetic waves.
The intensity of the electromagnetic radiation from an isotropic point source at a distance r is
P = Power of the source
3. An electromagnetic wave of intensity 100 watt/ m2 is propagating through vacuum. Find the average energy density.
Sol: u avg = 83 100 310 IJ Cm = × =0.333 µJ/m2
Try yourself:
3. A point source in vacuum is emitting energy at the rate P watt. Find the amplitude of electric field at a radial distance r from the source if wavelength of radiation is l.
Hint: 2 0 1 2 m IcE =∈
TEST YOURSELF
1. The electric field part of an electromagnetic wave in a medium is represented by E x =0
2.5cos21062 10 y Nradrad Etx CmS
E z = 0. The wave is
(1) moving along x direction with frequency 106 Hz and wavelength 100 m
(2) moving along x direction with frequency106 Hz and wavelength 200 m
(3) moving along x direction with frequency 106 Hz and wavelength 200 m
(4) moving along y direction with frequency 2π × 106 Hz and wavelength 200 m
2. For a transparent medium, relative permeability and permittivity, μr and r, are 1.0 and 1.44, respectively. The velocity of light in this medium would be,
(1) 2.5 × 108 m/s
(2) 3 × 108 m/s
(3) 2.08 × 108 m/s
(4) 4.32 × 108 m/s
3. A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, 6 ˆ .3 Ej =
The magnetic field B at this point is
(1) 8 4.210 kT × (2) 8 2.110 kT × (3)
18.9108 kT × (4)
4.2108 kT ×
4. A plane e.m. wave of frequency 40 MHz travels along x-axis. At same point, at same instant, the electric field E has maximum value of 750 N/C in y -direction. The magnitude and direction of magnetic field is
(1) 2.5 μT along x-axis
(2) 2.5 μT along y-axis
(3) 2.5 μT along z-axis
(4) 5 μT along z-axis
5. For plane electromagnetic waves propagating in the z direction, which one of the following combinations gives the correct possible direction for and EB field, respectively?
(1) ()() 23 and 32 ˆˆˆˆ ijij
(2) ()() 34 an4 ˆ 3 ˆ ˆˆ d ijij +−
(3) () (2) a ˆˆ nd2 ˆˆ ijij +−
(4) ()() 23 an ˆ 2 ˆ ˆˆ d ijij ++
6. The maximum electric field of a plane electromagnetic wave is 88 V/m. The average energy density is
(1) 3.4 × 10–8 Jm–3
(2) 13.7 × 10–8 Jm–3
(3) 6.8 × 10–8 Jm–3
(4) 1.7 × 10–8 Jm–3
7. The relative permeability of glass is 3/8 and the dielectric constant of glass is 8. The refractive index of glass is
(1) 1.5 (2) 1.1414
(3) 1.732 (4) 1.6
Answer Key
(1) 2 (2) 1 (3) 2 (4) 3
(5) 1 (6) 1 (7) 3
8.3 MOMENTUM AND RADIATION PRESSURE
i) Electromagnetic waves have linear momentum as well as energy. When electromagnetic waves strike a surface, a pressure is exerted on the surface, called radiation pressure.
ii) When electromagnetic waves are incident on a surface and the total energy transferred to the surface in a time t is U, then the magnitude of the momentum delivered to the surface is
= U p c (Complete absorption)
2 ;(); E EmcEmccmcp c ====
where c = velocity of light
iii) When the sun shines on the hands of a person, he feels the absorption of electromagnetic waves (hands get warm). Electromagnetic waves also transfer momentum to the hand but c is very large, the amount of momentum transferred is extremely small, and he does not feel the pressure.
iv) When the radiation incident on a surface is entirely reflected back along its original path, the magnitude of the momentum delivered to the surface is = 2U p c , where c = velocity of light
v) When radiation is incident on a surface, for total absorption case, the radiation pressure is given by r I P c = For total reflection back along the incident path the radiation pressure is 2 I p c µ =
vi) Electromagnetic waves obeys the principle of superposition.
vii) The electric vector of an electromagnetic field is responsible for all optical effects. For this reason, electric vector is also called a light vector.
Key Insights:
■ In 1903, the American scientists Nichols and Hull succeeded in measuring radiation pressure of visible light. It was found to be of the order of 7 × 10 –6 N/m2. Thus, on a surface of area 10 cm2, the force due to radiation is only about 7 × 10–9 N.
■ The great technological importance of electromagnetic waves stems from their
capability to carry energy from one place to another. The radio and TV signals from broadcasting stations carry energy. Light carries energy from the sun to the earth, thus making life possible on the earth.
4. Radiations of wavelength 200 nm, propagating in the form of a parallel beam, fall normally on a plane metallic surface. The power of the beam is 5 mW and its cross-sectional area is 1.0 mm 2. Find the pressure exerted by the radiation on the metallic surface, if the radiation is completely reflected.
Sol: Radiation pressure =
= 33.33×10–6 N/m2
Try yourself:
4. Electromagnetic wave of intensity I is incident normally on a plane surface. If 20% of incident energy is absorbed and 80% of incident energy is reflected back along the same path, then find the radiation pressure.
Ans: 1.8 I c
TEST YOURSELF
Momentum and Radiation Pressure
1. The rms value of the electric field of a plane electromagnetic wave is 314 V/m. The average energy density of electric field and the average energy density are
(1) 4.3 × 10–7 Jm–3 ; 2.15 × 10–7 Jm–3
(2) 4.3 × 10–7 Jm–3 ; 8.6 × 10–7 Jm–3
(3) 2.15 × 10–7 Jm–3 ; 4.3 × 10–7 Jm–3
(4) 8.6 × 10–7 Jm–3 ; 4.3 × 10–7 Jm–3
2. Light with energy flux of 18 W/cm2 falls on a non reflecting surface of area 20 cm2 at normal incidence. The momentum delivered in 30 minutes is
(1) 1.2 × 10–6 kg ms–1
(2) 2.16 × 10–3 kg ms–1
(3) 1.8 × 10–3 kg ms–1
(4) 3.2 × 10–3 kg ms–1
3. Light with energy flux of 24 Wm–2 is incident on a well polished disc of radius 3.5 cm for one hour. The momentum transferred to the disc is
(1) 1.1 μkg ms–1
(2) 2.2 μkg ms–1
(3) 3.3 μkg ms–1
(4) 4.4 μkg ms–1
4. The intensity of TV broad cast station of E = 800 sin(109t – kx ) V/m is......... and the Wavelength in meter is............
(1) 850 Wm–2 ; 0.6 π
(2) 425 Wm–2 ; 0.6 π
(3) 850 Wm–2 ; 0.3 π
(4) 425 Wm–2 ; 0.3 π
5. The sun delivers 103 W/m2 of electromagnetic flux incident on a roof of dimensions 8 m × 20 m, will be
(1) 6.4 × 103 W (2) 3.4 × 104 W
(3) 1.6 × 105 W (4) 3.2 × 105 W
6. A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is ( c = velocity of light)
(1) 2 2 E c (2) 2 E
(4)
(1) 2 (2) 2 (3) 2 (4) 1
(5) 3 (6) 4
8.4 ELECTROMAGNETIC SPECTRUM
The orderly distribution of electromagnetic waves (according to wavelength or frequency) in the form of distinct groups, having widely differing properties, is called the electromagnetic spectrum.
There is no sharp division between one kind of wave and the next. The classification is based roughly on how the waves are produced and/or detected.
8.4.1 Gamma Rays
i) Becquerrel and Curie discovered these rays in 1896.
ii) Their wavelength range is 10–14 to 10–10 m but they have high frequency.
iii) The main sources are the natural and artificial radioactive substances.
iv) They are used in medicine to destroy cancer cells.
v) These affect the photographic plates.
vi) This high frequency radiation is produced in nuclear reactions and also emitted by radioactive nuclei.
8.4.2.
X-rays
i) Roentegen discovered these rays in 1895.
ii) Their wavelength range is 10–13 m to 10–8m
iii) When high energy electrons bombard a metal target, these rays are produced.
iv) They are mainly used in detecting the fracture of bones and also used in the study of crystal structure.
v) X-rays are used in medicine and as a treatment for certain forms of cancer. X-rays damage or destroy living tissues and organisms. So, care must be taken to avoid unnecessary or over-exposure.
vi) They affect the photographic plates and can penetrate through transparent materials.
8.4.3
Ultraviolet Rays
Ultraviolet rays were discovered by Ritter in 1801. The ultraviolet rays are part of the solar spectrum. They can be produced by the arcs of mercury and iron. They can also be obtained by passing discharge through hydrogen and xenon.
i) Wavelengths range from 4 × 10 –7 m to 6 × 10–10 m.
ii) The sun is an important source of ultraviolet light. Most of it is absorbed in the ozone layer in the atmosphere at an altitude of about 40–50 km.
iii) Ozone layer in the atmosphere plays a protective role, and hence, its depletion by chlorofluorocarbons (CFCs) is a matter of international concern.
iv) Exposure to UV radiation induces the production of more melanin, causing tanning of the skin.
Properties
i) Ultraviolet rays are electromagnetic waves and they travel with the speed of 3 × 108 ms–1
ii) They also obey the laws of reflection and refraction.
iii) They can also undergo interference and can be polarised.
iv) When allowed to fall on metals, they cause the emission of photoelectrons.
v) They can affect a photographic plate.
vi) Ultraviolet rays can cause fluorescence in certain materials.
vii)Ultraviolet rays cannot pass through glass but quartz, fluorite and rock salt are transparent to them.
viii)Ultraviolet rays possess the property of synthesising vitamin D, when the skin is exposed to sunlight.
Applications
i) Ultraviolet rays are used for checking the mineral samples by making use of their property of causing fluorescence.
ii) Ultraviolet absorption spectra are used in the study of molecular structure.
iii) Ultraviolet rays destroy bacteria and hence, they are used for sterilising surgical instruments.
iv) As ultraviolet rays can cause photoelectric effect, they are used in burglar alarms.
Key Insights:
■ Welders wear special glass goggles or face masks with glass windows to protect their eyes from large amounts of UV rays produced by welding arcs.
■ Due to shorter wavelengths, UV radiations can be focused into very narrow beams for high precision applications, such as LASIK (laser-assisted in situ keratomileusis) eye surgery.
■ UV lamps are used to kill germs in water purifiers.
8.4.4 Visible Rays
i) These are the most familiar form of electromagnetic waves. They fall in a spectrum, which is the only spectrum that can be detected by the human eye.
ii) Their wavelength ranges from 4 × 10 –7 m to 7 × 10–7 m, from violet to red colour.
iii) These were studied by Newton.
iv) Visible light emitted or reflected from objects around us provides us information about the world.
Key Insights:
■ Different animals are sensitive to different ranges of wavelengths. Snakes can detect infrared waves. Many insects can also detect UV-rays along with visible rays.
8.4.5
Infrared Rays
Infrared rays were discovered by Herschell. Infrared rays are heat radiations and, therefore, all hot bodies are the sources of infrared rays. About 60% of the solar radiation is infrared in nature. Infrared waves, with frequencies lower than those of visible light, vibrate not only the electrons, but all the atoms or molecules of a substance. This vibration increases the internal energy and, consequently, the temperature of the substance. Infrared waves are sometimes referred to as heat waves.
Few Sources of Infrared Rays
Nernst lamp: The filament of Nernst lamp is made from the mixture of zirconium, thorium, and cesium. When a current is passed through such a filament, then at a temperature of about 1200 K, infrared rays are emitted.
Globar: It is basically a rod of silicon carbide, which when heated to a temperature of about 900 K by passing current, produces infrared rays.
LASER: It is used to produce monochromatic infrared rays. For example, He–Ne LASER provides infrared rays of wavelength 0.69 × 10 –6 m, 1.19 × 10–6m, and 3.39 × 10–6m. CO2 LASER provides infrared rays of wavelength 10.6 × 10–6m.
To detect infrared rays, thermocouples, thermopiles, bolometers, and photo conducting cells are used.
Properties
Infrared rays are electromagnetic waves and they travel with the speed of 3×108ms-1.
Infrared rays obey the laws of reflection and refraction.
Infrared rays can produce interference and can be polarised.
When allowed to fall on matter, infrared rays produce an increase in temp erature.
Infrared rays affect a photographic plate. When absorbed by molecules, the energy of infrared rays gets converted into molecular vibrations.
They are scattered less, compared to the visible light by the atmosphere. Hence, infrared rays can travel through longer distances through atmosphere under the conditions of smoke, fog, etc. Nitrogen and oxygen gases are found to be transparent media to all the wavelengths of infrared rays.
Applications
i) In frared rays from the sun keep the earth warm and hence, help to sustain life on earth through greenhouse effect. Incoming visible light is absorbed by the earth's surface and re-radiated as infrared radiations. This radiation is trapped by greenhouse gases, such as carbon dioxide and water vapour.
ii) The coal deposits in the interior of the earth are the result of conversion of forest wood into coal due to infrared rays.
iii) Infrared rays are used in solar water heaters and cookers.
iv) Infrared ray photographs are used for weather forecasting
v) Infrared rays are used for taking photographs during condition s of fog, smoke, etc.
vi) Infrared rays absorption spectra is used in the study of molecular structure and to check the purity of chemicals.
vii) Infrared rays are used for producing dehydrated fruits.
viii)Infrared rays are used to provide electrical energy to satellites by using solar cells.
ix) Infrared rays are used to treat muscular strains.
x) Infrared detectors are used in earth satellites, both for military purposes and to observe growth of crops.
xi) Electronic devices also emit infrared rays and are widely used in remote switches of house-hold electronic systems, such as TV sets, video recorders, and hi-fi systems.
8.4.6 Microwaves (Short Wavelength Radio Waves)
i. The frequency range is 1 GHz to 30 GHz.
ii. They were discovered by Hertz.
iii. They are used in radar and telecommunications.
iv. They are used to analyse the fine details of molecular structure.
v. They are produced by special vacuum tubes called klystrons, magnetrons, and Gunn diodes.
vi. Based on the microwaves, speed guns are designed, which are used to time fast balls, in tennis serves, and automobiles.
vii. Microwaves produce heat when absorbed by matter.
Microwave Oven
Microwave oven is a domestic appliance to cook food items. All food items contain water as a constituent. When the temperature of a body rises, the energy of the random motion of atoms and molecules increases and the molecules travel or vibrate or rotate with higher energies. Frequency of rotation of water molecules is 3GHz. The frequency of microwaves is selected to match the resonant frequency of the water molecule. When such waves pass through water, the KE of the waves is efficiently transferred to the water molecules. This increases the temperature of the water. This increases the temperature of food containing water. The food materials are to be placed in a porcelain container.
The molecules in porcelain vibrate with much smaller frequencies. Because its large molecules vibrate and rotate with much smaller frequencies, it cannot absorb microwaves as they are of smaller frequencies. Hence, the container is unaffected and it is cool. When a metal container is used, it may accumulate some charges and there is a danger of getting a shock, when touched. The heat developed in the container may sometimes melt it. The heat is directly delivered to water molecules, which is shared by the entire food, unlike in conventional heating.
In conventional heating, the vessel on the burner gets heated first, and then the food inside, because of transfer of energy from the vessel.
8.4.7 Radio Waves
i) Radio waves are produced by the accelerated motion of charges in conducting wires.
ii) They are used in radio and television communication systems.
iii) They are generally in the frequency range of 500 kHz to about 1000 MHz.
iv) The AM (amplitude modulated) band is from 530 kHz to 1710 kHz.
v) Higher frequencies, up to 54 MHz, are used for short wave bands.
vi) TV waves range from 54 MHz to 890 MHz.
vii) The FM (frequency modulated) radio band extends from 88 MHz to 108 MHz.
viii)Cellular phones use radio waves to transmit voice communication in the ultra-high frequency (UHF) band.
Low Frequency Waves The AM Band
i) Radio waves having wavelengths of 10 m or more (frequency less than 30 MHz) are said to constitute the AM band.
ii) The lower atmosphere is transparent to these waves, but the ionosphere reflects them back.
iii) A signal transmitted from a certain point can be received at another point in two possible ways, directly along the surface of the earth (called ground wave) and after reflection from ionosphere (called sky wave).
iv) Waves having frequencies up to about 1500 kHz (wavelengths above 200 m) are mainly transmitted through ground because low frequency sky waves lose their energy very quickly compared to the ground waves.
v) Therefore, higher frequencies are mainly transmitted through the sky.
vi) These two regions of the AM band are called medium wave and short wave bands, respectively.
High Frequency Waves Television Transmission
i) Above a frequency of about 30 MHz, the ionosphere does not reflect the wave towards the earth.
ii) The television signals have frequencies in the range 100−200 MHz. Therefore, TV transmission via the sky is not possible.
iii) Only direct reception via the ground is possible.
iv) Therefore, in order to have larger coverage, the transmission has to be done through very tall antennas.
v) The height of transmitting antenna for T V telecast is given by = 2 2 e d h R , where d is the radius of the area to be covered for TV telecast and R e is the radius of the earth.
Key Insights:
■ The basic difference between various types of electromagnetic waves lies in their wavelengths of frequencies since all of them travel through vacuum with the same speed. Consequently, the waves differ considerably in their mode of interaction with matter.
■ The wavelength of the electromagnetic wave is often correlated with the characteristic size of the system that radiates.
■ The centre of sensitivity of our eyes coincides with the centre of the wavelength distribution of the sun. This is because humans have evolved with a vision most sensitive to the strongest wavelengths from the sun.
TEST YOURSELF
1. The energy of the electromagnetic waves is of the order of 15 keV. To which part of the spectrum does it belong?
(1) Y-rays
(2) X-rays
(3) Infrared rays
(4) Ultraviolet rays
2. Which of the following rays has minimum frequency?
(1) UV rays
(2) X–rays
(3) microwaves
(4) infrared rays
3. Electromagnetic waves with wavelength λ1 is used in satellite communication, λ2 is used to kill germs in water purifies, λ3 is used to detect leakage of oil in underground pipelines, and λ4 is used to improve visibility in runways during fog and mist conditions. Arrange these wavelengths in ascending order of their magnitude.
(1) λ3> λ2> λ4 >λ1
(2) λ3< λ2< λ1<λ4
(3) λ3< λ4< λ1 <λ2
(4) λ3< λ2< λ4 <λ1
4. Ozone layer absorbs all such radiation whose wavelengths are
(1) < 3000 Å
(2) > 3000 Å
(3) 4000 Å to 8000 Å
(4) 103 to 106 Å
5. The altitude of the ozone layer in the atmosphere and the ‘type’ of the electromagnetic waves emitted by the sun
CHAPTER REVIEW
Displacement Current
■ Generalised Ampere's law is
c
■ Displacement current id = 0 dE dt φ ∈ , during charging of capacitor.
■ In a region, if there exists a time varying electric field, a magnetic field is induced.
Electromagnetic Waves
■ A charged particle moving with constant velocity produces both electric and magnetic field.
■ An accelerated charged particle radiates energy in the form of electromagnetic wave.
■ The energy associated with the propagating wave comes at the expense of the energy of the source – the accelerated charge.
■ In an electromagnetic wave, both electric field (E) and magnetic field( B) exist.
■ In an EM wave, and EB oscillate in phase. and they are perpendicular to the direction of propagation.
■ Planes of propagation of and EB are mutually perpendicular.
■ The direction of EB × gives the direction of EM wave.
being absorbed by it are, respectively,
(1) 80 km, infrared rays
(2) 40 km, infrared rays
(3) 90 km, ultraviolet rays
(4) 50 km, ultraviolet rays
Answer Key
(1) 2 (2) 3 (3) 4 (4) 1
(5) 4
■ EM wave travels in a transparent medium with the speed of light. In vacuum 00 1 c µ = ∈ and in a medium other than vacuum, 1 v µ = ∈
■ E/B = c in vacuum and E/B = v(<c) in other medium.
■ Refractive index of a transparent medium . rr c µµ v ==∈ , where µ r = relative permeability and ∈ r = relative permittivity of the medium.
■ An EM wave transports energy. Direction and magnitude of energy flow per unit area per second is given by Poynting vector
0 1 SEB µ =× (in vacuum).
■ Wave impedance in a medium is given by 0 0 . r r zµµ = ∈∈
■ For vacuum or free space, 0 0 376.6 zµ==Ω ∈
■ For conducting medium, the impedance zµω = σ , where, µ=permeability of medium, ω = angular frequency of incident
EM wave, and s = conductivity of medium.
■ For a perfect conductor, s is very high and z = 0.
■ When an EM wave travels in space, average energy density is given by 2 2 0 0 0 1 22 m B uE µ =∈=
■ If I is the average intensity of an electromagnetic wave and u is the average energy density, then I = uc
■ If P is the power of a point source in an isotropic medium, then intensity at a radial distance r is I = P/4 π r.2
Momentum and Radiation Pressure
■ When electromagnetic waves are incident on a surface and the total energy transferred to the surface in a time t is U, the magnitude of momentum delivered to the surface is p = U / c [when the incident energy is completely transferred] and P = 2U/c [when the radiation incident on a surface is entirely reflected back along its original path].
■ When radiation of intensity I is incident on a surface radiation pressure Pr= I/c [For total absorption] and P r = 2I/c [when the incident radiation is reflected back along the same path].
Electromagnetic Spectrum
■ Radio waves are produced by the accelerated motion of charges in conducting wires.
■ Frequency of radio waves generally lies between 500 kHz to 1000 kHz.
■ Frequency of microwaves lies between 1 GHz to 30 GHz.
■ Sources of infrared rays are all hot bodies. About 60% of solar radiation is infrared in nature.
■ Wavelength of visible radiation lies between 4000 Å to 7000 Å.
■ Ultraviolet radiation ranges from 4 × 10–7m to 6 × 10–10 m.
■ Wavelength of X-rays ranges from 10–13 m to 10–8 m.
Exercises
JEE MAIN LEVEL
Level-I
Displacement Current
Single Option Correct MCQs
1. A parallel plate condenser has conducting plates of radius 8 cm seperated by a distance of 2.0 mm. It is charged by an external source, with a constant charging current of 0.15 A. The displacement current is
(1) 0.10 A (2) 0.15 A
(3) 0.20 A (4) 0.30 A
Electromagnetic Waves
Single Option Correct MCQs
2. An electromagnetic wave in vacuum has the electric and magnetic fields E and B which are always perpendicular to each other. The direction of polarization is given by X and that of wave propagation by k , then
(1) and XEkEB ×
(2) and XBkEB ×
(3) and XEkBE ×
(4) and XBkBE ×
3. A linearly polarized electromagnetic wave given as () 0 c ˆ os r EEikzt =−ω is incident normally on a perfectly reflecting infinite wall at z = a. Assuming that the material of the wall is optically inactive, the reflected wave will be given as
(1) () 0 cos ˆ r EEikzt =−−ω
(2) () 0 c ˆ os r EEikzt =+ω
(3) () 0 cos ˆ r EEikzt =−+ω
(4) () 0 s ˆ in r EEikzt =−ω
4. An electromagnetic wave travelling in the x-direction has frequency of 3 × 1014 Hz and electric field amplitude of 27 Vm–1. From the options given below, which one describes the magnetic field for this wave ?
(1) ()() 8 6 14 ,910sin21.51021 ˆ 0 BxtTjxt =×π×−×
(2) ()() 8 6 14 ,310sin21.01031 ˆ 0 BxtTjxt =×π×−×
(3) ()() 8 8 14,910sin21.510310 BxtTixt =×π×−×
(4) ()() 8 6 14 ,910sin21.01031 ˆ 0 BxtTkxt =×π×−×
5. The electric field for an electromagnetic wave in free space is ()30cos5108 ˆ Ekzti =−× where magnitude of E is in Vm -1 . The magnitude of wave vector, k is (velocity of em wave in free space = 3 × 10 8 ms-1)
(1) 0.46 rad.m–1
(2) 3 rad.m–1
(3) 1.66 rad m–1
(4) 0.83 rad m–1
6. For a plane electromagnetic wave, the magnetic field at a point x and time t is ()() 7 3 ,1.210sin0.5101.51011 T ˆ Bxtxtk =××+× The instantaneous electric field E corresponding to B is (speed of light c = 3 × 108 ms–1)
(1) ()() 3 11 V ,36sin0.5101.51 m ˆ 0 Extxtj =−×+×
(2) ()() 3 11 V ,36sin1100.510 m ˆ Extxtj =×+×
(3) ()() 3 11 V ,36sin0.5101.510 m ˆ Extxtk =×+×
(4) ()() 3 11 V ,36sin1101.510 m ˆ Extxti =×+×
7. In a plane electromagnetic wave, the directions of, electric field and magnetic field are represented by ˆ k and 2 ˆ 2 ˆ ij respectively. What is the unit vector along direction of propagation the wave?
(1) () 1 2 ˆ ˆ ij + (2) () 1 2 ˆ ˆ jk + (3) () 1 5 ˆ 2 ˆ ij + (4) () 1 2 5 ˆ ˆ ij +
8. A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, 6.3V/m ˆ Ej = corresponding magnetic field B , at that point will be
(1) 8 18.910 ˆ T k × (2) 8 2.110 ˆ T k × (3) 8 6.310 ˆ T k × (4) 8 18.910T ˆ k ×
9. If an electromagnetic wave propagating through vacuum is described by E y = Eosin(kx–ωt), B z = Bosin(kx– ωt), then
(1) E o k = B o ω (2) E o B o = ωk (3) E o ω = B o k (4) 00EB k ω =
10. The figure gives the magnetic field of an electromagnetic wave at a certain point and at a certain instant. The wave transports energy in negative y-direction. Then the direction of electric field at this point and instant is y z x B →
(1) z - axis (2) – z - axis (3) x - axis (4) – x - axis
Numerical Value Questions
11. Electric field in a plane electromagnetic wave is given by E = 50 sin (500 x – 10 × 1010 t) Vm-1. If the velocity of electromagnetic wave in this medium is (Given c =speed of lig ht in vacuum) c N , then the value of N is____.
12. A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z -direction. At a particular point in space and time, the magnetic field is given
by 8 510T ˆ Bj =× and the corresponding electric field m ˆ isV/Exi (speed of light c = 3×108 ms–1). What is the value of x?
13. The electric field in an electromagnetic wave is given by E = (50 NC–1) sin ω(t – xc-1) the energy contained in a cylinder of volume V is 5.5 × 10–12 J. The value of V is _____ cm3 (Given εo = 8.8 × 10–12 C2N–1m–2)
Momentum and Radiation Pressure
Single Option Correct MCQs
14. A flood light is covered with a filter that transmits red light. The electric field of the emerging beam is represented by a sinusoidal plane wave E x= 36 sin (1.20 × 107 z – 3.6 × 10 15 t ) Vm -1. The average intensity of the beam will be (1) 0.86 Wm-2 (2) 1.72 Wm-2 (3) 3.44 Wm-2 (4) 6.88 Wm-2
15. The electric field intensity produced by the radiations coming from 100 W bulb at a 3 m distance is E. The electric field intensity produced by the radiations coming from 50 w bulb at the same distance is (1)
16. A desk lamp illuminates a desktop with a light of wavelength λ. The amplitude of this wave is E0. If the illumination is normal the number of photons striking the desk per unit area per second is
17. The intensity of solar radiation at the earth's surface is 1 kW/m 2 . What would be the power entering an eye through a pupil of diameter 0.5 cm ?
19.6 mW (2) 12.3 mW (3) 14.3 mW (4) 17.2 mW
8: Electromagnetic Waves
18. The amplitude of magnetic field at a region carried by an electromagnetic wave is 0.1 µT. The intensity of wave is (1) 4 nWm–2 (2) 1.2 Wm–2 (3) 4 Wm–2 (4) 1.2 µWm-2
Numerical Value Questions
19. The intensity of electromagnetic wave at a distance of 1 km from a source of power 12.56 kW is n × 10–1 Wm-2. Then the value of n is____.
20. The electric field intensity produced by the radiation coming from a 100 W bulb at a distance of 3 m is E. The electric field intensity produced by the radiation coming from 60 W at the same distance is 5 x E where the value of x is ____.
21. The electric field in a plane electromagnetic wave is given by
If this wave falls normally on a perfectly reflecting surface having an area of 100 cm2. If the radiation pressure exerted by the EM wave on the surface during a 10 minute exposure is 92 N 10m x , then find the value of x.
Electromagnetic Spectrum
Single Option Correct MCQs
22. Arrange the following electromagnetic radiations per quantum in the order of increasing energy.
A: Bluelight B: Yellowlight
C: X-ray D: Radiowave.
(1) C, A, B, D (2) B, A, D, C
(3) D, B, A, C (4) A, B, D, C
23. Microwave oven acts on the principle of (1) giving rotational energy to water molecules.
(2) giving translational energy to water molecules.
(3) giving vibrational energy to water molecules.
(4) transferring electrons from lower to higher energy levels in water molecule.
24. If microwaves, X–rays, infrared, gamma rays, ultra-violet, radio waves and visible parts of the electromagnetic spectrum are denoted by M, X, I, G, U, R and V, then which of the following is the arrangement in ascending order of wavelength ?
(1) R, M, I, V, U, X and G
(2) M, R, V, X, U, G and I
(3) G, X, U, V, I, M and R
(4) I, M, R, U, V, X and G
25. Photons of an electromagnetic radiation has an energy 11 keV each. To which region of electromagnetic spectrum does it belong?
(1) X-ray region
(2) Ultra violet region
(3) Infrared region
(4) Visible region
26. The frequency of X–rays, γ–rays and ultraviolet rays are respectively a, b and c then
(1) a < b; b > c (2) a > b; b > c
(3) a < b < c (4) a = b = c
27. T.V. waves have a wavelength range of 1 m to 10 m. Their frequency range in megahertz is
(1) 30 to 300 (2) 3 to 30
(3) 300 to 3000 (4) 3 to 3000
28. Which of the following is not an electromagnetic wave?
(1) X–rays (2) b –rays
(3) power waves (4) γ–rays
29. Which physical quantity is same for X–rays and γ–rays in vacuum?
(1) Frequency
(2) Wavelength
(3) Energy of one photon
(4) Speed
30. Which of the following EM radiations has least wavelength?
(1) γ–ray
(2) X–ray
(3) Infrared
(4) Radio waves
31. Which of the following is/are not used as diagnostic tool in medicine?
(1) X–ray
(2) γ–ray
(3) Ultrasonic wave
(4) None of these
32. Different types of electromagnetic waves are defined on the basis of (1) the source from which they are produced.
(2) frequency.
(3) wavelength.
(4) colour.
33. Which of the following can be used in cancer treatment?
(1) X–rays
(2) UV–rays
(3) γ–rays
(4) Both X–rays and γ–rays
34. One cannot get tanned or sunburn through glass window because
(1) glass absorbs UV radiation.
(2) glass reflects UV radiation.
(3) glass is transparent to UV radiation.
(4) None of these
35. Fingerprints on a pieces of paper may be detected by sprinkling fluorescent powder on the paper and then looking at it in
(1) red light
(2) sunlight
(3) infrared light
(4) ultraviolet light
Level-II
Displacement Current
Single Option Correct MCQs
1. The diameter of the plates of a condenser is 10 cm and it is charged by an external source with a steady current of 0.1 A. The maximum magnetic field induced in the gap is
(1) 4 μT (2) 8 μT (3) 40 μT (4) 0.4μT
Numerical Value Questions
2. A parallel plate condenser of capacity 1nF is connected to a battery of emf 2 V through a resistance of 1 MΩ. The displacement current after 10–3 s in µA is ____.
Electromagnetic Waves
Single Option Correct MCQs
3. The electric field of a plane electromagnetic wave is given by ()() 0 s ˆˆ in.EExykzt =+−ω Its magnetic field will be given by
(1) ()() 0 sin ˆˆ E xykzt c −+−ω
(2) ()() 0 sn ˆ i ˆ E xykzt c +−ω
(3) ()() 0 sn ˆ i ˆ E xykzt c −−ω
(4) ()() 0 cs ˆ o ˆ E xykzt c −−ω
4. The electric fields of two plane electromagnetic plane waves in vacuum are given by () 10 ˆ cos EEjtkx =ω− , () 20 ˆ cos EEktky =ω− At t = 0, a particle of charge q is at origin with a velocity .8 ˆ 0 vcj = (c is the speed of light in vacuum), the instantaneous force experienced by the particle is
(1) () 0 ˆ 0.8. ˆ 04 ˆ Eqijk −+
(2) () 0 0 ˆ ˆ ˆ 0.43.8 Eqijk −+
(3) () 0 ˆ ˆ ˆ 0.8 Eqijk −++
(4) () 0 ˆ 0.8. ˆ 02 ˆ Eqijk ++
5. A linearly polarized electromagnetic wave in vacuum is
()() 3.1cos1.85.4N61 ˆ 10CEzti =−× is incident normally on a perfectly reflecting wall at z = a. Choose the correct option.
(1) The wavelength is 5.4 m.
(2) The frequency of electromagnetic wave is 54× 104 Hz.
(3) The transmitted wave will be
()() 3.1cos1.85.410C61 ˆ N zti −×
(4) The reflected wave will be ()() 3.1cos1.85.410C61 ˆ N zti +× .
6. A plane electromagnetic wave is propagating along the direction ˆ ˆ 2 ij + with its polarization along the direction ˆ k The correct form of the magnetic field of the wave would be (here B o is an appropriate constant)
(1)
(2)
(3)
(4)
7. An EM wave from air enters a medium. The electric fields are 101 cos2 ˆ z EExvt c
in air and ( 202 co[2 ˆ s EExkzct = in medium, where the wave number k and frequency v refer to their values in air. The medium is nonmagnetic. If 12 and rr εε refer to relative permittivities of air and medium respectively, which of the following options is correct?
(1) 1 2
Numerical Value Questions
8. The velocity of electromagnetic wave in a medium is 2×10 8 ms -1 . If the relative permeability is 1, the relative permeability of medium is ____. (c = 3 × 10 8 ms-1)
9. In an electromagnetic wave in vacuum the electric and magnetic fields are 40 πV/m and 0.4 × 10–7 T. The Poynting vector in W/m is ___.
10. An electron is constrained to move along the y -axis with a speed of 0.1c (c is the speed of light) in the presence of electromagnetic wave, whose electric field is () 72 30sin1.510510V/m ˆ Ejtx =×−× The maximum magnetic force experienced by the electron will be 4.8 × 10 –x N. Find the value of x ? (Given c = 3×108 ms–1 and electron charge = 1.6 ×10 –19 C).
11. A plane electromagnetic wave, has frequency of 2.0 × 10 10 Hz and its energy density is 1.02 × 10–8 J/m3 in vacuum. The amplitude of the magnetic field of the wave is close to ( 2 9 2 0 1 910 4 Nm C =× πε and speed of light = 3×108 ms–1), 160 × 10–x T. What is the value of x?
Momentum and Radiation Pressure
Single Option Correct MCQs
12. The incident intensity of an a horizontal surface at sea level from sun is about 1 kW/ m 2 . Assuming that 50% of this intensity is reflected and 50% is absorbed, then the radiation pressure on his horizontal surface is (1) 5 × 10-6 Pa (2) 10 × 10-6 Pa (3) 2.5 × 10-6 Pa (4) 1.25 × 10-6 Pa
13. Light with energy flux 36 w/cm2 is incident on a well polished metal square plate of side 2 cm. The force experienced by it is (1) 0.96 µN (2) 0.24 µN (3) 0.12 µN (4) 0.36 µN
14. In an electromagnetic wave the electric and magnetising fields are 100 V m–1 and 0.265 A m–1.The maximum rate of energy flow per unit area in W m–2 is (1) 26.5 (2) 36.5 (3) 46.7 (4) 765
15. The electric field of a plane electromagnetic wave is given by () 0 ˆ co 2 ˆ s ij EEkzt + =+ω At t = 0, a positively charged particle is at the point (),,0,0, xyzk π = If its instantaneous velocity at (t= 0) is 0 ˆ vk the force acting on it due to the wave is
(1) parallel to ˆ ˆ 2 ij + (2) zero
(3) anti parallel to ˆ ˆ 2 ij + (4) parallel to ˆ k
16. An electromagnetic wave of intensity 50 Wm –2 enters in a medium of refractive index n without any loss. The ratio of the magnitudes of electric fields, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by
(1) 11 ,
(2) () , nn (3) 1 , n n
(4) 1 ,
Numerical Value Questions
17. If 2.5 ×10–6 N average force is exerted by a light wave on a non-reflecting surface of 30 cm2 area during 40 minutes of time span, the energy flux of light just before it falls on the surface is ___ W cm -2. (Round off to the nearest integer) (Assume complete absorption and normal incidence conditions are there)
18. Light is incident normally on a completely absorbing surface with an energy flux of 25W cm–2. If the surface has an area of 25 cm 2 , the momentum transferred to the surface in 40 minutes time duration will be A × 10–3 N.s. What is the value of A?
19. 50 W . m -2 energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy (25%) is reflected from the surface and the rest is absorbed. If the force exerted on 1 m2 surface area will be close to (c = 3×108 m.s-1) 20 × 10–x N, then what is the value of x?
Level-III
1. E.M. radiation with energy flux 50 W/cm2 is incident on a totally absorbing surface has an area of 0.05 m2, then the average force due to the radiation pressure on it is n × 10–5 , where the value of n is ____. (Round off te value of n to nearest integer)
2. A radiation is emitted by 1000 W bulb and it generates an electric field and magnetic field at P, placed at a distance of 2 m. The efficiency of the bulb is 1.25%. The value of peak electric field at P is x × 10 –1 V/m. The value of x is ____. (Rounded–off to the nearest integer)[Take εo = 8.85 × 10–12 C2N–1m–2, C = 3 × 108 m/s]
3. The peak electric field produced by the radiation coming from the 80 W bulb at a distance of 10 m is 0 V 10m xc µ π The efficiency of the bulb is 10% and it is a point source. The value of x is ____.
4. A parallel plate capacitor consists of two circular plates of radius 0.05 m. If electric field between the plates is changed as 1010 dEV dtms = ,then the displacement current between the plates is x mA. What is the value of x ? (Round off to the nearest integer)
5. A point source of electromagnetic radiation has average power output of 800 watt, then maximum value of electric field at a distance 3.5 m from the source is 0.626 ×10 x N/C. The value of x is ____.
THEORY-BASED QUESTIONS
Single Option Correct MCQs
1. Displacement current is
(1) due to flow of free electrons.
(2) due to flow of positive ions.
(3) due to flow of both positive and negative free charge carriers.
(4) due to time varying electric field.
2. The displacement current was named as current because
(1) it is similar to conduction current.
(2) it produces a magnetic field.
(3) it is a time varying electrical field.
(4) it is current due to uniformly moving charges.
3. The displacement current is found
(1) between the plates of a condenser when it is being charged.
(2) between the plates of a condenser when it is being discharged.
(3) between the plates of a condenser when AC is applied to the condenser.
(4) in the above 1, 2, 3 cases.
4. The time varying electric and magnetic fields in space
(1) produce EM wave which is propagated with a velocity less than velocity of light.
(2) do not produce EM wave.
(3) produce EM wave which propagates with a velocity greater than the velocity of light.
(4) produce EM waves which propagates with the velocity of light.
5. The time varying magnetic and electric fields
(1) can be created independently.
(2) can exist independently.
(3) are interdependent.
(4) sometimes they are dependent and sometimes they are independent.
6. Time varying electrical field is produced along with time varying magnetic field, by (1) a charge moving with uniform velocity.
(2) a conductor carrying electrical current.
(3) an oscillating neutron.
(4) an oscillating charge.
7. A charged particle oscillates about its mean equilibrium position with a frequency of 10 9 Hz. What is the frequency of the eletromagnetic waves produced by the oscillator ?
(1) 109 Hz (2) 10–9 Hz
(3) 105 Hz (4) 107 Hz
8. Out of the following options which one can be used to produce a propagating electromagnetic wave?
(1) A charge moving at constant velocity.
(2) A stationary charge.
(3) A chargeless particle.
(4) An accelerating charge.
9. Maxwell’s equation m d Edl dt
is a statement of
(1) Ampere’s Law.
(2) Faraday’s law of induction.
(3) Gauss’s law of electricity.
(4) Gauss’s law of magnetism.
10. Maxwell’s equation 0 Bds ∫⋅= is a statement of
(1) Faraday’s law of induction. (2) modified Ampere’s law.
(3) Gauss’s law of electricity.
(4) Gauss’s law of magnetism.
11. The Maxwell’s equation 00 d BdlI dt
is a statement of
(1) Faraday’s law of induction.
(2) modified Ampere’s law.
(3) Gauss’s law of electricity.
(4) Gauss’s law of magnetism.
12. According to Maxwell’s hypothesis, a changing electric field gives rise to (1) an emf.
(2) electric current. (3) magnetic field. (4) pressure gradient.
13. The phase and orientation of the electric field vector linked with electromagnetic wave differ from those of the corresponding magnetic field vector, respectively by (1) zero and zero (2) 2 π and 2 π
(3) zero and 2 π (4) 2 π and zero
14. The velocity of electromagnetic wave is parallel to (1) BE × (2) EB × (3) E (4) B
15. Electromagnetic waves are transverse in nature. It is evident by (1) polarization (2) interference (3) reflection (4) diffraction
16. Which of the following statement is true ?
(1) Velocity of light is constant in all media. (2) Velocity of light in vacuum is maximum. (3) Velocity of light is not same in all reference frames.
(4) Velocity of light in denser medium is greater than that in rarer medium.
17. If uE and uB be the time average of the electric and magnetic field energy densities at a point due to electromagnetic wave, then (1) uE = uB (2) uE = 2uB (3) 2uE = uB (4) none of these
18. The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is (1) c : 1 (2) c2 : 1 (3) 1 : 1 (4) c :1
19. An EM wave radiates outwards from a dipole antenna, with E 0 as the amplitude of its electric field vector. The electric field E 0 which transports significant energy from the source, falls off as
(1) 3 1 r (2) 2 1 r
(3) 1 r (4) remains constant.
20. A cube of edge a has its edges parallel to x, y and z - axes of rectangular coordinate system. A uniform electric field E is parallel to y - axis and a uniform magnetic field is parallel to z - axis. The rate at which energy flows through each face of the cube is
(1) 2 0 2 aEB µ parallel to x - y plane and zero in others.
(2) 2 0 aEB µ parallel to y - z plane and zero in others.
(3) 2 0 2 aEB µ from all faces.
(4) 2 0 2 aEB µ parallel to y - z faces and zero in others.
21. Which of the following rays are not electromagnetic wave ?
(1) X-rays
(2) γ-rays
(3) β-rays
(4) heat rays
22. Which one of the following electromagnetic radiations have the smallest wavelength ?
(1) Ultraviolet waves
(2) X-rays
(3) γ-rays
(4) Microwaves
23. X–rays are known as to be electromagnetic radiation. Therefore the X–ray photon has (1) electric charge but no magnetic moment. (2) magnetic moment but no electric charge. (3) both electric charge and magnetic moment.
(4) neither electric charge nor magnetic moment.
24. The condition under which a microwave oven heats up a food item containing water molecules most efficiently is
(1) the frequency of the microwave must match the resonant frequency of the water molecules.
(2) the frequency of the microwaves has no relation with natural frequency of water molecules.
(3) microwaves are heat waves, so always produce heating.
(4) infra-red waves produce heating in a microwave oven.
25. An electromagnetic radiation has an energy 14.4 eV. To which region of electromagnetic spectrum does the radiation belong to
(1) UV region
(2) Visible region
(3) Infrared region
(4) γ region
26. If the earth had no atmosphere, the average surface temperature would have been (1) the same.
(2) less than the present.
(3) more than the present.
(4) –273°C (considering the absorption of EM waves by ozone layer).
27. Ozone layer absorbs
(1) visible light.
(2) UV radiation.
(3) infrared radiation.
(4) microwave radiation.
28. A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
(1) 20m–25m
(2) 75m–90m
(3) 40m–20m
(4) 40m–25m
29. The type of electromagnetic wave propagation which is effected as a result of reflection in the upper stratosphere of atmosphere is known as (1) air wave propagation.
(2) space wave propagation.
(3) sky wave propagation.
(4) tropospheric propagation.
30. The atmosphere above the height of 80 km is called
(1) Stratosphere.
(2) Troposphere.
(3) Mesosphere.
(4) Ionosphere.
31. The frequency from 3Hz to 30MHz is known as
(1) audio band
(2) medium frequency band
(3) very high frequency band
(4) high frequency band
Assertion and Reason Questions
In each of the following questions, a statement of Assertion (A) is given, followed by a corresponding statement of Reason (R). Mark the correct answer as
(1) if both (A) and (R) are true and (R) is the correct explanation of (A)
(2) if both (A) and (R) are true but (R) is not the correct explanation of (A)
(3) if (A) is true but (R) is false
(4) if both (A) and (R) are false
32. (A) : Induced electric fields are produced by time varying magnetic field.
(R) : According to Faraday’s law, d d d ElB t
33. (A) : The radiation force on an absorbing surface is twice that on a reflecting surface.
(R) : The radiation force on a reflecting surface is the same as that on an absorbing surface.
34. (A) : A light beam or a radio beam of same intensity will have same values of E and B .
(R) : A light beam or a radio beam of same intensity will have different values of E and B .
35. (A) : Electromagnetic waves are transverse in nature.
(R) : The Poyntings vector which gives the power density is given by PEB =× .
36. (A) : In microwave oven the energy is not wasted in heating up the vessel.
(R) : The principle of microwave oven is to generate microwave radiation with magnetron source of appropriate frequency in the work space of the oven where we keep food.
37. (A) : Infrared waves are often called heat waves.
FLASHBACK (P revious JEE Q uestions )
JEE Main
1. In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5 × 1010 and an amplitude of 50 Vm–1. The total average energy density of the electromagnetic field of the wave is:
[Use ε 0 = 8.85 × 10–12 C2 /Nm2] (2024)
(1) 83 1.10610 × Jm
(2) 83 4.42510 × Jm
(3) 83 2.21210 × Jm
(4) 103 2.21210 × Jm
2. A parallel plate capacitor has a capacitance C = 200 pF. It is connected to 230 V ac supply with an angular frequency 300 rad/s. The rms value of conduction current in the circuit and displacement current in the capacitor respectively are : (2024)
(1) 1.38 μA and 1.38 μA
(2) 14. 3 μA and 143 μA
(3) 13. 8 μA and 138 μA
38. (A) : Speed of E.M. wave in a medium depends on electrical permittivity and magnetic permeability of the medium.
(R) : E.M. wave transport energy in the form of oscillating electric and magnetic fields.
39. (A) : Gamma rays are more energetic than X-rays.
(R) : Gamma rays are of nuclear origin but X-rays are produced due to sudden deceleration of high energy electron while falling on a metal of high atomic number.
40. (A) : The small ozone layer on top of the stratosphere is crucial for human survival.
(R) : Infrared waves, involve vibrations of electrons, atoms or molecules of a substance. This increases the internal energy of the substance.
(R) : It absorbs ultraviolet radiations from the sun and prevents it from reaching the earth’s surface and causing damage to life.
(4) 13. 8 μA and 13.8 μA
3. A plane electromagnetic wave propagating in x-direction is described by ()17 y E200Vmsin1.510t0.05x = ×−
The intensity of the wave is : 12212
0 (Use 8.8510C N m ) ε=× (2024)
(1) 35.4 Wm–2
(2) 53.1 Wm–2
(3) 26.6 Wm–2
(4) 106.2 Wm–2
4. An object is placed in a medium of refractive index 3. An electromagnetic wave of intensity 6 × 108 W/m2 falls normally on the object and it is absorbed completely. The radiation pressure on the object would be (speed of light in free space = 3 × 10 8 m/s): (2024)
(1) 36 Nm–2
(2) 18 Nm–2
(3) 6 Nm–2
(4) 2 Nm–2
5. If the total energy transferred to a surface in time t is 6.48 × 105 J, then the magnitude of the total momentum delivered to this surface for complete absorption will be : (2024)
(1) 2.46 × 10−3 kgm/s
(2) 2.16 × 10−3 kgm/s
(3) 1.58 × 10−3 kgm/s
(4) 4.32 × 10−3kgm/s
6. Match List I with List II
(I) Gauss’ law for electricity
(II) Gauss' law for magnetism
(C)
(III) Faraday law (D)
(III) Ampere –Maxwell law
Chose the correct answer from the options given below (2024)
(A) (B) (C) (D)
(1) IV I III II
(2) II III I IV
(3) IV III I II
(4) I II III IV
7. If frequency of electromagnetic wave is 60 MHz and it travels in air along z direction then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other and the wavelength of the wave (in m) is (2024)
(1) 2.5 (2) 10 (3) 5 (4) 2
8. A plane electromagnetic wave of frequency 35 MHz travels in free space along the X-direction. At a particular point (in space and time) 6/ ˆ 9. = EjVm The value of magnetic field at this point is : (2024)
(1) 8 3.210 × kT
(2) 8 3.210 × i T
(3) 9.6j T
(4) 8 9.610 × kT
9. The electric field of an electromagnetic wave in free space is represented as () 0 ˆ EEcosi. =ω− tkz The corresponding magnetic induction vector will be (2024)
(1) () 0 Bj ˆ E =ω− Ccostkz
(2) () 0E Bcosj C ˆ =ω− tkz
(3) () 0 Bj ˆ E =ω+Ccostkz
(4) () 0E Bcosj C ˆ =ω+ tkz
10. Given below are two statements.
Statement I : Electromagnetic waves carry energy as they travel through space and this energy is equally shared by the electric and magnetic fields.
Statement II : When electromagnetic waves strike a surface, a pressure is exerted on the surface.
In the light of the above statements, choose the most appropriate answer from the options given below: (2024)
(1) Statement I is incorrect but Statement II is correct
(2) Both Statement I and Statement II are correct.
(3) Both Statement I and Statement II are incorrect.
(4) Statement I is correct but Statement II is incorrect.
11. Match List I with List II
List I List II
∮
(A) Gauss’s law of magnetostatics (I) 0 1 E dadV ⋅=∫ρ ε
∮
(B) Faraday’s law of electro magnetic induction (II) B0 da ⋅=−
14. The energy of an electromagnetic wave contained in a small volume oscillates with (2023)
(1) zero frequency.
(2) half the frequency of the wave.
(3) double the frequency of the wave.
(4) the frequency of the wave.
∮
(C) Ampere’s law (III) d EBdlda dt ⋅=∫⋅
(D) Gauss’s law of electrostatics (IV) 0 BI dl ⋅=−µ
Chose the correct answer from the options given below (2024)
(A) (B) (C) (D)
(1) I III IV II
(2) III IV I II
(3) IV II III I
(4) II III IV I
12. For the plane electromagnetic wave given by E = E osin(ωt – kx), B = B osin(ωt – kx). The ratio of average electric energy density to average magnetic energy density is (2023)
(1) 1 2 (2) 2 (3) 4 (4) 1
13. The energy density associated with electric field E and magnetic field B of an electromagnetic wave in free space is given by ( ε o–permittivity of free space, µ o –permeability of free space) (2023)
(1) 2 2 0 0 , 22EB EB UU µ == ε
(2) 22 00 , 22EB EB UU== εµ
(3) 22 00 , 22EB EB UUεµ ==
(4) 2 2 0 0 , 22EB EB UU ε == µ
15. The amplitude of magnetic field in an electromagnetic wave propagating along y- axis is 6.0 × 10–7 T. The maximum value of electric field in the electromagnetic wave is (2023)
(1) 2×1015 Vm–1 (2) 180 Vm–1 (3) 15×1014 Vm–1 (4) 6.0 ×10–7 Vm–1
16. The electric field in an electromagnetic wave is given as -1 20sinNC x Etj c
, where ω and c are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of 5 × 10–4 m3 will be____.(Given εo = 8.85 × 10–12 C2/ Nm2) (2023)
(1) 8.85×10–13 J (2) 28.5×10–13 J (3) 17.7×10–13 J (4) 88.5×10–13 J
17. A plane electromagnetic wave of frequency 2 MHz propagates in free space along xdirection. At a particular space and time, m ˆ 6.6 V/ Ej = . What is B at this point? (2023)
(1) 8 2.210 ˆ T i × (2) 8 2.210T ˆ k −× (3) 8 2.210 ˆ T k × (4) 8 2.210T ˆ i −×
18. Which of the following Maxwell’s equation is valid for time varying conditions but not valid for static conditions? (2023)
(1) . EdlB t
(2) 0 . BdlI
(3) .0Edl
(4) BdAQ∫=
19. The source of time varying magnetic field may be.
A) A permanent magnet
B) An electric field changing linearly with time
C) Direct current
D) A decelerating charge particle
E) An antenna fed with a digital signal
Choose the correct answer from the options given below (2023)
(1) A only
(2) D only
(3) C and E only
(4) B and D only
20. In an electromagnetic wave, at an instant and at a particular position, the electric field is along the negative z -axis and magnetic field is along the positive x-axis. Then the direction of propagation of electromagnetic wave is (2023)
(1) positive z-axis
(2) positive y-axis
(3) negative y-axis
(4) at 45° angle from positive y-axis
21. Match List I with List II of Electromagnetic waves with corresponding wavelength range (2023)
List I List II
(a) Microwave (i) 400 nm to 1 nm
(b) Ultraviolet (ii) 1 nm to 10–3 nm
(c) X-Ray (iii) 1 mm to 700 nm
(d) Infra-red (iv) 0.1 m to 1 mm
(a) (b) (c) (d)
(1) iv ii i iii
(2) iv i iii ii
(3) i iv ii iii
(4) iv i ii iii
22. In E and K represent electric field and propagation vectors of the E.M. waves in vacuum, then magnetic field vector is given by (ω is angular frequency) (2023)
(1) () 1 KE × ω (2) () KEω×
(3) KE × (4) () EKω×
23. An electromagnetic wave is transporting energy in the negative z – direction. At a certain point and certain time the direction of electric field of the wave is along positive y direction. What will be direction of the magnetic field of the wave at that point and instant? (2023)
(1) Positive direction of z
(2) Negative direction of y
(3) Positive direction of x
(4) Negative direction of x
24. Match List I with List II
List I List II
(a) Gauss’s Law in Electrostatics (i) EdldB dt φ
(b) Faraday’ Law (ii) 0 BdA ∫⋅=
(c) Gauss’s Law in Magnetism (iii) 000 E c d Bdli dt
(d) Ampere –Maxwell Law (iv) 0 Edsq ⋅= ∫ε
Choose the correct answer from the options given below (2023)
(a) (b) (c) (d)
(1) i ii iii iv
(2) iv i ii iii
(3) ii iii iv i
(4) iii iv i ii
25. In a medium the speed of light wave decreases to 0.2 times to its speed in free space. If the ratio of relative permittivity to the refractive index of the medium is x :1, then the value of x is ____. (Given speed of light in free space = 3 × 108 ms–1 and for the given medium μr=1) (2023)
26. A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 × 10 –2 Am –1 at a point, what will be the approximate magnitude of electric field intensity at that point? (Given : permeability of free space
μo = 4π × 10–7 NA–2, speed of light in vacuum c = 3×108 ms–1) (2022)
(1) 16.96 Vm–1 (2) 2.25×10–2 Vm–1
(3) 8.48 Vm–1 (4) 6.75×106 Vm–1
27. Light wave traveling in air along x-direction is given by E y = 540 sin (π × 104 (x – ct)) Vm–1. Then, the peak value of magnetic field of wave will be (Given c = 3×10 8 ms–1) (2022)
(1) 18×10–7 T (2) 54×10–7 T (3) 54×10–8 T (4) 18×10–8 T
28. The electromagnetic waves travel in a medium at a speed of 2.0 × 10 8 m/s. The relative permeability of the medium is 1.0. The relative permittivity of the medium will be (2022)
(1) 2.25 (2) 4.25
(3) 6.25 (4) 8.25
29. The oscillating magnetic field in a plane electromagnetic wave is given by B y = 5 × 10–6 sin 1000 π (5x – 4 × 108 t)T. The amplitude of electric field will be (2022)
(1) 15×102 Vm–1 (2) 5×10–6 Vm–1
(3) 16×1012 Vm–1 (4) 4×102 Vm–1
30. If electric field intensity of a uniform plane electromagnetic wave is given as ()() 301.6 sin 452.4 sin ˆˆ xy V Ekztakzta m =−−ω+−ω
Then, magnetic intensity H, of this wave in Am–1 will be[Given: Speed of light in vacuum c = 3 × 10 8 ms –1, permeability of vacuum μo = 4π × 10–7 NA–2] (2022)
(1) ()() 0.8sin 0.8sin ˆ co ˆ t yx kztakza+−ω+−
(2) ()()1.010sin66 1. ˆˆ 510 yx kztakzta+×−ω+×−ω
(3) ()() 0.8sin 1.2si ˆˆ n yx kztakzta−−ω−−ω
(4) ()()66 c ˆ o ˆ 1.010sin 1.510sint yz kztakza−×−ω−×−
31. In free space, an electromagnetic wave of 3 GHz frequency strikes over the edge of an object of size 100 λ where λ is the wavelength of the wave in free space. The phenomenon, which happens there will be (2022)
(1) Reflection (2) Refraction
(3) Diffraction (4) Scattering
32. Identify the correct statements from the following descriptions of various properties of electromagnetic waves.
A. In a plane electromagnetic wave electric field and magnetic field must be perpendicular to each other and direction of propagation of wave should be along electric field or magnetic field.
B. The energy in electromagnetic wave is divided equally between electric and magnetic fields.
C. Both electric field and magnetic field are parallel to each other and perpendicular to the direction of propagation of wave.
D. The electric field, magnetic field and direction of propagation of wave must be perpendicular to each other.
E. The ratio of amplitude of magnetic field to the amplitude of electric field is equal to speed of light.
Choose the most appropriate answer from the options given below. (2022)
(1) D only (2) B and D only
(3) B, C and E only (4) A, B and E only
33. Match List – I with list – II
List I List II
(a) UV rays (p) Diagnostic tool in medicine
(b) X- rays (q) Water purification
(c) Microwave (r) Communication, Radar
(d) Infrared wave (s) Improving visibility in foggy days
Choose the correct answer from the options given below. (2022)
8: Electromagnetic Waves
(a) (b) (c) (d)
(1) r q p s
(2) q p r s
(3) q s r p
(4) r p q s
34. As shown in the figure, after passing through the medium 1. The speed of light v 2 in medium 2 will be (Given c = 3 × 108 ms–1) (2022)
(1) () 3 8 -1 60sin10310 ˆ Vm 4 y Extj
()38 2sin10310T ˆ 4 z Bxtk π =×−×
(2)
() 3 8 -1 60sin10310 ˆ Vm 4 y Extj π =×−×
()738 210sin10310T ˆ 4 z Bxtk π
Medium 1
Medium 2 Air
(1) 1.0 ×108 ms–1
(2) 0.5 ×108 ms–1
(3) 1.5 ×108 ms–1
(4) 3.0 ×108 ms–1
35. A beam of light travelling along x –axis is described by the electric field
900sin y x Et c =ω−
. The ratio of electric force to magnetic force on a charge q moving along y – axis with a speed of 3 × 107 ms-1 will be [Given speed of light = 3 × 108 ms–1] (2022)
(1) 1 : 1
(2) 1 : 10
(3) 10 : 1
(4) 1 : 2
36. An E.M. wave propagating in x -direction has a wavelength of 8 mm. The electric field vibrating y –direction has maximum magnitude of 60 V.m-1. Choose the correct equations for electric and magnetic fields if the E.M. wave is propagating in vacuum. (2022)
(3) () 7 3 8 -1 210sin1031m 4 ˆ 0V y Extj π
()38 60sin1030 4 ˆ 1T z Bxtk π =×−×
(4) () 7 4 8 -1 210sin1041m 4 ˆ 0V y Extj
()48 60sin1040 4 ˆ 1T z Bxtk π
37. Nearly 10% of the power of a 110 W light bulb is converted to visible radiation. If the change in average intensities of visible radiation, at a distance of 1 m from the bulb to a distance of 5 m is a × 10–2 W/m2, then the value of a will be (2022)
38. The intensity of the light from a bulb incident on a surface is 0.22 W/m2. The amplitude of the magnetic field in this light–wave is ______ × 10 –9 T (Given : Permittivity of vacuum ε o =8.85 × 10–12 C2 N–1, speed of light in vacuum c = 3×108 ms–1) (2022)
39. The displacement current of 4.425 μA is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 106 Vs -1. The area of each plate of the capacitor is 40 cm 2. If the distance between each plate of the capacitor is x × 10 −3 m, then the value of x is____. (Permittivity of free space, ε o = 8.85 × 10–12 C2 N–1 m–2) (2022)
CHAPTER TEST – JEE MAIN
Section – A
1. A parallel plate condenser consists of two circular plates each of radius 2 cm seperated by a distance of 0.1 mm. A time varying potential difference of 5 ×1013 V/s is applied across the plates of the condenser. The displacement current is
(1) 5.50 A (2) 5.56×102A
(3) 5.56×103A (4) 2.28×104A
2. The relation between electric field E and magnetic field H in an E.M. wave is
(1) E = H (2) 0 0 EH µ = ε
(3) 0 0 EH µ = ε (4) 0 0 EH ε = µ
3. The electric field of a plane electromagnetic wave 3 propagating along the x–direction in vacuum is () 0 ˆ cos EEjtkx =ω− . The magnetic field B as the moment t = 0 is
(1) () 0 00 s ˆ co E Bkxk = µε
(2) () 000 s ˆ co BEkxj =µε
(3) () 000 s ˆ co BEkxk =µε
(4) () 0 00 s ˆ co E Bkxj = µε
4 An electromagnetic radiation point source operates at a 250 W power with a 10% efficiency. The average energy density at a point distant 10 m from the source is
(1) 0.85 Jm–3 (2) 1.7 ×10–11 Jm–3 (3) 3.3 × 10–11 Jm–3 (4) 6.6 × 10–11 Jm–3
5. The sun radiates electromagnetic energy at the rate of 3.9 × 1026 W. Its radius is 6.96 × 108 m. The intensity of sun light at the solar surface will be (in W/m 2)
(1) 1.4 × 104 (2) 2.8 × 105 (3) 64 × 106 (4) 5.6 × 107
6. The maximum electric field of a plane electro-magnetic wave is 88 V/m. The average energy density of electric field is (1) 3.4 × 10-8 Jm–3 (2) 13.7 × 10-8 Jm–3 (3) 6.8 × 10-8 Jm–3 (4) 1.7 × 10-8 Jm–3
7. The rms value of the electric field of the light coming from the sun is 720 N/C. The average total energy density of the electromagnetic wave is (1) 3.3 × 10–3 J/m3 (2) 4.58 × 10–6 J/m3 (3) 6.37 × 10–9 J/m3 (4) 81.35 × 10-12 J/m3
8. The energy associated with electric field is (UE) and with magnetic fields is (UB) for an electromagnetic wave in free space. Then
(1) 2 B E U U = (2) UE > UB
(3) UE < UB (4) UE = UB
9. Select the correct statement from the following.
(1) Electromagnetic waves cannot travel in vacuum.
(2) Electromagnetic waves are longitudinal waves.
(3) Electromagnetic waves are produced by charges moving with uniform velocity.
(4) Electromagnetic waves carry both energy and momentum as they propagate through space.
10. Choose the correct option relating wavelengths of different parts of electromagnetic wave spectrum :
(1) visiblemicrowavesradiowavesXrays λ<λ<λ<λ
(2) radiowavesmicrowavesvisibleXrays λ>λ>λ>λ
(3) Xraysmicrowavesradiowavesvisible λ<λ<λ<λ
(4) visibleXraysradiowavesmicrowaves λ>λ>λ>λ
11. Photon of frequency v, has a momentum associated with it. If c is the velocity of light, the momentum is
(1) c v (2) hvc (3) 2 h c v (4) h c v
12. If a source of power 4 kW produces 10 20 photons/second, the radiation belong to a part of the spectrum called (1) X–rays. (2) Ultraviolet rays. (3) Microwaves. (4) γ–rays.
13. The part of electromagnetic spectrum referred as heat wave is (1) Microwaves. (2) Radio waves. (3) X–rays. (4) Infrared.
14. The following can be arranged in decreasing order of wave number
A) AM radio B) TV and FM radio
C) Microwave D) Short radio wave
(1) A > B > D > C (2) C > D > B > A
(3) A > B > C > D (4) D > C > B > A
15. The law which states that the variation of electric field causes magnetic field is
(1) Faraday's law
(2) Biot-Savart law
(3) Modified Amperes law
(4) Lenz's law
16. Given below are two statements. One is labelled Assertion (A) and the other is labelled Reason (R).
Assertion(A) : Out of the radio waves and microwaves, the radio waves undergo more diffraction.
Reason(R) : Radio waves have greater frequency compared to microwaves.
In light of the above statements, choose the correct option from the options given below.
(1) Both (A) and (R) are true and (R) is the correct explanation of (A).
(2) Both (A) and (R) are true but (R) is not the correct explanation of (A).
(3) (A) is true but (R) is false.
(4) (A) is false but (R) is true.
17. The energy of electromagnetic wave in vacuum is given by the relation (1) 22 2200 EB + εµ
18. A plane polarized monochromatic E.M. wave is travelling in vacuum along z–direction such that at t = t1, it is found that the electric field is zero at a spatial point z 1. The next zero that occurs in its neighborhood is at z2.The frequency of the electromagnetic wave is
(1) 21 1 8 1 310 zz t + × (2) 8 21 310 zz × (3)
Section – B
19. Displacement current through a capacitor of capacitance 2 µF is 4 mA. If the rate of potential difference across that capacitor is N × 103 V/s, then what is the value of N?
20. In a plane electromagnetic wave, the electric field oscillates at a frequency of 2 × 10 10 Hz and amplitude 48 V/m. The amplitude of oscillating magnetic field will be n ×10 –8 Wb/m2, then the value of n is ____.
21. An electromagnetic wave of frequency 5 GHz, is travelling in a medium whose relative electric permitiivity and relative magnetic permeability both are 2. Its velocity in this medium is ____ × 10 7 m/s.
22. A plane electromagnetic wave with frequency of 30 MHz travels in free space. At particular point in space and time, electric field is 6 V/m. If the magnetic field at this point will be x × 10–8 T, then the value x is _____.
23. A light beam is described by 800sin x Et c
=ω−
An electron is allowed to move normal to the propagation of light beam with a speed of 3 × 107 ms–1. If the maximum magnetic force exerted on the electron is 12.8 × 10 –x N, then what is the value x?
ANSWER KEY
JEE Main Level
Theory-Based Questions
Flashback
Chapter Test – JEE Main
