Issuu

Networks Chapter 1 N etwork Elements and Basic Laws

1.3

Chapter 2 Network Theorems

1.21

Chapter 3

Transient Analysis (Ac and Dc) 1.40

Chapter 4 Two-Port Networks

1.73

Chapter 5 Resonance 1.93

Unit 01.indd 1

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5:10:50

Hints/Solutions | 1.3

Chapter 1 Network elemeNts aNd BasiC laws exerCises Practice Problems 1 Directions for questions 1 to 21: Select the correct alternative from the given choices. 1. The current in the 2 W resistor ‘I’ is 1Ω

1Ω

V I

10V –

1Ω

2Ω

1 (A) A 8

3 (B) A 8

+ I – 8V

21 (C) A 8

23 (D) A 8

2. The voltage eo in the figure is

10 Ω

16 V

12 Ω

6Ω

R1 R2

(B) 4 A

3 Ω 2

(D)

8 A 3

(A)

25 A 3

– 2Ω

2 Vx

1Ω

(B)

25 A 6

(D) 3 A

8. In the circuit shown in the figure, R2

 2 (C) 3 +  Ω (D)  3

2 Ω 3

(B)

+ 25 V –

(A)

4 4

A R2

3Ω

o+ eo o–

(A) 48 V (B) 24 V (C) 36 V (D) 28 V 3. The driving point impedance of the infinite ladder network shown in the figure is _____. Given R1 = 3 W and R2 = 2 W R1

(A) 2 A

7. In the circuit shown in the figure, the current I is

2Ω

(A) 132 J (B) 98 J (C) 144 J (D) 168 J 6. In the circuit of the given figure, the source current ‘I’ is

2Ω

2Ω 6Ω

4. The current wave formed in a pure resistor of 5 W is shown in the figure. The power dissipated in the resistor is IA

2Ω 6Ω

RAB =? (A)

6Ω

21 W 4

(B)

5 W 6

(D) 8 W

9. What should be the value of current ‘I’ to have zero current flowing through AB? 1Ω 0

+ – 5V

6 t sec

(A) 20 W (B) 45 W (C) 60 W (D) 90 W 5. Figure shows the waveform of the current passing through an inductor of resistance 1 W and inductance 2 H. The energy absorbed by the inductor in the first four seconds is _____. I

A 1Ω

– + 1 A 1Ω

(A) 2 A (B) 4 A (C) −4 A (D) −2 A 10. The two electrical sub-networks N1 and N2 are connected through three resistors, as shown in the figure. The voltage across the 2 W resistor is 8 V and the 3 W resistor is 6 V. The voltage across the 4 W resistor is +

6A N1

8Ω – 2Ω

4Ω

N2 3Ω

t in sec

(A) 24 V

Unit 01.indd 3 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 5

6Ω –

(B) −24 V

(D) −8 V

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1.4 || Part Networks 3.6 III • Unit 1 • Networks 11. In the circuit shown in the figure, the voltages V1, V2, and V3 are 3Ω

6Ω 1Ω

6Ω

16. The current i4 in the circuit of the figure equal to

6Ω

i0 = 7 A

1Ω V3

8 16 , ,4 3 3

16 8 (C) , ,4 3 3

17.

5∠ 0 ° A

12. In the network shown in the figure, the current in resistor R is

B 2Ω 10 V

3Ω

(A) 2 A

(B) 3 A

3Ω

10 ∠ 60° A

10 3 10 3 ∠ − 90°Amp ∠90° A (B) 2 2 (C) 5∠60° A (D) 5∠ − 60° A 18. A delta-connected network with its Wye equivalent is shown in the given figure. The resistance R1, R2, and R3 (in Ohm) are, respectively, (A)

R C

For the circuit shown in the figure, the instantaneous current i1(t) is

−j 2 Ω

j 2Ω

16 8 (D) 4, , 3 3

(B) −12 A (D) None of these

(A) 12 A (C) 4 A

16 8 8 , , 3 3 3

(B)

i3 = 4 A

i 4 =?

12V

(A)

i2 = 3 A

i1 = 5 A

(D) 9 A

13. In the following circuit, the voltage AB is A

50 V

5Ω

40 V

4Ω

30V

3Ω

a B

5Ω b

R1 30Ω

⇒ R2

15 Ω

R3 c

b 2Ω

(A) 35 V

(B) 28.2 V

(D) 42.6 V

(A) 1.5, 3, and 9 (B) 3, 9, and 1.5 (C) 9, 3, and 1.5 (D) 3, 1.5, and 9 19. The voltage e0 in the figure 2Ω

14. A network contains linear resistors that are connected in series across an ideal voltage source. If all the resistances are halved and the voltage is doubled, then the voltage across each resistor becomes (A) doubled (B) halved (C) not changed (D) None of these 15. Obtain the equivalent capacitance of the network given

+ 10 Ω

(A) 48 V 20.

5Ω

1µF

12 Ω

1µF

(B) 0.8 mF

Unit 01.indd 4 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 6

(D) 28 V

x 2Ω

I +

6Ω

3Ω

50 V y

1µF

(A) 1mF

–

1µF 1µF

8Ω

12 Ω e 0 −

6Ω

(B) 24 V

a 1µF

16 V

(D) 2.6 mF

The current I supplied by the source 50 V is (A) 25 A (B) 13.7 A (C) 9.8 A (D) 3.66 A 21. The voltages VC1, VC2, and VC3 across the capacitors in the circuit in the given figure, under steady state are, respectively,

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Hints/Solutions Chapter 1 • Network Elements and Basic Laws || 1.5 3.7 2F

(A) 80 V, 32 V, 48 V (C) 20 V, 8 V, 12 V

25 k Ω

+ − VC2 40 k Ω

10 kΩ VC1 + −

+ VC3 −

Practice Problems 2

6. The current ‘I’ supplied by the source in the figure is

Directions for questions 1 to 19: Select the correct alternative from the given choices. 1. A network contains linear resistors that are connected in series across an ideal voltage source. If all the resistances are halved and the voltage is doubled, then the voltage across each resistor becomes (A) doubled (B) halved (C) not changed (D) None of these 2. Twelve similar conductors of 1 W resistance form a cubical framework. Then, the resistance between two adjacent corners, two opposite corners of one face, and two opposite corners of the cube are (A)

3 5 7 , , 4 6 12

(B)

7 5 3 , , 4 6 4

(C)

7 3 5 , , 12 4 6

(D)

12 4 6 , , 7 3 5

3. In the network shown in the figure, the voltage at node 2 is

10 A

2Ω

1Ω

4Ω

2 5A 4Ω

(A) 2 V (B) 10 V (C) 6 V (D) 4 V 4. In the network shown in the figure, the voltage AF is 3Ω

10 v

2Ω

20 v

2Ω

5Ω F

1Ω

(A) 2 A

1Ω

(B) 3 A

Unit 01.indd 5 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 7

4Ω 4Ω

4Ω

3Ω

1Ω

3 2 A (D) A 2 3 7. A resistance of 10 W is connected in series with two resistances of 20 W arranged in parallel. What resistance should be shunted across this parallel combination so that the total current taken shall be 2 A with 30 V applied? (A) 5 W (B) 10 W (C) 20 W (D) 25 W 8. The resistance of a strip of conductor is RW. If the strip is elongated such that its length is doubled, the resistance of the strip is given by R (A) 4 R (B) 2 R (C) (D) R 2 9. The current i(t) through a 10 W resistor in series with an inductance is given by i(t) = 3 + 4 sin(100t + 45°) + 4 sin(300t + 60°) A. The RMS value of the current and the power dissipated in the circuit are (A) 2 A

(A)

(B) 3 A

(C)

41A, 410 W, respectively

(B) 35A, 350 W, respectively (C) 5 A, 250 W, respectively (D) 11 A, 1210 W, respectively 10. The nodal method of circuit analysis is based on (A) KVL and Ohm’s law (B) KCL and Ohm’s law (C) KCL and KVL (D) KCL, KVL, and Ohm’s law 11. In the given circuit, the voltage v(t) is 1Ω

2Ω I 3Ω

6Ω

(A) 4 V (B) −4 V (C) 6 V (D) 2 V 5. For the circuit shown in the figure, the current ‘I’ is given by 4Ω

(B) 80 V, 48 V, 32 V (D) 20 V, 12 V, 8 V

1Ω e at

+ v (t ) −

e bt

(A) eat − ebt (B) eat + ebt at bt (C) a e − b e (D) a eat + b ebt 12. The rms value of the voltage defined by v(t) = 5 + 5 sin (314t + p 6) is

+ 3V –

(D) zero

(A) 5 V

(B) 2.5 V

(D) 10 V

4/25/2019 10:19:27 AM 03/05/2017 15:10:57

1.6 || Part Networks 3.8 III • Unit 1 • Networks 13. If R1 = R2 = R4 = R and R3 = 1.1R in the bridge circuit shown in the figure, then the reading in the ideal voltmeter connected between a and b is

16. + V1

10 V

−

V −

If V2 = 1 V in the abovementioned network, the value of V1 will be (A) 2.5 V (B) 4 V (C) 5 V (D) 8 V

(A)

(B) 0.138 V (D) 1 V

(B) 3Z

8Ω

+ −

E=?

25 A

30 Ω

The current I in the circuit is (A) 20 A (B) −20 A (C) 16.2 A (D) −16.12 A 15.

− V1

+ 4V

R ab

1.5

R ac R bc

Previous Years’ Questions 1. If R1 = R2 = R4 = R and R3 = 1.1 R in the bridge circuit shown in figure, then the reading in the ideal voltmeter connected between a and b is [2005]

+ R2

1.5V

(A) 0.6 V

(B) 1.8

(D) 0 V

(A) 0.238 V (B) 0.138 V (C) −0.238 V (D) 1 V 2. A fully charged mobile phone with a 12 V battery is good for a 10−min talk time. Assume that, during the talk time, the battery delivers a constant current of 2 A and its voltage drops linearly from 12 V to 10 V, as shown in the figure. How much energy does the battery deliver during this talk time? [2009] 12V 10V 0

Unit 01.indd 6 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 8

v(t) b

−

2Ω

3Ω

+ −

10 V

The value of Rab, Rac, and Rbc are (A) 30 W, 15 W, 5 W (B) 5 W, 30 W, 15 W (C) 5 W, 30 W, 15 W (D) 15 W, 5 W, 30 W

+ 5V −

(A) −16 V (B) 4 V (C) −6 V (D) 16 V 19. What is the current through resistor 2 W in the circuit given? 0.2 µF

Z 3

0V + 1V

−

I 12 Ω

(D)

18. In the given circuit, the value of the voltage source E is V2

14.

2 Ω V2 = 1 V −

17. If each branch of a Delta circuit has impedance 3Z , then each branch of the equivalent Wye circuit has impedance

(A) 0.238 V (C) −0.238 V

2Ω

− R1

2Ω

10min

4/25/2019 10:19:28 AM 03/05/2017 15:10:59

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5:10:59

Hints/Solutions Chapter 1 • Network Elements and Basic Laws || 1.7 3.9 (A) 220 J (C) 13.2 kJ

7. If VA − VB = 6 V, then VC − VD is

(B) 12 kJ (D) 14.4 kJ

3. In the interconnection of ideal sources shown in the figure, it is known that the 60 V source is absorbing power +−

+ − 10V 2A

1Ω

(A) 0 W (C) 10 W

(B) 5 W (D) 100 W

5. In the following circuit, the current through the inductor is [2012]

∼ −

+ 1∠ 0A

∼ −

(A) 2.8 and 36 (B) 7 and 119 (C) 2.8 and 32 (D) 7 and 80 10. Consider the configuration shown in the figure, which is a portion of a larger electrical network [2014]

1Ω

−1 1 A (C) A 1+ j 1+ j

(D) 0A i5 i2

6. The impedance looking into nodes 1 and 2 in the given circuit is [2012] ib

1kΩ 9k Ω 100Ω

(A) 50 W (C) 5 kW

Unit 01.indd 7 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 9

1∠ 0A 1∠ 0V 1∠ 0V

j1Ω

(B)

(A) k2 (B) k (C) 1/k (D) k 9. Three capacitors C1, C2, and C3, whose values are 10 μF, 5 μF, and 2 μF, respectively, have breakdown voltages of 10 V, 5 V, and 2 V, respectively. For the interconnection shown in the figure, the maximum safe voltage in Volts that can be applied across the combination and the corresponding total charge in μC stored in the effective capacitance across the terminals are, respectively, [2013]

j1Ω

1Ω

2 A 1+ j

10V

(A) −5 V (B) 2 V (C) 3 V (D) 6 V 8. Consider a delta connection of resistors and its equivalent star connection as shown in the following figure. If all elements of the delta connection are scaled by a factor k, k > 0, the elements of the corresponding star equivalent will be scaled by a factor of [2013]

1Ω

12A

1Ω

−

1Ω

20V

Which of the following can be the value of the current source I? [2009] (A) 10 A (B) 13 A (C) 15 A (D) 18 A 4. In the following circuit, the power supplied by the voltage source is [2010]

(A)

VB R

+ − 5V

+ 60V −

2Ω

[2012]

99i b 1 2

(B) 100 W (D) 10.1 kW

i3 i4

For R = 1 W and currents i1 = 2 A, i4 = −1 A, and i5 = −4 A, which one of the following is true?

4/25/2019 10:19:29 AM 03/05/2017 15:11:01

1.8 || Networks 3.10 Part III • Unit 1 • Networks (A) i6 = 5 A (B) i3 = −4 A (C) Data is sufficient to conclude that the supposed currents are impossible. (D) Data is insufficient to identify the currents i2, i3, and i6 11. A Y-network has resistances of 10 W each in two of its arms, while the third arm has a resistance of 11 W. In the equivalent ∆-network, the lowest value (in W) among the three resistances is _________. [2014]

(A) voltage-controlled voltage source (B) voltage-controlled current source (C) current-controlled current source (D) current-controlled voltage source 16. The magnitude of current (in mA) through the resistor R2 in the following figure is __________ [2014] R2

12. In the following figure, the value of the current I (in Amperes) is __________ [2014]

1 kΩ

5Ω

10 mA

10Ω

13. In the circuit shown in the figure, the value of node voltage V2 is [2014]

V2 4Ω −j 3Ω

6Ω

j 6Ω

(B) 2 + j 22 V (D) 2 − j 22 V

R R R

R = 300Ω R

19. In the given circuit, the values of V1 and V2, respectively, are [2015] 4Ω

R1 + V2 −

3Ω

7.5Ω

15. The circuit shown in the figure represents a

Unit 01.indd 8 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 10

Rab

14. For the Y-network shown in the figure, the value of R1 (in W) in the equivalent ∆-network is _____. [2014]

5Ω

(A) 22 + j 2 V (C) 22 − j 2 V

3 kΩ

The value of Re/R is _________ 18. In the network shown in the figure, all resistors are identical with R = 300 Ω. The resistance Rab (in Ω) of the network is ______. [2015]

+ −

4 0° A

2 mA

17. The equivalent resistance in the infinite ladder network shown in the figure is Re. [2014]

10 0° V

R3 4 kΩ

2 kΩ

I + −

AiIi

[2014]

I 5A

4Ω

+ V1 −

(A) 5 V, 25 V (B) 10 V, 30 V (C) 15 V, 35 V (D) 0 V, 20 V 20. A dc voltage of 10 V is applied across an n-type silicon bar having a rectangular cross-section and a length of 1 cm as shown in figure. The donor doping

4/25/2019 10:19:30 AM 03/05/2017 15:11:05

M02_

concentration ND and the mobility of electrons mn and 1016 cm−3 and 1000 cm−2V−1s−1, respectively. The average time (in ms) taken by the electrons to move from one end of the bar to other end is ______. [2015] 10V

n −Si

concentration ND and the mobility of electrons mn and −1 −1 1016 cm−3 and 1000 cm−21Vcm s , respectively. The average time (in ms) taken by the electrons to move from 21. one For end the of circuit shown in end the isfigure, the Thevenin’s the bar to other ______. [2015] equivalent voltage (in volts) across terminals a-b is 10V _______. [2015] 3Ω a

21. For the circuit shown in the figure, theb Thevenin’s equivalent voltage (in volts) across terminals a-b is 22. In the circuit shown, the voltage Vx (in volts) is _______. [2015] 3Ω ________. [2015] 0.5Vx 12V

3 5 25. In the figure shown, the current i(in ampere) is + _________ . [2016] R3 + _

60 V 1A

5Ω 1Ω

+ 0.25V x −

8Ω

− 22. In the circuit shown, the voltage Vx (in volts) is ________. [2015] 0.5V

23. In the given circuit,x each resistor has a value equal to 1W [2016] a

10Ω + Vx 20Ω −

1 W 3 38 (D) W 15 R3 (B)

0.04 Vx

8V ±

b 20Ω

1 1 W 6 5 9 (C) W 20

(A)

6Ω

+ 0.25V x −

8Ω

23. In the given circuit, each resistor has a value equal to 1W [2016] a b

1Ω

± 8V

1Ω

5Ω

1 W 6 9 (C) W 20 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 11 (A)

1 W 3 8 (D) W 15 (B)

8V ±

1 Ω consisting of resistance A in 26. A connection is made series with a parallel combination of resistances B and C. Three resistors of value 10 Ω ,15ΩΩ and 2 Ω are pro1Ω ± 8 Vpermutations of the given vided. Consider all possible resistors into the positions A, B, C, and iidentify the configurations with maximum possible overall resistance and the ones with 1a Ωminimum possible overall resistance. The ratio of maximum to minimum values of the resistances (up to second decimal place) is _________. [2017] 27. Consider the network shown below with R1 = 1 W, R2 = 2 W and R2 = 3 W. The network is connected to a constant voltage source of 11 V.

Chapter 1 • Network Elements and Basic Laws | 3.11 What is the equivalent resistance across the terminals a and b?

1Ω

25. In the figure shown, the current i(in ampere) is i _________ . [2016]

of electrons mn and pectively. The averrons to move from ____. [2015]

R1 R2

R3 R2

+ – 11 V

The magnitude of current (in amperes, accurate to two decimal places) through the source is _______. [2018]03/05/2017 15:11:15

24. In the circuit shown in the figure, the magnitude of the current (in amperes) through R2 is __________. [2016]

re, the Thevenin’s s terminals a-b is [2015]

5:11:05

10Ω

24. In the iscircuit shown in resistance the figure,across the magnitude of thea What the equivalent the terminals current and b? (in amperes) through R2 is __________. [2016]

6Ω

1 cm 1A

1 1 W (B) W 3 6 9 8 (C) W (D) W Hints/Solutions 1.9 Chapter 1 20 • Network Elements and Basic Laws || 3.11 15 (A)

Vx 60 V _ 5 24. In the circuit shown in the figure, the magnitude of the 0.04 V _ current (in amperes) through R2 xis __________. [2016]

n −Si

12V

What is the equivalent resistance across the terminals a and b?

01.indd 9 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 11 a Unit

4/25/2019 10:19:31 AM 03/05/2017 15:11:15

3.12 | Part III • Unit 1 • Networks

hiNts/solutioNs Practice Problems 1 1. Convert current source into its equivalent voltage source and solve by applying super position theorem. Hence, the correct option is (C). 2. Using super position theorem, considering 8 A current source only 10 Ω

+ 80 V –

eIIo o–

12 Ω

(6 ||12) = 20 V (6 ||12) + 10 + 2 Considering 16 V voltage source only

e0′ = 80 ×

2Ω o+

16 V

e0|| = 16 ×

6Ω

12 Ω

= 20 + 8 = 28 V Hence, the correct option is (D).

(R + R ) R eq

4Ω

2Ω 4Ω

4Ω

Observing the circuit, the conducting path consists of 2 W resistance. 8 I= =4A 2 Hence, the correct option is (B). 7. Converting the current source of 1 A with a parallel resistance of 1 W into voltage source 3Ω

(10 + 2)|| 12 =8V 6 + (10 + 2)||12

2Ω

eo o–

Hence, total output voltage = e0 = e0| + e0||

3. Req = R1 +

2Ω

6Ω

10 Ω

Total energy = 108 + 36 = 144 J Hence, the correct option is (C). 6. Redrawing the given circuit

R1 + R2 + Req

or Req . R1 + Req . R2 + Req2 = R12 + R1R2 + Req R1 + R1R2 + Req . R2 Hence, the correct option is (D).

+ 25V

−

2Vx

2Ω

−

1Ω 1V

Applying KVL for the loop in the above circuit 3I + 2(3I) + 2I + I − 1 = 25 V 12I = 24 24 I= =2A 12 Hence, the correct option is (C). 8. (A) Redrawing the circuit

4. I 2rms =

1 (3t ) 2 dt 2 ∫0

2Ω

9 t3    2  3 0

3 = × 8 = 12 2

6Ω

Converting Delta formed by three 6 W resistors into star

Power I2 R = 12 × 5 = 60 W Hence, the correct option is (C). 5. Total energy absorbed = Energy dissipated in resistor + Energy stored in inductor Energy dissipated in resistor =

2Ω

A 2Ω 2Ω

2Ω 2Ω

2Ω

B 2Ω

∫ (3t ) × 1dt + 6 × 1 × 2 2

6Ω

2Ω

= 36 + 72 = 108 J

1 1 Energy stored in inductor = LI 2 = × 2 × 36 = 36 2 2

Unit 01.indd 10 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 12

2Ω

4Ω

2Ω

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Hints/Solutions Chapter 1 • Network Elements and Basic Laws || 1.11 3.13 6Ω

Reducing 6Ω

2Ω A

6Ω

2Ω 2Ω B

Reducing further Req = RAB =

3Ω

10 Ω 3

21 W 4

6Ω

V1 = 3 ×

Applying KCL at node ‘B’

12.

B 2Ω

5A 4A

3Ω

R C

N2 6V –

Taking cutset XX, the total current at the junction is zero, i.e., I2W + I4W + I3W = 0 4 + I4W + 2 = 0 I4W = − 6 A Voltage across 4 W resistance = − 6 × 4 = − 24 V Hence, the correct option is (B). 11. Redraw the given circuit as shown in the figure. given below.

Unit 01.indd 11 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 13

+ 8V – 2Ω

16 16 = V 9 3

8 8 = V 3 3 V3 = 1 × 4 = 4 V Hence, the correct option is (C).

2VA − VA = 5 ⇒ VA = 5, VB = 5,

4Ω

12V

8A 3

V2 = 1 ×

(1)

1Ω

3Ω

VA − 5 VA + 1 − VB + =1 1 1

10.

6Ω

16 A 9

2VB − VA = I + 1 2VB − VB = I + 1 VB = I + 1 = 5 I=5−1=4A Hence, the correct option is (B).

12V

6Ω

Hence, the correct option is (A). 9. Current through AB will be zero if VA = VB Applying KCL at node A,

V B V B −V A −1 + =I 1 1

1Ω

Total resistance seen through supply terminals is 2 W. 12 Total current = =6A 2 Current distribution among branches is indicated in the figure shown below.

Alternate method: From the figure, we can conclude that equivalent resistance will be greater than 4 W (as two 2 W resistors are in series with other section of network) and less than 6 W (the section of the network excluding two 2 W series resistors cannot be more than 2 W). 21 So, the answer is W. 4

2VA − VB = 5

1Ω

On applying KCL at node D Current through R is = 5 + 4 = 9 A Hence, the correct option is (D). 1 1 1 50 × + 40 × + 30 × 5 4 3 13. VAB = 1 1 1 + + 5 4 3 Hence, the correct option is (C). 14. Let source voltage Vs and n no.of resistors are connected in series voltage across each Resistor (R) is VR let all resistors are equal so VR= Vs/n if Vs is doubled then VR doubled

4/25/2019 10:19:33 AM 03/05/2017 15:11:42

1.12 || Part Networks 3.14 III • Unit 1 • Networks 15. The circuit can be reprinted as

i1′ = 5∠0°

with 10∠60°A alone 2µF

1µF

j2 Ω

0.5µF

− j2 Ω

1µF

10∠60 ° A

3Ω

i 1" 1µF

2 µF

i1|| = 10 ∠60°

0.5 µF

10 3 ∠90° 2 Hence, the correct option is (A).

i1 = i1|| − i1| = 10 ∠60° − 5∠0° =

Applying l → ∆

1µF

0.5µF

0.25µF

30 × 5 = 3Ω 5 + 30 + 15

R2 =

15 × 5 = 1.5 Ω 5 + 30 + 5

0.5µF

C2 2µF

18. R1 =

30 × 15 = 9Ω 5 + 30 + 5 Hence, the correct option is (D). R3 =

1× 2 1+ 2 +1 1×1 C3 = 4 1× 2 C2 = 1+ 4 Further simplifying C1 =

19. 2Ω

10 Ω

16 V 12 Ω

80 V

6Ω

+ e0

−

80 = I1(10 + 2 + 6) − I2 6 + 16 64 = I1(18) − I26

1µF

0.9µF

− I2(12 + 6) + I16 + 16 = 0 16 = I16 + I2(18)

i1 = 5 A

18 64 −6 16 228 + 384 672 I2 = = = 324 − 36 288 (18) 2 − 36

i2 = 3 A

I0 = 7 A

7+5 = 12 A

I2 = 2.33 E0 = 12 × 2.33

I3 = 4 A

I 4 = −12 A

Hence, the correct option is (B). 17. With 5∠0° A alone

= 28 V Hence, the correct option is (D). 20.

5Ω

j2 Ω 5∠0°

Unit 01.indd 12 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 14

(2)

64   18 −6   I1  16  =  −6 18   I   2   

Ceq = 1.9 mF. Hence, the correct option is (C). 16.

(1)

i 1’

3Ω

2Ω 6||3

12||8 + − 50 V

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Hints/Solutions Chapter 1 • Network Elements and Basic Laws || 1.13 3.15 10 k Ω

12 × 8 + 5 = 9.80 W 12 + 8 6×3 =2W 6+3 Req = 2 +

100 V

9.8 × 2 = 3.66 W 9.8 + 2

+ VC1 −

+ −

40 = 80 V 40 + 10 C3 3 VC2 = 80 × = 80 × = 48 V C 2 + C3 5

VC1 = 100 ×

50 = 13.7 A 3.66 Hence, the correct option is (B). 21. In steady state, the inductors are short circuited and the capacitors are open circuit. I=

C2 = 16 × 2 = 32 V C 2 + C3 Hence, the correct option is (B).

VC3 =

Practice Problems 2

Current in loop L1 is 4 A.

2. The resistance measured between two adjacent corners, two opposite corners of one face and two opposite cor7 3 5 ners of the cube are , , . 12 4 6 If the cube is formed by R W resistance conductors, 7 3 5 then the resistances will be R , R , R . This is a 12 4 6 standard question in which all the resistors have equal resistance. Solution method is not important. Just remembering the answer is suggested. Hence, the correct option is (C). 3. Two 4 W resistors that are in parallel can be replaced by equivalent resistance of 2 W and the simplified network is shown below after converting the current source with a parallel resistance into equivalent voltage source.

Current in loop L2 is 3 A.

VAF = 8 − 10 + 6 = 4 V Hence, the correct option is (A). 5. Solve the problem by application of the superposition theorem. I1 4Ω

+ −

20V

I1 = 6 ×

10 − 10 =0A 5

1Ω

1 =2A 3

Considering 3 V source I2 4Ω

2Ω

+ 3V

3Ω –

10V

2Ω

5Ω

2Ω

A F + − 8V − 10V + + − 6V F

Unit 01.indd 13 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 15

2Ω

4Ω

1Ω

3Ω

2Ω + 10V −

Current flowing in the circuit is = Voltage at node 2 is = 10 V Hence, the correct option is (B). 4.

2Ω

1Ω

2Ω 1Ω

10V

40 kΩ

1Ω

Current supplied by 3 v Source = I2 = 2 ×

3 =2A 1.5

3 =1A 3+ 3

Current through 2 W resistor is 3 A. Hence, the correct option is (B).

4/25/2019 10:19:35 AM 03/05/2017 15:12:09

1.14 || Part Networks 3.16 III • Unit 1 • Networks R1 = 4R Hence, the correct option is (A).

6. Redraw the given circuit 4Ω

4Ω

2 Power = I rms R = 25 × 10 = 250 W Hence, the correct option is (C).

4Ω

3Ω

1Ω

11. V(t) = L

di dt

d [e at + e bt ] dt = a eat + b ebt Hence, the correct option is (D). =1

4Ω 6Ω

4Ω

3Ω

4Ω

 5  12. RMS value = 5 +  = 6.12 V  2  2

Hence, the correct option is (C).

1Ω

2Ω 6V 1Ω

6 = 2A 2 +1 Hence, the correct option is (A). 7. Combined resistance of parallel combination of two 20 20 W resistors = = 10 W 2 If the parallel resistance required is R W, then 10 + 10 R 30 = 10 + R 2 I=

10 R =5 10 + R ⇒ R = 10 W Hence, the correct option is (B). l 8. R ∝ a R corresponds to l, a R → l, a

a Length becomes 2l, area becomes , and new resist2 2l ance R1 ∝ a 2

i.e., R1 ∝

 4   4  9. R.M.S value = 32 +  + =5A  2   2 

6Ω

4l a

Unit 01.indd 14 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 16

10 10 ×R− × 1.1 R 2R 2.1 R 5 − 5.238095 = −0.238095 Hence, the correct option is (C). 14. By superposition theorem 30 I1′ = 25 × = 15 A 30 + 20 12 I1′′ = 5 × = 1.2 A 30 + 20 I = 15 + 1.2 = 16.2 A Hence, the correct option is (C).

13.

15. Rab =

3 × 1.5 + 9 × 1.5 + 9 × 3 =5W 9

Rac =

3 × 1.5. + 9 × 1.5 + 9 × 3 = 30 W 1.5

3 × 1.5 + 9 × 1.5 + 9 × 3 = 15 W 3 Hence, the correct option is (C).

Rbc = 16.

2Ω

+ I1

I 1′

V i′

2Ω

−

V2 = 1 V 1 1 I1′ = + = 1 A 2 2 V1′ = 1 × 2 + 1 = 3 V

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Hints/Solutions Chapter 1 • Network Elements and Basic Laws || 1.15 3.17 V 1′ 3 = 2 2 3 5 I1 = + 1 = = 2.5 A 2 2 5 V1 = × 2 + 3 2 V1 = 8 V Hence, the correct option is (D). 17.

3Z 2

( )

Z 3 3

Z 3

Hence, the correct option is (A). 18. By KVL − 1 − E − 5 − 10 = 0 E = − 16 V Hence, the correct option is (A). 19. For dc, capacitor will not pass current \ Current through 2 W = 0 Hence, the correct option is (D).

3Z . 3Z 3Z + 3Z + 3Z

1   = 2 100 + × 2 × 10  × 60 2   = 13,200 = 13.2 kJ Hence, the correct option is (C).

Previous Years’ Questions 1. 3.

20 V

+ −

+–

I′

+ 60 V −

Given that 60 V source is absorbing power so I + I1 = 12 ⇒ I < 12 Hence, the correct option is (A). Given that R1 = R2 = R4 = R and R3 = 1.1 R R1 1.1 Vb = × 10 = = 5.23V R1 + R2 2.1 Va =

4. 1Ω

I+2

1Ω

= 2 × shaded area

Unit 01.indd 15 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 17

+ − 10V

1Ω

2. Energy delivered in 10 minutes is E = ∫ VI dt

⇒ Va − Vb = 5 − 5.23 V = − 0.23 V Hence, the correct option is (C).

3A 1Ω

1Ω

R1 × 10 = 5V R1 + R2

= 2 ∫ V dt

I+3

Consider I current is flowing from voltage source write loop equation. 10 = 2(I + 3) + 2(I + 2) ⇒I=0 So, power supplied by voltage source (P) = VI = 0 W. Hence, the correct option is (A). 5. The part of given network can be drawn as IL = 1 ∠ 0 ×

1 1 + j1

4/25/2019 10:19:37 AM 03/05/2017 15:12:30

1.16 || Part Networks 3.18 III • Unit 1 • Networks

1∠ 0 A

1Ω

j1Ω

Alternatively, by using superposition theorem also we can solve this problem, there are four sources, take each source at a time and calculate the current in desired branch and add all these four effects, to get the final current in inductor. Hence, the correct option is (C). 6. The circuit can be drawn as 1

Ra Rc K 2 Ra Rc = K⋅ Ra + Rb + Rc ( Ra + Rb + Rc ) K

RC =

Ra Rb K 2 Ra Rb = K⋅ Ra + Rb + Rc ( Ra + Rb + Rc ) K

\ Elements corresponding to star connection are also scaled by a factor ‘K’. Hence, the correct option is (B). 9. VC1 = Voltage across C1 = V Let ‘V’ be applied voltage

VC2 = Voltage across (C2) =

C3 2 .V = .V C 2 + C3 7

VC3 = Voltage across (C3) =

C2 5 .V = .V C 2 + C3 7

From the breakdown voltages

99I b

100

10K

RB =

VC1 max 10 Volts ⇒ Vmax1 = 10 Volts 2

To find out the impedance looking into nodes 1 and 2, connect voltage source Vx, current drawn by the circuit is Ix, V So, impedance = X IX By applying KCL at node 1. V 99Ib + Ib + Ix = X 100 −V Ib = X Vx is voltage across 10 kW. 10 k V By solving these two equations, X = 50Ω IX Hence, the correct option is (A). 7. VA − VB = 6 V, current through 2 W is 3 A, the same current leaving at VB will enter at VD. So the current flowing through 1 W resistor is 2A + 3A from VD to VC. That is, voltage VC − VD = 1 × −5 = −5 V. Hence, the correct option is (A). 8. After scaling the elements of delta configuration network is given by KRa

KRb

RA =

KRc

Rb Rc K 2 Rb Rc = K⋅ Ra + Rb + Rc ( Ra + Rb + Rc ) K

Unit 01.indd 16 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 18

VC2 max =

2 V = 5 Volts ⇒ Vmax 2 = 7.5 Volts 3

VC3 max =

5 V = 2 Volts ⇒ Vmax 3 = 2.8 Volts 7

Maximum safe voltage = minimum {Vmax1, Vmax2, Vmax3} = 2.8 V Effective capacitance = (5m//2m) + 10m = 11.42 mF Total charge (Q) = C.V = 11.42 × 2.8 mC = 32 mC Hence, the correct option is (C). 10. i5 i2 R

R i3

R i1

Given R = 1 W i1 = 2A, i4 = −1A, i5 = −4A According to KCL i3 + i6 = i1 i4 + i1 = i2 i2 = 1A i3 = i2 + i5

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Hints/Solutions Chapter 1 • Network Elements and Basic Laws || 1.17 3.19 = −3 A \ i6 = i1 − i3 =2+3=5A Hence, the correct option is (A). 11. From the given data

V1 − V2 = 10 V Ideal voltage source existing between two non-reference nodes. Sos apply super node at node V1 and V2. −4 +

V1 V1 − V2 V2 − V1 V2 V2 + + + + =0 − j3 4 4 6 j6

−4 −

V1 V2 V2 + + =0 j3 6 j6

 V V2{1 + j1} =  1 + 4 j 6 j 3   V2{1 + j1} = 2V1 + j24

RR 100 ZA = R1 + R3 + 1 3 = 21 + = 32Ω R2 11

But V1 = 10 + V2 V2{1 + j1} = 20 + 2V2 + j24

RR 100 ZB = R1 + R2 + 1 2 = 20 + = 29.09Ω R3 11

V2{−1 + j1} = (20 + j24)

\ From all the above, lowest value of Z is ZB = 29.09 W Hence, the correct answer is 29.08 to 29.10. 12. 5Ω

5Ω

I 5V

+ −

{20 + j 24} = (2 − j22) volts ( −1 + j1) Hence, the correct option is (D). 14. Y − ∆ conversion From the given data 5×3 R1 = 5 + 3 + 7.5 = 8 + 2 = 10 W Hence, the correct answer is 9 to 11. 15. Dependent current source depends on Ii value \ It is a current controlled current source Hence, the correct option is (C). 16. Applying source transformation to the given network it becomes V2 =

RR ZC = R2 + R3 + 2 3 = 32Ω R1

10Ω

Applying nodal analysis at node V1. V1 − 5 V −1+ 1 = 0 5 15 3V1 − 15 − 15 + V1 = 0 4V1 = 30 V1 = 7.5 volts V 7.5 I= 1 = = 0.5Amp 15 15

AQ1

R + V -

V R 1k

2k 20V +-

↑ IN

− + 8V

13. 3k Ω

10 ∠ 0 V +−

V1 o

4∠0 A

Unit 01.indd 17 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 19

4Ω

−j 3 Ω

I =

V2 6Ω

j6 Ω

28 mA = 2.8 mA 10

Hence, the correct answer is 2.79 to 2.81. 18. The given network is in balanced mode, so redraw equivalent network.

4/25/2019 10:19:39 AM 03/05/2017 15:12:43

1.18 || Part Networks 3.20 III • Unit 1 • Networks a R

Rab

R R

b a

Vd = mE =

mV 1000 × 10 = = 104 cm/sec L 1

Distance L We know drift velocity = = Time τ 1cm L −4 \ t= = = 10 sec V d 10 4 cm sec

\ t = 100 m sec Hence, the correct answer is 95 to 105. 21. 3Ω

b Rab 12V

\ Rab = (2R||2R)||(R||R) = (R||R||R) = R/3 But give R = 300 Ω \ Rab = 100 Ω Hence, the correct answer is from 99.5 to 100.5. 19. Apply nodal analysis to the given network. −5 +

Apply nodal analysis: V th − 12 V − 1 + th = 0 3 6 2(Vth − 12) − 6 + Vth = 0 3Vth − 24 − 6 = 0 Vth = 10 volts Hence, the correct answer is 10. 22.

0.5Vx

But I=

V1 4

V1 V1 + = 5 ⇒ V1 = 5 volts 2 2 Apply KVL in 1st loop V2 − 4 × 5 − V1 = 0 V2 = 20 + 5 = 25 Volts Hence, the correct option is (A). 20.

Vth

6Ω

V1 V1 + + 2I = 0 4 4 V1 + 2I = 5 2

10Ω + Vx 20Ω −

+ 0.25V x −

8Ω

Redraw the given network 5Vx

− + 20Ω

10Ω V /4 x

8Ω

+ Vx − 4

10V

Applying nodal analysis, n − Si 1 cm

ND = 1016 cm−3 and mn = 1000 cm2/V-sec

Unit 01.indd 18 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 20

V x V x + 5V x − 0.25V x + =0 20 10 100 = Vx + 2{5.75 Vx} Vx = 8 V Hence, the correct answer is 7.95 to 8.05. −5 +

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Hints/Solutions Chapter 1 • Network Elements and Basic Laws || 1.19 3.21 23. Apply ∆ ↔ Y conversion to the given network.

4 − 3

1Ω

24.

•

5Ω

3Ω

5Ω

−

−60 + i1 (5Ω) + i 2 (3Ω) + i 2 (5Ω) = 0 5i1 + 8i 2 = 60 Vx = i 2 (5Ω) i1 – i2 = −0.04Vx v i1 = x − 0.04 Vx 5 i1 = 0.16vx i2 = 0.2vx 5(0.16vx) + 8(0.2vx) = 60 Vx [0.8 + 1.6] = 60 Vx = 25 volts 25 i2 = = 5A 5 i R = 5A

1Ω 1Ω 4Ω

4Ω b •

Mesh equation

1Ω

4Ω

0.04 V

60V

4 − 3 1Ω

a•

1Ω

−

4 − 3

•

• 1Ω

(i)

4 −Ω 5

[Answer: 5A] 25. 1A

4 −Ω 5

O 1Ω

 4   8 \ Rab =   //    5   5 5A

4 8 × Rab = 5 5 4 8 + 5 5

Unit 01.indd 19 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH01.indd 21

8 Ω 15

± 8V

1Ω

4A O 1Ω

± 8V

i 1Ω

4A 1A [Answer: 6.656 mA]

Apply nodal analysis at node V1

8×4 5 Rab = × 25 12 =

5Ω

V1 − 0 V1 − 8 V1 − 0 V − 8 + + + =0 1 1 1 1 Choice (D)

4 V1 = 16 ⇒ V1 = 4 volts \ i = −1 Amp

[Answer: −1 Amp]

4/25/2019 10:19:41 AM 03/05/2017 15:13:16

1.20 | Networks 26. From the given data

R1 = 1 W R2 = 2 W

Given A

R3 = 3 W B

I = ?

Req

Rev = A + (B||C)

The given network is symmetrical, so we can redraw the given circuit. V1 = V2 and V3 = V4 I

Let A= 10 Ω, B = 5 Ω and C = 2 Ω Rmax = 10 + (5||2) = 10 + =

5× 2 7

80 Ω 7

Let A = 2 Ω, B = 5 Ω and C = 10 Ω Rmin = 2 + (10||5) 10 × 5 =2+ 15 =

6 + 10 3

1Ω

3Ω

1Ω

3Ω

+ 11 V –

1Ω

Req = {(1 || 1) + {1.5 || 0.5} + 0.5} = 0.5 + 0.5 + 0.375 = I=

16 Ω Rmin = 3

1Ω

11 Ω 8

11 = 8 Amp 11 8

Hence, the correct answer is 7.9 to 8.1.

Rmax 80 3 15 = × = Rmin 7 16 7 = 2.142. Hence, the correct answer is 2.12 to 2.16. I 27. R1

1 R1

R3 R2

Unit 01.indd 20

+ 11 V –

4/25/2019 10:19:45 AM

M02_

Hints/Solutions | 1.21

chaptEr 2 NEtwork thEorEms ExErcisEs Practice Problems 1 Directions for questions 1 to 28: Select the correct alternative from the given choices. 1. The maximum power transferred to the load in the circuit is given as 0.5 W. Get the values of R and RL.

6. Find the Thevenin’s resistance associated with the circuit.

10V + −

3VAB +−

4K Ω

4KΩ

2K Ω B

(A) 1 kW (C) 2 kW

(A) 15 W, 10 W (B) 12.5 W, 12.5 W (C) 10 W, 15 W (D) 10 kW, 10 kW 2. Find the efficiency of the circuit given for RL = 50 W.

(B) 0.45 kW (D) 0.22 kW

Common Data for Questions 7 and 8: Select the correct alternative from the given choices.

5Ω V

10V

(A) 99% (B) 91% (C) 80% (D) 87% 3. Current in the circuit is given by the equation i(t) = 10cos(20p t + 50) and the impedance of the load is given as ZL = 5 + j3. Find the average power delivered to the load. (A) 353.5 W (B) 291.5 W (C) 250 W (D) 176.7 W 4. In the following circuit, the Norton equivalent current (in A) across A - B is j5

A +

− j 50

10 Ω

− 20 Ω B

1kΩ

+ ~ – 50V

V 1k Ω

−

Unit 01.indd 21 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 23

15 Ω

10Ω

10 Ω

30Ω

30 Ω

15 Ω

15Ω

7. Find Thevenin’s equivalent voltage of the circuit. (A) 100 V (B) 120 V (C) 125 V (D) 150 V 8. The resistance across A − B is 10 W. Find the current through the 10 W resistor. (A) 5.1 A (B) 6.45 A (C) 3.35 A (D) 13.9 A 9. When a resistor R is fed from an electrical network, ‘N’ consumes a power of ‘P’ W, as shown in the Figure (a). If an identical network is added as shown in Figure (b) the power consumed by R will be _____. N

Figure (a) N

V ab

Figure (b)

20mA

(A) 25 V (C) 49 V

15 Ω

+ −

(A) 19.45 + j3.24 (B) 6.48 - j1.08 (C) 12.97 - j2.16 (D) 20 + j0 5. Find the Thevenin’s equivalent voltage external to the load RL. 5K

7Ω

50 Ω 200V

20∠0°

8Ω

(B) 50 V (D) 45 V

(A) 2P (C)

P 2

(B) P (D) between P and 4P

4/25/2019 10:19:45 AM 29/04/2017 14:23:24

1.22 || Part Networks 3.24 III • Unit 1 • Networks Statement for Linked Answer Questions 10 and 11: Select the correct alternative from the given choices.

–j1 Ω

1Ω

j4 Ω

10. Find the value of ZL at which maximum power is transferred to ZL (A) (1.24 - j0.676) W (B) (1.24 + j0.676) W (C) 1.31 W (D) 1.24 W 11. The maximum power transferred is (A) 201.6 W (B) 617 W (C) 2016 W (D) 6170 W Directions for questions 12 to 15: Select the correct alternative from the given choices. 12. For the circuit shown in the figure, the Thevenin’s voltage and resistance looking into x - y are _____. 1Ω

X o

+ –

1F Po

1Ω 1A

(B) 1 + S +

(A) 1 1 S

(D)

1 S

S2 + S +1 S 2 + 2S + 1

14. The short-circuit test of a two-port p network is shown in Figure (a). The voltage across the terminals 111 in the network shown in Figure (b) will be

+ –

5Ω

15 Ω + 20 V –

1 1′

2 N1

1A 2′

Figure (a)

Unit 01.indd 22 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 24

+ 5Ω

VX –

10 Ω

– 100 V x + VX

85 Ω 12 (D) 500 A, 3.33 W

(A) 10 A, 3.33 W

(B) 100 A,

Statement for Linked Answer Questions 16 and 17: Select the correct alternative from the given choices. 16. j8Ω

6Ω

j8 Ω

110∠0° ∼

o Y

(A) 4/3 V, 2 W (B) 4 V, 2/3 W (C) 4/3 V, 2/3 W (D) 4 V, 2 W 13. The Thevenin’s equivalent impedance ZTH between the nodes P and Q in the following circuit is _____.

(A) 2 V (B) 5 V (C) 10 V (D) 1 V 15. The Norton’s equivalent circuit at terminals PQ has a current source and a Norton’s resistance of _____.

1Ω

2′

2Ω

1Ω o

10V

6Ω

+ –

Figure (b)

3Ω

100 ∠ 0° ∼

∼ 90∠ 0°

In the circuit shown in figure, under the maximum power transfer condition, the value of RL is _____. (A) 5 W

(B) 20 W

(C)

25 Ω 3

(D) 6 W

17. The power absorbed by RL at maximum power transfer condition is _____. (A) 1,000 W (B) 500 W (C) 625 W (D) 2,000 W Directions for questions 18 to 22: Select the correct alternative from the given choices. 18. The Thevenin’s voltage at the terminals AB of the network shown in the figure is 1Ω + 2V –

(A) 4 V

–

2 VX

(B) 2 V

2Ω

+ 2Ω

2Ω

VX –

(C)

3 V 2

(D)

1 V 2

19. The value of the resistance R, connected across the terminals A and B, which will absorb the maximum power, is

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4:23:26

Hints/Solutions Chapter 2 • Network Theorems || 1.23 3.25

4 kΩ

3 kΩ

(B) 0 A, 20 W (D) 0 A, 12 W

24. For the following circuit, the values of Rth and Vth are

(A) 2 A, 12 W (C) 0.5 A, 20 W

R 4 kΩ

6 kΩ

–2ix

100Ω

a +

0.01Vx

(A) 4 kW (B) 4.11 kW (C) 8 kW (D) 9 kW 20. For the circuit shown in figure, the Thevenin’s voltage and resistance looking into X - Y are 1Ω

i 2i

1Ω

2Ω

100 Ω

800 Ω

4 2 (A) V, 2 W (B) 4 V, W 3 3 4 2 V, W (D) 4 V, 2 W (C) 3 3 21. An AC source of RMS voltage 20 V with internal impedance ZS = (1 + 2j) W feeds a load of impedance ZL = (7 + 4j) W shown in the figure. The reactive power consumed by the load is Z s = (1 +2 j ) Ω

+ −

(A) 0 V, 100 W (C) 5 V, 100 W 4A

Ix N

4Ω 3A

20Ω

(B) 4

(D) 16

10 Ω

5Ω

25Ω

5mH

2 sin 50t

+ ~ –

+ –

~ 2 sin100t

2µF

10 Ω

(A) Thevenin’s Theorem (B) MPTT (Maximum Power Transfer Theorem) (C) Milliman’s Theorem (D) Superposition Theorem 27. Consider the following network

4Ω

2Ω 3V + −

1Ω + 6V −

5Ω

IN, RN

Unit 01.indd 23 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 25

30V

26. In the circuit shown in figure, which one of the following theorem can be more conveniently used to evaluate the responses in the 10 W resistors?

9V + −

Figure (a) Figure (b) The network ‘N’ contains only resistances. Use the data given in Figure (a) and find the current i in Figure (b). (A) 0 A (B) 12 A (C) -6 A (D) 6 A

23. For the following circuit, the value of iN and RN are

15i1

2A 15V

10 Ω

5Ω

(A) 2

Rth

(B) 0 V, 1200 W (D) None of the above

Z L = (7 + 4 j )

(A) 8 VAR (B) 16 VAR (C) 28 VAR (D) 35 VAR 22. The value of R (in Ohms) required for maximum power transfer in the following network is

− 25 V

–

25. Consider the following circuits

20 ∠ 0° V

300 Ω

The current I is (A) 0.23 A (C) 2.25 A

(B) -0.23 A (D) -0.5 A

4/25/2019 10:19:48 AM 29/04/2017 14:23:27

1.24 || Part Networks 3.26 III • Unit 1 • Networks 28. Consider the following circuit 1Ω

The current ix would be (A) ix = 1.6 A (C) ix = - 1.5 A

2Ω

(B) ix = 1.3 A (D) ix = 0.8 A

12V ±

± 2 ix

Practice Problems 2 Directions for questions 1 to 16: Select the correct alternative from the given choices. 1. Find the Thevenin’s equivalent of the circuit given 4I2

10Ω + − 20V

(A) 5W

+ − 0.2 Vx +

di2 = 0.7Vx − 1.5i2 R + .5V dt di (B) L 2 = −0.7Vx + 1.5i2 R − .5V dt di (C) L 2 = 0.7Vx − 1.5i2 R + .5V dt di (D) L 2 = 0.7Vx − 1.5i2 R − .5V dt 3. Find the transfer function of the following network. (A) L

I2 + R V0 −

V 1 (A) 0 = Vi 5

V 1 (B) 0 = Vi 13

V (C) 0 = 0 Vi

V −1 (D) 0 = Vi 13

Unit 01.indd 24 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 26

(D) 2W

2A A 2Ω

2Ω

+ 20 V –

–

I1 R

1Ω

R Vx

(B) 7W

5. The Thevenin’s resistance across the terminals AB of the figure is _____.

i1 V + −

−

5Ω

2A ↑

(A) VTH = 20 V, RTH = 3.3 W (B) VTH = 16 V, RTH = 5 W (C) VTH = 20 V, RTH = 5 W (D) VTH = 4 V, RTH = 10 W 2. Find the state equation for the circuit given. R

2Ω

− + 5Ω

4. A network is shown in the following figure with an unknown load R. Find the value of R so that maximum power is delivered to the load.

(A)

2 Ω 3

(B) 2 W

(C)

7 Ω 3

(D) 5 W

6. In the following circuit, the power consumed by RL is 1Ω

2Ω 10 V +–

(A)

1Ω

25 W 16

(B)

25 W 3

(C)

+ – 5V 2Ω

R L= 2 Ω

25 W 8

(D)

5 W 2

7. In the following circuit, the Thevenin’s impedance between terminals A and B is _____. 10 ∠ 0°

(A) 6 W

∼

3Ω A j3Ω

(B) 3 W

–j3 B 3Ω ∼

(C)

3 j3 − 2 2

(D) 6 + j6

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4:23:30

Hints/Solutions Chapter 2 • Network Theorems || 1.25 3.27 8. In the following circuit, the current through resistance RL is _____. 2Ω + 1V

2Ω

RL =

–

13. The Thevenin’s equivalent voltage VTH appearing between the terminals A and B of the following network is given by

1 Ω 2

A + VTH

3Ω ~

100 ∠ 0°

− j6

− B

2Ω

(A)

2 A 3

(B)

3 A 2

(C)

4 A 3

(D)

1 A 3

9. A source of angular frequency 1 rad/s has source impedance consisting of 1 W resistance in series with 1 H inductance. The load that will obtain the maximum power transfer is (A) 1 W resistance (B) 1 W resistance in parallel with 1 H inductance (C) 1 W resistance in series with 1 F capacitor (D) 1 W resistance in parallel with 1 F capacitor 10. Superposition Theorem is not applicable to networks containing (A) non-linear elements (B) dependent voltage sources (C) dependent current sources (D) transformers 11. The maximum power that can be transferred to the load resistor RL from the voltage source in the figure is 100 Ω + 10 V −

(A) 1 W

(B) 10 W

(D) 0.5 W

12. The Thevenin’s equivalent impedance ZTH between the nodes P and Q in the following circuit is 1H

1Ω ±

1 s

Unit 01.indd 25 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 27

14. 5Ω +

20Ω

(D)

1 s

s2 + s + 1 s 2 + 2s + 1

50 V –

In the following circuit, the adjustable resistor R is set such that the power in the 5 W resistor is 20 W. The value of R is (A) 6 W (B) 25 W (C) 4 W (D) 16 W 15. 4Ω 20 V

6Ω 8Ω 3A

The Norton equivalent of the abovementioned circuit is (A) IN = 8 A, RN = 10 W (B) IN = 0.8 A RN = 10 W (C) IN = 3 A RN = 8 W (D) IN = 8 A RN = 3 W 16. In the following circuit, VAB = 48.3∠30°. The applied voltage V is

V −

(B) 1 + s +

(A) 1

(B) j16 (3+j4) (D) 16 (3 - j4)

P 1F 1Ω

10 Ω

(A) j16 (3 - j4) (C) 16 (3 + j4)

(A) 40 ∠ 90° (C) 50 ∠ 135°

4Ω

5Ω B

− j4Ω

j8.66Ω

(B) 100 ∠ 130° (D) 100 ∠ 135°

4/25/2019 10:19:51 AM 29/04/2017 14:23:32

1.26 || Part Networks 3.28 III • Unit 1 • Networks

Previous Years’ Questions

(A) 0

(B) 5 ∠ 30°

1. The maximum power that can be transferred to the load resistor RL from the voltage source in figure is [2005]

(D) 17∠ 30°

100 Ω

10V

+ −

Zs = (1+2j) Ω

(A) 1 W (B) 10 W (C) 0.25 W (D) 0.5 W 2. For the following circuit, Thevenin’s voltage and Thevenin’s equivalent resistance at terminals a - b is [2005] 1A

5Ω

+ 5Ω −

−

10V

ZL = (7+4j) Ω

(A) 8 VAR

(B) 16 VAR

(D) 32 VAR

7. In the following circuit, what value of RL maximizes the power delivered to RL? [2009] Vx

(A) 5 V and 2 W (B) 7.5 V and 2.5 W (C) 4 V and 2 W (D) 3 V and 2.5 W 3. An independent voltage source in series with an impedance ZS = RS + jXS delivers a maximum average power to a load impedance ZL when [2007] (A) ZL = RS + jXS (B) ZL = RS (C) ZL = jXs (D) ZL = RS − jXs 4. For the following circuit, the Thevenin’s voltage and resistance looking into X - Y are [2007] X + −

+ −

20∠0°V

0.5l1

6. An AC source of RMS voltage 20 V with internal impedance Zs = (1 + 2j) W feeds a load of impedance ZL = (7 + 4j) W in the following figure. The reactive power consumed by the load is [2009]

i 1Ω

2Ω Y

4 2 V, 2W (B) 4 V, Ω 3 3 4 2 (C) V, Ω (D) 4 V, 2W 3 3 5. In the following AC network, the phasor voltage VAB (in Volts) is [2007]

−

4Ω

+ 4Ω

4Ω − Vx + Vi

+ −

100V

(A) 2.4 W

(B)

8 Ω 3

(D) 6 W

8. In the following circuit, the Norton equivalent current in amperes with respect to the terminals P and Q is [2011]

(A)

j30 Ω P 16 ∠ 0°A

25 Ω

− j50Ω Q

15 Ω

A 5Ω

5Ω

−j 3 Ω

j3Ω

5 30°A

(A) 6.4 - j4.8

(B) 6.56 - j7.87

(D) 16 + j0

9. In the following circuit, the value of RL such that the power transferred to RL is maximum. [2011]

Unit 01.indd 26 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 28

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M02_

4:23:33

Hints/Solutions Chapter 2 • Network Theorems || 1.27 3.29 10 Ω

10 Ω

+ –

(A) 5 W (B) 10 W (C) 15 W (D) 20 W 10. Assuming both the voltage sources are in phase, the value of R for which maximum power is transferred from circuit A to circuit B is [2011]

VL1

−

15. In the following circuit, the angular frequency w (in rad/s), at which the Norton equivalent impedance as seen from terminals b–b’ is purely resistive, is ____________. [2014]

10 cos ωt (Volts)

−

Circuit B

5Ω

RL = 10 Ω

10VL1 I2

(A) 100 ∠ 90° (B) 800 ∠ 0° (C) 800 ∠ 90° (D) 100 ∠ 60° 13. For maximum power transfer between two cascaded sections of an electrical network, the relationship between the output impedance Z1 of the first section to the input impedance Z2 of the second section is [2014] (A) Z2 = Z1 (B) Z2 = −Z1 (C) Z2 = Z*1 (D) Z2 = −Z*1 14. Norton’s Theorem states that a complex network connected to a load can be replaced with an equivalent impedance [2014] (A) in series with a current source (B) in parallel with a voltage source (C) in series with a voltage source (D) in parallel with a current source

Circuit A

−

−j1Ω

j6Ω

j40I2

+ ~ 3V –

+ ~ 10V –

(A) 0.8 W (B) 1.4 W (C) 2 W (D) 2.8 W 11. A source Vs(t) = V cos 100p t has internal impedance of (4 + j3) W. If a purely resistive load connected to this source has to extract the maximum power out of source, its value in W should be [2013] (A) 3 (B) 4 (C) 5 (D) 7 12. In the following circuit, if the source voltage Vs = 100 ∠ 53.13°V, then the Thevenin’s equivalent voltage (in Volts) as seen by the load resistance RL is [2013]

j4 Ω

3Ω

1Ω

2Ω

2Ω + −

a 4Ω

2Ω

18. In the circuit shown in the figure, the maximum power (in watt) delivered to the resistor R is __________ . [2016] 3 kΩ

10 kΩ

+ +

b′

17. In the circuit shown, the Norton equivalent resistance (in Ω) across terminals a – b is _____. [2015]

0.5H

j2Ω

4 ∠0 Vnms

−

2 kΩ

+ −

100n

40 kΩ

−

16. In the given circuit, the maximum power (in watts) that can be transferred to the load RL is _______. [2015]

Unit 01.indd 27 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 29

4/25/2019 10:19:53 AM 29/04/2017 14:23:35

1.28 || Part Networks 3.30 III • Unit 1 • Networks 18. I In the circuit shown below, VS is a constant voltage source and IL is a constant current load +

R IL

20. The Consider valuethe of circuit IL that shown maximizes in thethe figure. power absorbed by the constant current load is: [2016] – + Vref Vs (A) 3 i0 (B) 2R 2R (C)

Vs R 10 V + –

−

1Ω 1Ω

The value of IL that maximizes the power absorbed by the constant current load is: [2016] Vref Vs (A) (B) 2R 2R (C)

Vs R

Unit 01.indd 28 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 30

P i0

1Ω 1Ω

constant voltage d

(D) ∞

The Thevenin equivalent resistance (in Ω ) across P-Q is __________. [2017]

(D) ∞

4/25/2019 10:19:55 AM 29/04/2017 14:23:36

M02_

Chapter 2 • Network Theorems | 3.31

hiNts/solutioNs Practice Problems 1 1. P =

0.5 =

3V AB

E2 when R = RL 4 RL 52 4 RL

+ −

2k Ω

R = RL = 12.5 W Hence, the correct option is (B).

i = i1 + i2 2 × i1 = 3VAB + 2VAB

50 × 10 2. VL = = 9.09V 50 + 5

i1 = 2.5VAB i2 = 2VAB, i = 4.5VAB

9.092 P h = out × 100% = 50 = 90.9% 10 2 Pin 55 Hence, the correct option is (B).

RTH =

VAB = 0.22 kΩ 4.5VAB

Hence, the correct option is (D). 7.

1 1 3. P = I m 2 RL = 10 2 × 5 = 250 W 2 2 Hence, the correct option is (C).

15 15

4. When A and B are shorted, j5

20∠0

By current division rule, Isc =

Hence, the correct option is (B).

V ′ 1k Ω + 20V − i

Apply ∆ to l transformation in the network, RC1 = RB1 = RA1 =

15 × 15 =5W 15 + 15 + 15

−

30 × 30 = 10Ω 3 × 30 200V

−

VOC = Vab

50 − 20 = 5mA 5 +1 voltage across 5 kW V = 5 × 5 = 25 V voltage across 1 kW V′ = 5 V ∴ Vab = -20 - 5 + 2 × 25 = 25 V Hence, the correct option is (A).

5Ω

10Ω

10 Ω

5Ω

Unit 01.indd 29 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 31

RC2 = RB2 = RA 2 =

5. The equivalent circuit is

50V

20 × 10 10 + j 5 + 20

1k Ω

= 6.48 − j1.08A

5k Ω

+ 200V −

ISC ISc

10Ω

V AB

10 Ω

VTH =

10 Ω

200 × 25 = 125 V 40

Hence, the correct option is (C).

4/25/2019 10:19:55 AM 29/04/2017 14:23:39

1.30 || Part Networks 3.32 III • Unit 1 • Networks 8.

200V

5Ω

25 Ω

10 Ω

15 × 25 125 75 = = 40 8 8

RTH = 15||25 = = 9.375 W IL =

10 Ω

+−

5Ω

ZL = 1.24 + j0.676 Hence, the correct option is (B). V2 V2 = 11. The maximum power transferred = 4 RL 4 (1.24 ) 100 2 = 2016.13 W = 4 (1.24 ) Hence, the correct option is (C). 12. The Thevenin’s theorem in a network with dependent source is as follows: V = VOPEN VTH = VOPEN and ZTH = TH I SHORT For the calculation of VTH:

VTH 125 = RTH + 10 9.375 + 10

1Ω

= 6.45 A Hence, the correct option is (B). 9. Suppose, network N contains a resistance R and voltage V, then Fig (a) is redrawn as follows:

+ –

i2 i2

i1 2A

1Ω

o (i 2 + 2) 2Ω o

R V

Taking two loops as shown above, Applying KVL for loop 1 2i - i1 - (i1 - i2) = 0 i = i1 - i2 and applying KVL for loop2 (i1 - i2) - 2(i2 + 2) = 0 i1 - 3i2 = 4 from which it is found as i2 = 0 and i1 = 4 Hence, VTH = VOPEN = 2 × 2 = 4 V For the calculation of ZTH: The equivalent circuit is as follows:

V2 V  . Now, redrawing Fig (b) power and P =   R = 4R  2R  dissipated in R = P = V (2I)2 R R

1Ω

2i 2

+ –

o i

1Ω 2A

2Ω

4V 16  2V  P1 =   R = = P 9R 9  3R  Hence, P < P1 < 4P Hence, the correct option is (D). 10. 1Ω

From the network shown above, it can be stated that 2 A current will pass through only XY shorted terminal as the current i through 1 W resistor is zero [as voltage across 1 W resistor = VXY = 0], dependent voltage source has no effect as its voltage 2i = 0. Hence, Ishort = 2 A Vopen 4 V = ZTH = =2W Ι short 2A

–j1 Ω 3Ω j4 Ω

Hence, the correct option is (D). 13.

Z th 1

Impedance seen through load terminals is

(3 + j 4) (1 − j1) = 1.24 - j0.676 (3 + j 4) + (1 − j1)

Unit 01.indd 30 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 32

Po 1 S Q

o o

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M02_

4:23:42

Hints/Solutions Chapter 2 • Network Theorems || 1.31 3.33 1  ZTH between nodes P and Q = ( s + 1)||  + 1 s  1  ( s + 1).  + 1 s  = 1  ( s + 1) +  + 1 s 

17. The voltage across RL =

18.

1Ω

–

I 5 14. Y11 = 1 = = 0.5 V1 V = 0 10 –

I2 −1 = = − 0.1 = Y12 V1 V = 0 10

–

5Ω – 500V 10Ω +

5V –

–

Figure 1

Short circuit current through terminals PQ is = 500 = 100 A [from Q to P] 5 5 × 10 50 From Fig (2), Nortons resistance = = 5 + 10 15 = 3.33 Ω Hence, the correct option is (C). 16. Condition for maximum power transfer is obtained when RL is equal to magnitude of impedance. 6Ω

j8 Ω

6Ω

A + VX –

I1 B

20 ×5=5V 15 + 5

2Ω

−Y V 0.1 × 5 or V1 = 12 2 = = 1 Volt 0.5 Y11 Hence, the correct option is (D).

5Ω

2Ω

15 Ω

Now, in Fig (b) I1 = Y11 V1 + Y12 V2 = 0

20 V

2VX

2Ω

15. VX =

 1   1   6 + j8  +  6 + j8 

V = 100 ∠0° Maximum power transferred to RL is V2 100 2 = = 500 W 4 RL 4 × 5 Hence, the correct option is (B).

 1   s + + 2 s = 1    s + + 2 s Hence, the correct option is (A).

Y21 =

 1   1  110  + 90   6 + j8   6 + j8 

j8Ω

Assign the mesh currents I1 and I2 as shown in figure. VX = 2I1 Applying KVL for mesh 1 2(I1 - I2) + 2I1 - 2 - 4I1 = 0 ⇒ -2I2 - 2 = 0 ⇒ I2 = - 1 Applying KVL for mesh 2 I2 - 1 + 2I2 + 2(I2 - I1) + 4I1 = 0 5I2 + 2I1 = 1 Substituting for I2 -5 + 2I1 = 1 ⇒ 2I1 = 6 VX = 2I1 = 6 V A

+ – 2V + VX = 6 V –

VAB = 6 - 2 = 4 V Hence, the correct option is (A). 19. To find RTH, 3 kΩ

4 kΩ B

( 6 + j 8) ( 6 + j 8) = 3 + j 4 ( 6 + j 8) + ( 6 + j 8)

Maximum power will be transferred when RL = z = 5 Ω Hence, the correct option is (A).

Unit 01.indd 31 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 33

6 kΩ

4 kΩ

RTH = 6k || 3k + 4k || 4k = 2k + 2k = 4 kW Hence, the correct option is (A).

4/25/2019 10:19:59 AM 29/04/2017 14:23:45

1.32 || Part Networks 3.34 III • Unit 1 • Networks 10 i1 - 10 + 10 i1 = 0 20 i1 = 10 ⇒ i1 = 0.5 A Vx = 5 + 30 i1 Vx = 5 + 15 = 20 Volts V RN = Rth = x = 20 W 1 Hence, the correct option is (B). 24. In the given circuit, there is no independent sources. So connect the source.

20. To find VTH 2i − VTH V = i − 2 + TH 1 2 V i = TH 1 3V − 2VTH 2 = TH 2 To find RTH 1Ω I +

i 2i

1Ω

2Ω

−

2ix

0.01Vx

V − 2i V V +i + = I; i = 1 2 1 3 V −V = I 2 V =I 2 V = RTH = 2 Ω I Hence, the correct option is (D). 20 21. I = = 2A 2 8 + 62 Reactive power = 4 × 2 = 8 VAR Hence, the correct option is (A). 22.

100 Ω Vx

100 Ω

300Ω

800Ω

Vx ; 1 Vx = 100 (1 − 2ix) + 300(1 − 2ix − 0.01 Vx) + 800 ix But ix = 1 A Vx = 100 V Vth = 0 V for dependent circuit and Rth =

100Ω

4Ω

5Ω

R th R TH

⇒

20 Ω

20 × 5 + 4= 8 W 20 + 5 Hence, the correct option is (C). 23. The circuit is as follows: RTH =

1–i1

10Ω

5Ω i1

15i1 ±

25 Ω

+ Vx ↑ 1A –

There are no independent sources. So Vth = 0 V and IN = 0 A 15 i1 = 25 i1 - 10 (1 - i1)

Unit 01.indd 32 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 34

Vx 100 = = 100 W ix 1 Hence, the correct option is (A). 25. By reciprocity theorem V = constant I ∴ Rth =

V1 V2 = I1 I 2

5 15 = 4 i1 i1 = 12 A and consider 30 V source 5 30 = 2 −i2 ⇒ i2 = -12 A So i = i1 + i2 = 12 A - 12

4/25/2019 10:20:01 AM 29/04/2017 14:23:48

M02_

4:23:48

Hints/Solutions Chapter 2 • Network Theorems || 1.33 3.35 =0A Hence, the correct option is (A). 26. From the above problem w1 = 50 rad/s and w2 = 100 rad/s ∴ w1 ≠ w2 ⇒ So, Millman’s theorem is not applicable. ∴ In this case, superposition (SPT) theorem is applicable. ∴ In w1 = w2. both Millman’s and SPT are applicable. But Millman’s theorem is suitable. Note: If the n/w does not having any memory elements (L and C) in these case millman’s theorem conveniently used to evaluate the response (w1≠ w2 also). Hence, the correct option is (C). 27. By using Millman’s theorem 9 3 6 V1G1 + V2G2 + V3G3 4 + 2 − 1 −2.25 1 = V = = = -1.2857 1 1 1 G1 + G2 + G3 1.75 + + 4 2 1 1 1 1 1 + + and 1 = R R1 R2 R3

= -0.23 A Hence, the correct option is (B). 28. By using SPT at a time, only one independent source is activated Case 1: Only 12 V voltage source is activated. 12 − 2 ix1 ix1 = 3 ⇒ 5ix1 = 12 ix1 = 2.4 A Case 2: current source is activated and voltage source deactivated. 1Ω

2Ω

ix ±

2ix

V A − 2 ix 2 2ix + 4 = VA - 2ix But VA = -ix 4ix + ix = -4 5ix = - 4 ix2 = −0.8 Α ix + 2 =

= R = 0.571 1

V1 ±

5Ω

∴ ix = ix1 + ix2 I=

= 2.4 - 0.8 = 1.6 A Hence, the correct option is (A).

V1 −1.2857 = 5.571 R1 + 5

From I1 loop, V = (i1 + i2)R + i1R + Vx di From I2 loop, Ri2 + L 2 = 0.2Vx + Vx + i1 R dt From Eq. (1), we get V − Vx − i2 R i1 = 2R Substituting Eq. (3) in Eq. (2), we get di V − Vx − i2 R Ri2 + L 2 = 0.2Vx + + Vx dt 2 Ldi ∴ 2 = 0.7Vx - 1.5i2R + 0.5V dt Hence, the correct option is (A).

Practice Problems 2 1. a-b short circuited, ∴ I2 = 0 ∴ 4I2 = 0 5Ω

10 Ω

+ 20V −

20 = 4A 5 VTH = 20 V, RTH = 10||5 = 3.3 W Hence, the correct option is (A). Isc =

2. i

i1 + −

i = i1 + i2

Unit 01.indd 33 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 35

R V

I4 2

− I2

0.2Vx

R I2

(1) (2)

(3)

+ R V0

4/25/2019 10:20:03 AM 29/04/2017 14:23:51

1.34 || Part Networks 3.36 III • Unit 1 • Networks I1 = I3 + I4 Vi − V1 V1 V1 − V2 = + R R R Vi = 3V1 - V2 I4 = I5 - I2 V1 − V2 V2 V0 − V2 = − R R R V0 = 3V2 - V1

(1)

(2)

V0 − V2 − V0 = R R V2 = 2V0 (3) By solving Eqs (1), (2), and (3) V1 = 5V0 Vi = 15V0 - 2V0 V0 1 = Vi 13 Hence, the correct option is (B). 4. Apply source transformation to the given circuit it becomes I2 =

V2 − V1 V2 − 5 V2 + =0 + 1 2 2 5 2V2 - V1 = 2 On solving Eqs (1) and (2), we get 45 V2 = V 16 Power consumed by RL 2

 45    V2V 16 = 3.95 W = = 2 RL 25 W 8 Hence, the correct option is (C). 7. Removing ZL and drawing the given circuit Nearest answer is

3Ω j 3Ω

–j 3 Ω A

3Ω −

A j 3Ω

∴Rth = 5 + 2 = 7 W Hence, the correct option is (B).

ZAB =

1Ω

From the above figure, RAB = 2 W Hence, the correct option is (B). 6. Solution is obtained using nodal analysis. 2Ω

+ –

1Ω

3 ( − j 3) 3− j3

2Ω 2A

2Ω

↑

2 B

Ω

2Ω

Figure 1

2Ω

–

2Ω

Due to the presence of short circuit between A and B, no current flow through the load resistance. Considering 1 V voltage source

2Ω

V1 − 10 V1 V1 − V2 + + =0 2 1 1

Unit 01.indd 34 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 36

3+ j3

j 27 + 27 − j 27 + 27 54 = = 3Ω 18 32 + 32

Applying KCL at node 1

5V1 − V2 = 5 2 Applying KCL at node 2

3 ( j 3)

3Ω

Hence, the correct option is (B). 8. To find out current through the RL, apply super position theorem. Considering 2A current source

2Ω

10 V

−j 3Ω

Rth

3Ω

2Ω

5Ω

10V

(2)

2Ω

(1)

A + 1V –

2Ω B

RL =

1 2

Ω

2Ω

Figure 2

4/25/2019 10:20:05 AM 29/04/2017 14:23:54

M02_

I1 + I2 = 2.5

The Thevenin’s equivalent of the above circuit is shown in Figure 3.

40 = 16 W 2.5 Hence, the correct option is (D).

R Th = 1Ω + RL =

V Th = 0.5 V –

1 2

15. 4Ω

6Ω +

Current through the RL =

0.5 1 1+ 2

20 V

20 −VTH =3 4 VTH = 8 V 4Ω

VTh2 10 2 = = 0.25 W 4 RTH 4 × 100 Hence, the correct option is (C).

11. Pmax =

IN = 1 s

8Ω

6Ω RTH

RTH = 10 W

P s

VTH

3A −

1 = A 3 Hence, the correct option is (D). 9. Source impedance RS = R + jwL = 1 + j For maximum power transfer, RL = Rs* = 1 - j which indicates that a 1 W resistance in series with 1 F capacitor. Hence, the correct option is (C).

12.

8Ω

⇒

4:23:54

Hints/Solutions Chapter 2 • Network Theorems || 1.35 3.37

VTH 8 = = 0.8 A RTH 10

RN = RTH = 10 W Hence, the correct option is (B).

16. Q

 1 ( s + 1) 1 +   s  ( s + 1) ( s + 1) = ZTH = s 1 s +1+ +1 s s2 + s + 1 + s s =1 Hence, the correct option is (A). 13. VTH = 100∠0° ×

j4 100 ∠0° × j 4 (3 − j 4) = j4 + 3 32 + 4 2

= j16(3 - j4) Hence, the correct option is (A). 14. I12 R = 20 I12 × 5 = 20 I1 = 2 2 × 5 = 10 V VR = 50 - 10 = 40 V 10 I2 = = 0.5 20

Unit 01.indd 35 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 37

+ V

4Ω

5Ω

−

B − j 4Ω

j 8.66 Ω

VAX =

− j4 1 V= V 4 − j4 1+ j

VBX =

j8.66 V 5 + j8.66

 1 j 8.66  VAB = VAB − VBX =  − V  1 + j 5 + j 8.66  VAB =

1 V −0.268 + j 1

V = (- 0.268 + j1) VAB V = (1.035 ∠105°) (48.3 ∠30°) V = 50.0 ∠135° Hence, the correct option is (C).

4/25/2019 10:20:06 AM 29/04/2017 14:23:58

1.36 || Part Networks 3.38 III • Unit 1 • Networks

Previous Years’ Questions

Substitute in Eq. (1) 1  Vth  + 1 + 1 − 2 = 2 2   Vth = 4V To find ISC, short circuit x and y node

1. For maximum power transfer, RL = RS = 100 W Maximum power transform =

2 m

V 4 RL

10 2 = 0.25W 4 × 100 Hence, the correct option is (C). =

1Ω

a 5Ω

0.5I1 –

1Ω

↑

2Ω

ISC = 2A

5Ω

Vth

i=0

2i = 0V + −

Rth =

+ – 10V

Vth 4 = = 2Ω I SC 2

Hence, the correct option is (D). 5. Apply KCL at VA ⇒ 5∠30o =

Vth Vth − 10 + =1 5 5 2 Vth = 15 ⇒ Vth = 7.5 V For Rth make independent sources dead

V V + (5 − j 3) (5 + j 3)

 10  =V    25 + 9  5 × 34 ∠30 o = 17∠30 o 10 Hence, the correct option is (D). ⇒V =

1+2j 5Ω

0.5I1

5Ω

– + –

20∠0 °

7+j 4

Rth = 5 // 5 = 2.5 W Hence, the correct option is (B). 3. Voltage source delivers maximum average power if ZL = Zs* ZL = RS - jXs Hence, the correct option is (D). 4. Vth

1Ω

↑ 2A

20 20 = = 2∠ − 36.87° 1 + 2 j + 7 + j 4 8 + j6

Reactive power consumed by load = I2XL = 22 × 4 = 16 VAR Hence, the correct option is (B). 7.

Vx −+

i 2i + –

4Ω

2Ω

− Vx + y

Apply KCL at X node Vth Vth Vth − 2i + + =2 2 1 1 V and th = i ⇒ Vth = i 1

Unit 01.indd 36 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 38

4Ω Va = 1V

4Ω A

+ 1V −

1 i Nodal analysis at node A ⇒ Req =

(1) (2)

Va Va − VX + = i [∵ Vx = A(i/2) = 2i, Va = 1] 8 4

4/25/2019 10:20:09 AM 29/04/2017 14:24:02

4:24:02

Hints/Solutions | 1.37 Chapter 2 • Network Theorems | 3.39 1 1 1 3i + = ⇒i= 2 8 4 4 ⇒ Req = 4 W Hence, the correct option is (C). 8.

j 30 Ω

16 ∠ 0° A

25 Ω

↑

dP ( R + 2) 2 70 − (70 R + 42)2( R + 2) = =0 dR ( R + 2) 4

–j 50 Ω

70(R + 2)2 = 2(70R + 42)(R + 2) 5(R + 2) = 2(5R + 3)

5R + 10 = 10R + 6

By shorting PQ, we can find out Norton current. By converting current source to voltage source, and shorting PQ, the circuit is P

25 Ω 400 ∠ 0° V

Hence, the correct option is (A). 11. For the maximum power to be transferred to the load, load resistance must be equal to the absolute value of source internal impedance: RL = 4 2 + 32 = 5Ω

The Norton current =

RL =

10 Ω 10 Ω

12. j4 Ω

3Ω

j6 Ω

5Ω

V2 Vs ~

10V2 + –

Vth I2

From the above circuit, I2 = 0 ⇒ Vth = 10V2 V2 = j4 × I1

VS [∵ I2 = 0] 3 + j4

I1 =

Vth = 10 × j4 ×

So the Thevenin’s resistance across AB, = 15 W = RL for max power. Hence, the correct option is (C). 10. The power transferred from circuit A to circuit B is VI. 2Ω

Rs2 + ( X L + X S ) 2

But given XL = 0 ⇒ RL = ZS Hence, the correct option is (C).

400 ∠0°V 40 + 30 j

= 6.4 - j 4.8A Hence, the correct option is (A). 9. Maximum power will be transferred to RL when RL = output resistance of the circuit. The circuit resistance is Thevenin’s equivalent resistance. By shorting voltage sources and opening current sources, the equivalent circuit is

R = 0.8 W

∼ 15 Ω

10 Ω

70 R + 42 ( R + 2) 2

P = VI =

15 Ω

j 30 Ω

7  10 R + 6  +3=   R + 2  R+2

V = R.I + 3 = R

100 ∠53.13° = 800 ∠90° 3 + j4

Hence, the correct option is (C). 13. Z th

RΩ

+ V th −

+ 10 V ∼

– 1jΩ

∼ 3V

Z2 = Z1* ⇒ for maximum power transfer

–

10 − 3 7 I= = 2+ R R+2

Unit 01.indd 37

Pmax =

Vth2 V2 = th 4 Rth 4 R1

Hence, the correct option is (C).

4/25/2019 10:20:10 AM

1.38 || Part Networks 3.40 III • Unit 1 • Networks D 14. Thevenins ← → Norton’s

R th

↑ I N

V th −

w r2 = 4 ⇒ wr = 2 rad/s Hence, the correct answer is 1.9 to 2.1. 16. Find the Thevenin’s equivalent circuit at the terminal of RL Zth

Hence, the correct option is (D). 15.

1Ω

1F b

10 cos ωt volts

Vth

−

+ 0.5H

~ -

Find Rth or RN (i) Deactivate all independent sources V → short circuit I → open circuit

0.5s

Z(s)

1 1 × 0.5s  Z ( s) = +   s 1 + 0.5s 

RL = |Zth| = 2 Ω V th rms =

4 × j2 = 2 + 2i = 2 2 ∠ 45°V 2 + j2

I rms =

V th 2 2 ∠45° 2 2 ∠45° = = Z th + R th 1 + j1 + 1.414 2.414 + j1

2 2∠45° = 1.082 ∠ 22.5° Amp 2.613∠22.5 1 Pmax = |I|2.RL = |Irms|2 .RL 2 = 1.655 watts Hence, the correct answer is 1.6 to 1.7. 17. The given network consists only dependent source, connect one test source and find the equivalent resistance across the test source. a

2Ω

1 0.5s = + s 1 + 0.5s

4I + −

2Ω

Z ( jw ) = Z ( jw ) =

Req = RN =

1 jw + jw 2 + jw

Applying Nodal analys is

2 − w 2 + jw

( −w + j 2w ) 2

At resonance imag, {Z(s)} is equal to zero.

(2 − w + jw ) (w + j 2w ) ∴Z ( jw ) = ( −w + j 2w ) ( −w − j 2w ) 2

∴ jw3 − j2w3 + j4w = 0 w2 - 4 = 0

Unit 01.indd 38 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 40

+ V − t

1 j 0.5w + jw 1 + j 0.5w

4Ω

Substituting s = jw Z(jw) =

2 ∠45° Ω

I rms =

1/s

1Ω

Zth = (2 || j2) = (1 + j1) Ω =

Req

Vt It

V t − 4I V t + + I = It 2 2

Vt – 4I + Vt + 2I = 2It ⇒ 2Vt – 2I = 2It V V but I = t ⇒ 2Vt - t = 2It 4 2 3Vt = 4It Vt 4 = It 3 RN = 1.33 Ω Hence, the correct answer is 1.3 to 1.35.

4/25/2019 10:20:12 AM 29/04/2017 14:24:09

Ω

VOC = 200 ×

VS −

40 50

= 160V ISC = 20mA 160 R th = = 8 kΩ 20

160 V

+ −

200 V

±p =  160  ×40 RkΩ

10 kΩ 2

8+ R   

dp =0 dR  ( 8 +RR )2=−V2OC R (8 + R )  th  =0 ISC 2   (8 + R ) 40 VOC = 200 × 50 = 160V ISC = 20mA 160 R th = = 8 kΩ 20

R = 8Ω dPL 2 = 0 (160 ) 160 ×160 = 800 = 0.8W p max = dI L = 4×8 4×8 1000 VS - 2ILR = 0 Hence, the correct answer is 0.8. V IL = s 19. 2R R Hence, the correct option is (B). + 20. Thevenin equivalent resistance+of VS IL VL (i) Deactive − − the independent sources (ii) consider one test source. Load voltage The equivalent circuit becomes VL = VS - ILR Load power = V–L . I+L PL = (VS - ILR) IL 3 i0 To get IL that maximize the power observed by the load

8kΩ 160 V

+ −

IL VL

Load voltage VL = VS - ILR Load power = VL . IL PChapter = (VS - 2 ILR) L L Hints/Solutions • INetwork Theorems || 1.39 3.41 To get IL that maximize the power observed by the load

8kΩ

18.

−

dPL 1Ω =0 dI L 1Ω

 160  p=  ×R 8+ R 

i0 1Ω

VS - 2ILR = 0

+ Vt –

V dp Rth IL = s =0 2R dR Applying nodal analysis Hence, the correct option is (B).  ( 8 + R ) 2 − 2R ( 8 + R )  (i) At node Vx   = 0 2 Chapter  Theorems | 3.41 (8 2+ R•) Network Vx + 3io = Vt (i) R = 8Ω

(160 ) = 160 ×160 = 800 = 0.8W 2

p max = R

4×8

1000

io =

R + VS −

+ −

Vx + 3Vt = Vt IL VL

Load voltage VL = VS - ILR Load power = VL . IL PL = (VS - ILR) IL To get IL that maximize the power observed by the load

4:24:09

Vt 1

sub io in equation (i)

M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 41

Vx + Vt = It (ii)

Hence, the correct answer is 0.8. 19.

Vx Vx − Vt Vt − Vx Vt + + + − It = 0 1 1 1 1

Vx = – 2Vt (iii) – 2Vt +Vt = It – Vt = It

29/04/2017 14:24:11

Vt = Rth = −1Ω It Hence, the correct answer is -1.01 to -0.99.

dPL =0 dI L VS - 2ILR = 0 Vs 2R Hence, the correct option is (B). IL =

Unit 01.indd 39 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH02.indd 41

4/25/2019 10:20:15 AM 29/04/2017 14:24:11

1.40 | Networks

chaptEr 3 transiEnt analysis (ac and dc) ExErcisEs Practice Problems 1 Directions for questions 1 to 32: Select the correct alternative from the given choices. 1. In the following circuit, the switch S is closed at t = 0. di The rate of change of current (0+ ) is given by dt

I(s) 0.002s

5. A2 mH inductor with some initial current can be represented as shown in the figure, where S is the Laplace transform variable. The value of initial current is

RS i(t )

RS LS L (D) ∞

(A) 0 (R + RS ) I S (C) L

(B)

− 1 mV +

(A) 0.5 A (B) 2.0 A (C) 1.0 A (D) 0 A 6. A square pulse of 3 V amplitude is applied to C-R circuit shown in the figure. The capacitor is initially uncharged. The output voltage V0 at time t = 2 s is 0.1 µF Vi

2. In the given figure, A1, A2, and A3 are ideal ammeters. If A2 and A3 read 3 A and 4 A, respectively, then A1 should read A2

~ Sinusoidal Voltage source

(A) 1 A (C) 7 A

(B) 5 A (D) None of these

10 V

−

1 µF

Z(s) i

4. An input voltage V(t) = 10 2 cos(t + 10°) + 10 5 cos(2t + 10°) V is applied to a series combination of resistance R = 1 W and an inductance L = 1 H. The resulting steady state current i(t) in ampere is (A) 10 cos(t + 55°) + 10 cos(2t + 10° + tan-12) 3 cos(2t + 55°) 2 (C) 10 cos(t - 35°) + 10 cos(2t + 10° - tan-1 2) (B) 10 cos(t + 55° ) + 10

Unit 01.indd 40 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 42

1 kΩ

The current in the circuit when the switch is closed at t=0 (A) 10 e-100t (B) 0.01 e-1,000t -1,000t (C) 0.1 e (D) 10 e-0.1t

(D) 10 cos(t - 35°) + 10

−

2 sec

1 kΩ

(A) 3 V (B) -3 V (C) 0 (D) -4 V 7. The driving point impedance of the following network 0.2s is given by Z(s) = 2 . The component values 0.1s + 2 s + are

t=0 +

3 cos(2t - 35°) 2

(A) L = 5 H, R = 0.5 W, C = 0.1 F (B) L = 0.1 H, R = 0.5 W, C = 5 F (C) L = 5 H, R = 2 W, C = 0.1 F (D) L = 0.1 H, R = 2 W, C = 5 F 8. 1 2 + 50 V −

40 Ω

+ 10 V

−

i(t )

20 mH

The switch has been in position 1 for a long time, it is moved to position 2 at t = 0. The expression for i(t) for t > 0 is (A) 0.25 e-2,000t (B) 0.25 + e-2,000t (C) 0.5 e-2,000t (D) 0.5 + e-2,000t

4/25/2019 10:20:18 AM 29/04/2017 14:24:34

M02_

4:24:34

Hints/Solutions Chapter 3 • Transient Analysis (AC and DC) || 1.41 3.43 9.

14. In the following circuit: iL(0−) = 0, Vc (0−) = 0.

t = 0°

+ VR −

40 µf

−

Switch S is closed at t = 0 i(0+) = 20 mA and Vab = 0 for t≥0

400 Ω

At t = 0-, just before the switch is closed, VC = 100 V. The current i(t) for t > 0 is (A) 100 e-62.5t (B) 50 e-160t -62.5t (C) 0.25 e (D) 50 e-62.5t 10. A series RL circuit with R = 5 W and L = 2 mH has an applied voltage V = 150 sin5,000t, the resulting current i is (A) 13.4 sin5,000t (B) 10 cos(5,000t - 36.4°) (C) 13.4 cos(5,000t - 63.4°) (D) 13.4 sin(5,000t - 63.4°) 11. Driving point impedance of the network shown in the figure is 1H

I(t ) + 5V

b C = 10 µF

The value of R is _____. 1 kΩ (B) 250 W (A) 4 (C) 350 W (D) 100 W 15. For the circuit shown in the following figure, if the switch is closed at t = 0, then i(t) for t ≥ 0 will be

+ –

8Ω

t=0

32 V +–

s2 +1 (B) s ( s 2 + 2)

s 4 + 3s 3 + 2s 2 + 1 s2 +1 (D) s +1 s 3 + 2s 12. Initially, the circuit shown in the given figure was relaxed. If the switch is closed at t = 0, the values of d i ( 0 + ) d 2 i ( 0+ ) , i(0+), will, respectively, be _____. dt dt 2 (C)

10Ω

100 Ω

–

16 V

s 4 + 3s 2 + 1 (A) s 3 + 2s

Vab + –

16 Ω

(A) 4 + 2e−4t (B) 4 - 2e−4t 4t (C) 4 + 2e (D) 4 - 24t 16. The network shown in the figure consists of only two elements. The response for unit step excitation is i(t) = e−5t, then the elements are v(t )

i( t ) N

+ 20

RΩ L

5µF

–

(A) 0, 10, and 100 (B) 0, 10, and 50 (C) 0, 10, and -100 (D) 0, 10, and -50 13. Transient response of the following circuit _____. i(t )

R U(t )

–

(A) rises exponentially (B) decays exponentially (C) is oscillatory and the oscillations dies down with time (D) will have sustained oscillations

Unit 01.indd 41 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 43

(A) R = 1 W, L = 5 H in series 1 (B) R = 1 W, C = F in series 5 (C) R = 1 W, L = 5 H in parallel 1 (D) R = 1 W, C = F in parallel 5 17. The capacitor in the circuit is initially charged to 15 V with S1 and S2 open. S1 is closed at t = 0, while S2 is closed at t = 4 s. The waveform of the capacitor current is 2Ω

+ 15 V

–

1Ω S2

4/25/2019 10:20:20 AM 29/04/2017 14:24:36

1.42 || Part Networks 3.44 III • Unit 1 • Networks (A)

1mH

I I 1Ω 1

V t

(B)

(C)

I 5A

(A)

cos( 200t + 45)

→t

1 2

b +

−

(D)

(B)

2Ω

a 4

5mH i (t)

(C) 0 (D) ∞ 21. For the network shown, the switch is at position ‘a’ initially. At steady state, the switch is thrown to position ‘b’. Now, i(0 − ) = 2 A and Vc (0 − ) = 2 V are the initial conditions. Find the circuit current.

4 → t

2µF

e−2t cos 200 t

+ 1F

−

V1 = 100∠45° ∼

−3 t 2

−3 t

18. The phase angle of the current I with respect to the voltage V2 in the circuit shown in the figure is

+j 5 Ω

↑

5Ω

0.5H

0.5F

+ V1 −

V2 = 100∠–45° ∼

V2 ↓

(A) 0° (B) -45° (C) +45° (D) +90° 19. In the circuit of figure, the switch S has been opened for a long time. It is closed at t = 0. The values of VL(0+) and IL(0+) are _____. 20 Ω

200V – t=0 IL

(A)

s3 + s2 + 1 (s + 1) s

(B)

(C)

s 4 + s 3 + 2s 2 + 1 (s + 1) s

(D)

20 Ω

(A) 200 V, -5 A (B) 0 V, 5 A (C) 100 V, 5 A (D) 100 V, -5 A 20. Obtain the value of current i(t) in the given circuit at steady state.

4Ω

3µF

A 10

Unit 01.indd 42 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 44

s4 + s2 + s 1 + 2s 2 s4 + s2 +1

(1 + 2s ) 2s 2

23. For the given circuit switch S is at position ‘A’ when t < 0. At t = 0, the switch is thrown to position B. What will the value of current ‘i’ in the circuit at the instant t = 4 s?

2mH

(B) -3e

(A) e-3t

(A) 0.1 mA (C) 10 mA

6Ω

3Ω + 5i −

(B) 0.01 mA (D) 200 mA

4/25/2019 10:20:23 AM 29/04/2017 14:24:39

M02_

4:24:39

Hints/Solutions Chapter 3 • Transient Analysis (AC and DC) || 1.43 3.45 The switch is closed at time t = 0. The current i(t) at a time t after the switch is closed is

24. In the given network, the switch K is closed for a long time and circuit is in steady state. Now, at t = 0, the switch is opened. Find Vc(0+) and i(0+) in the circuit.

10Ω

+K−

10V

10Ω Vc

(A) 1 A, 0 V (B) 0 A, 5 V (C) 0, a (D) 0, 0 25. In the following circuit, the 25 V source has been applied for a long time and the switch is opened at t =1 ms. t =1ms

5kΩ

25u(–t) ±

0.5µf

–

100V

2mH

1kΩ

t=0

i(t )

25mF

50µF

(A) i(t) = 15 exp (-2 × 103 t) A (B) i(t) = 5 exp (-2 × 103 t) A (C) i(t) = 10 exp (-2 × 103 t) A (D) i(t) = -5 exp (-2 × 103 t) A 29. In the following circuit, the switch is closed at t = 0. What is the initial value of the current through the capacitor? 2Ω

25kΩ

At t = 5 ms, the value of Vc is (A) 1.23 V (B) 20.16 V (C) 1.69 V (D) -1.23 V 26. In the following circuit, capacitor is initially uncharged. dV c d 2V c At t = 0+, the value of and dt dt 2

t=0

1Ω

12V

1Ω C

(A) 0.8 A

(B) 2.4 A

(D) 3.2 A

30. t=0 S

10Ω

2Ω

2V ± –t

e .u(t)V ±

1 F 20

20Ω

+ –

(A) 0 V/s, 8 V/s2 (B) -2 V/s, 8 V/s2 (C) 2 V/s, -8 V/s (D) None of these 27. In the following circuit, the current ix is 4Ω

± 10∠30°V

4Ω +

jx –j 2Ω

j 3Ω

(A) 3.94 ∠46.28° A (B) 4.62 ∠97.38° A (C) 7.42 ∠92.49° A (D) 6.78 ∠49.27° A 28. In the following circuit, the initial charge on the capacitor is 2.5 mC, with the voltage polarity as indicated.

Unit 01.indd 43 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 45

The time constant of the circuit after the switch is opened would be (A) 2 s (B) 0.5 s (C) 1 s (D) None of these 31. The power factor seen by the voltage source is

0.5ix +–

1 F 4

5cos2tV +−

(A) 0.8 (lagging) (C) 36.9 (lagging) 32. An input voltage

1Ω

−

1Ω 3/4V1

1/3F

(B) 0.8 (leading) (D) -36.9 (leading)

V(t) = 10 2 cos (t + 10°) + 10 5 . cos ( 2t + 10°)V is

4/25/2019 10:20:25 AM 29/04/2017 14:24:41

1.44 || Part Networks 3.46 III • Unit 1 • Networks applied to a series combination of resistance R = 1 W and an inductance L = 1 H. The resulting steady state current i(t) is (A) 10cos (t + 55°) + 10 cos(2t + 10° + tan−12) A (B) 10 cos (t + 55°) + 10

Directions for questions 1 to 24: Select the correct alternative from the given choices. 1. The condition on R, L, and C such that the step response y(t) in the figure has no oscillations is

(A) R ≥

(D) 10 cos(t − 35°) + 10

6. In the following AC network, the phasor voltage VAB (in volt) is A 5Ω

5Ω 5∠30°A

3 cos( 2t − 35°) A 2

3 .cos ( 2t + 55°) A 2

Practice Problems 2

+ u(t ) −

j3Ω

− j 3Ω

y(t ) C

1 L 2 C

(B) R ≥

L C

(D) R =

(A) 0 (B) 5 ∠ 30° (C) 12.5 ∠ 30° (D) 17 ∠ 30° 7. In the circuit shown, VC is 0 volt at t = 0 V at t = 0 s. For t > 0, the capacitor current iC(t), where ‘t’ is in seconds, is given by

L C 1 LC

2. A series circuit consists of two elements has the following current and applied voltage i = 4 cos(2,000t + 11.32°) A u = 200sin(2,000t + 50°) V. The circuit elements are (A) resistance and capacitance (B) capacitance and inductance (C) inductance and resistance (D) both resistance 3. Transient current of an RLC circuit is oscillatory when (B) R = 0 (A) R = 2 L C (C) R > 2 L C

(D) R < 2 L C

4. The transient response occurs (A) only in resistive circuits (B) only in inductive circuits (C) only in capacitive circuits (D) both in (B) and (C) 5. I dc C

20 kΩ 10 V + −

+ VC −

20 kΩ

4 µf

(A) 0.50 exp (- 25t) mA (B) 0.25 exp (- 25t) mA (C) 0.50 exp (-12.5t) mA (D) 0.25 exp (-6.25t) mA 8. A series RL circuit, with R = 10 W and L = 1 H, has a 100 V source applied at t = 0. The current for t > 0 is (A) 10 e-10t (B) 10 (1 - e-10t) -100t (C) 100 e (D) 100 (1 - e-100t) 9. The current in the circuit when the switch is closed at t = 0 is t =0

10 V

+ −

(A) 10 e-100t (C) 0.1 e-1,000t

1 kΩ

1 µF

i(t )

(B) 0.01 e- 1,000t (D) 10 e0.1t

10. The initial voltage across the capacitor when the switch S is opened at t = 0 I (A) zero (B) C . dc s 1 (C) I dc (D) Cs Idc Cs

Unit 01.indd 44 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 46

1Ω

v(t )= δ(t )

1F i(t )

4/25/2019 10:20:27 AM 29/04/2017 14:24:42

M02_

4:24:42

Hints/Solutions Chapter 3 • Transient Analysis (AC and DC) || 1.45 3.47 The circuit shown in the figure is initially relaxed. The Laplace transform of the current i(t) is 20 V

s +1 (A) 2 s + 2s + 1

s +1 (B) 2 s + s +1

s (C) 2 s + s +1

s (D) 2 s + 2s + 1

–

R 1 = 10 Ω

2 4H

R 2 = 10 Ω

11. The time constant of the network shown in the figure is R + –

3R 10V

3 RC (C) 3RC (D) RC 4 12. The time constant of the network shown in the figure is _____. (A) 4RC

(B)

The switch is moved from position 1 to position 2 at t = 0. 15. The current through inductor immediately after switching is _____. 1 A (A) 2 A (B) (C) 1 A (D) 5 A 2 16. The expression for current i(t) is _____. 2 (A) e−5t (B) 2e−5t (C) e −5t (D) 5e−2t 5 17. The switch in the circuit shown in the figure closes at t = 0. Find current ic for all times.

+ V –

20 V

2 3 RC (B) RC 3 2 (C) 2RC (D) RC 13. The network shown in the figure draws a current of ‘I’

+ −

(A)

I + j15

– j10

(B) -200

3k Ω 5t

(A) 5 × 10−3 × e −10 A

(B) 15 × 10−2 × e −10 A 5t

t S

V ± L

(A)

V R

(B) infinity

(C)

V R + jwL

(D) 0

19. Determine the current i for t ≥ 0, if Vc(0) = 1 V for the circuit shown

(D) 100

Statement for Linked Answer Questions 15 and 16: The circuit shown in the figure is initially under a steady state condition.

Unit 01.indd 45 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 47

t=0

If the supply frequency is doubled, then the current drawn by the circuit is _____. I I 2I (A) (B) 2I (C) (D) 2 5 5 14. In the circuit shown in the figure, the switch is thrown from position 1 to 2 at t = 0, after being at position 1 for d 2i ( 0 + ) a long time. The value of is _____. dt 2

(A) 200

0.01 µF

(C) 15 × 10−3 × e −10 A (D) 10 × 10−3 × e −10 A 18. For the given circuit, the current passing through inductor ‘L’ at the instant t = 0+ is

V(t ) = Vm sinωt

20 V + –

1k Ω

0.1F

10V

+ −

i(t ) 5Ω

4/25/2019 10:20:29 AM 29/04/2017 14:24:45

1.46 || Part Networks 3.48 III • Unit 1 • Networks 0.9e −5t 2 -2t (C) 5e

(B) 1.8 e-2t

(A)

(D) 3.6

 V (s )  20. Find the transfer function of the given system  0   V 1 (s )  100 Ω

↑ I(t ) 2A

↑

↑ Vi(S)

V0(s) 20F

10F

↓

(A)

↓

1 2000s + 1

(B)

100 (C) 1000s + 1

3s 3 + 2s 2s + 6s 3 + 7s 2 + 4s + 2 3s 2 + 2 (B) 4 2s + 5s 2 + 4s + 2 (3s 2 + 2)s (C) 2s 4 + 7s 3 + 6s 2 + 4s 5s 3 + 2s (D) 4 s + 7s 3 + 6s 2 + 4s + 2 (A)

23. Find the voltage Vab across the impedance of (2+5j)W in the network. The supply voltage e(t) = 10 sin (2pt + 45)

10s 2000s + 1

1 (D) 1000 s + 1

5j Ω

2Ω

→ e (t ) + −

− 2j Ω

3j Ω

21. Obtain the transfer function of the following system. 10 Ω

↑

5Ω

0.4s

Vin

0.2s

↓

(A) (C)

(A) 20 ∠ 44° (B) 24 ∠-40° (C) 4.48 ∠ -108° (D) 2.25 ∠63.43° 24. Find the current i(t) through the circuit given

↑

2Ω

↓

0.4s 0.2s + 1s s ( 2 + .08s )

(B)

s ( 2 + 0.4s )

0.4s + 8s + 100 2 + .04s (D) 2 s + 8s + 50

0.08s 2 + 8s + 50

22. Find the driving point admittance of the following network. 2Ω

1F 1H

Z →

+ 10 −

1H 0.5F

e t cos 2t

(A) 10e−t cos t

(B)

(D)

(A) 5sin(2t + 53.1°)A (C) 25sin(2t + 53.1°)A

(B) 5sin(2t - 53.1°)A (D) 25sin(2t - 53.1°)A

2 10 2

e t sin 2

Previous Years’ Questions 1 1 1. The circuit shown in figure, with R = W, L = H, and 3 4 C = 3 F has input voltage V(t) = sin2t, the resulting current i(t) is [2004]

2. For the circuit shown in figure, the time constant RC =1 ms. The input voltage is Vi(t) = 2 sin 103 t . The output voltage V0(t) is equal to [2004] R

i(t ) v(t )

Unit 01.indd 46 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 48

V1(t )

V0(t )

4/25/2019 10:20:32 AM 29/04/2017 14:24:48

M02_

4:24:48

Hints/Solutions Chapter 3 • Transient Analysis (AC and DC) || 1.47 3.49 (A) sin(103t - 45°)V (B) sin(103t + 45°)V 3 (C) sin(10 t - 53°)V (D) sin(103t + 53°)V 3. For the R–L circuit shown in figure, the input voltage Vi(t) = u(t). The current i(t) is [2004]

5. The condition on R, L, and C such that the step response y(t) in figure has no oscillations is [2005] L + u(t ) −

1H i(t )

V1(t )

(A) R ≥

1 L 2 C

(B) R ≥

L C

(D) R =

0.5 0.31

1 LC

−j 2Ω

j 2Ω

i(t )

(B)

L C

6. For the circuit in figure, the instantaneous current i1(t) is [2005]

t(sec)

y(t )

2Ω

i(t )

(A)

5∠0°A

10∠60°A

3Ω

0.63

10 3 10 3 ∠90° A (B) ∠ − 90° A 2 2 (C) 5 ∠ 60° A (D) 5 ∠-60° A 7. A square pulse of 3 volts amplitude is applied to C – R circuit shown in figure. The capacitor is initially uncharged. The output voltage V0 at time t = 2 s is [2005]

t(sec)

i(t ) 1/2

(C)

(A)

0.5 0.31

t(sec)

1/2

(D)

i(t )

0.1µF

3V 1

1kΩ

0.63 2sec t(sec)

4. The circuit shown in figure, initial current iL(0 ) = 1 A through the inductor and an initial voltage Vc(0−) = 1 V across the capacitor. For input V(t) = u(t), the Laplace transform of the current i(t) for t ≥ 0 is [2004] −

1Ω +

I(s) SL

i(t )

+ −

−

− +

(A)

s s + s +1

(B)

s+2 s + s +1

(C)

s−2 s2 + s +1

(D)

s−2 s2 + s −1

Unit 01.indd 47 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 49

(A) 3 V (B) -3 V (C) 4 V (D) -4V 8. A 2 mH inductor with some initial current can be represented as shown in the following figure, where s is the Laplace Transform variable. The value of initial current is [2006]

V(t )

3 mV

(A) 0.5 A (B) 2.0 A (C) 1.0 A (D) 0.0 A 9. In the following figure, assume that all the capacitors are initially uncharged. If Vi(t) = 10 u(t) Volts, V0(t) is given by [2006]

4/25/2019 10:20:35 AM 29/04/2017 14:24:51

1.48 || Part Networks 3.50 III • Unit 1 • Networks 1K +

+ 4µF

Vi(t )

1µF Vo(t )

−

(A) 8e−t/0.004t Volts (B) 8(1 - e−t/0.004t) Volts (C) 8u(t) Volts (D) 8 Volts 10. In the circuit shown, VC is 0 Volts at t = 0 s. For t > 0, the capacitor current iC(t), where t in seconds is given by [2007] 20 kΩ

10 V

+ −

20 kΩ

−

The value of the load resistance RL is (A) R/4 (B) R/2 (C) R (D) 2R 12. The switch in the circuit shown was on at position a for a long time, and is moved to position b at time t = 0. The current i(t) for t > 0 is given by [2009]

0.2 µF

+ −

5 kΩ 0.5 µF

i(t )

(C)

(B)

( R + R s )l s L

R s ls L

(D) ∞

15. The circuit shown in the figure is used to charge the capacitor C alternately from two current sources as indicated. The switches S1 and S2 are mechanically coupled and connected as follows: For 2nT ≤ t < (2n + 1) T (n = 0, 1, 2, …..) S1 to P1 and S2 to P2 For (2n + 1) T ≤ t < (2n + 2) T (n = 0, 1, 2, …..) S1 to Q1 and S2 to Q2 Q1 P1

Q2 P2 +

0.5Ω

1Ω 1A

Vc(t )

1Ω

−

1Ω

b i(t )

100 V

(A) 0

−

4 µF

10 kΩ

(B) i (t ) →

14. In the following circuit, the switch S is closed at t = 0. di The rate of change of current (0 + ) is given by dt [2008]

+ Vc −

2V 0 R 2V 0 (D) i (t ) → (1 + B ) R

V0 R V0 (C) i (t ) → (1 + B ) R

(A) i (t ) →

(A) 0.50exp(−25t) mA (B) 0.25exp(−25t) mA (C) 0.50exp(−12.5t) mA (D) 0.25exp(−6.25t) mA 11. If the transfer function of the following network is V 0 (s ) 1 = [2009] V i (s ) 2 + sCR +

V0 , the steady state R value of the current is given by [2009] For an initial current of i (0) =

0.3 µF

Assume that the capacitor has zero initial charge. Given that u(t) is a unit step function, the voltage Vc(t) across the capacitor is given be [2008] ∞

(A) ∑ ( −1) n tu (t − nT ) n=0

∞

(A) 0.2e−125t u(t)mA (B) 20e−1,250t u(t)mA (C) 0.2e−1,250t u(t)mA (D) 20e−1,000t u(t)mA 13. The time domain behaviour of an RL circuit is represented by L

(

)

di + Ri = V 0 1 + Be − Rt / L sin t u (t ). dt

Unit 01.indd 48 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 50

(B) u (t ) + 2∑ ( −1) n u (t − nT ) n =1

∞

(D) ∑ 0.5 − e − (t − 2 nT ) + 0.5e − (t − 2 nT −T )  n=0

4/25/2019 10:20:38 AM 29/04/2017 14:24:54

M02_

4:24:54

Hints/Solutions Chapter 3 • Transient Analysis (AC and DC) || 1.49 3.51 Common Data for Questions 16 and 17: The following series RLC circuit with zero initial conditions is excited by a unit impulse function d(t) 1H

1Ω

δ(t ) + −

(C) i (t) = 10exp(-2 × 103t)A (D) i (t) = -5exp(-2 × 103t)A 20. In the following circuit, the current I is equal to [2011]

(A)

(C)

− 3  t 2  − 12 t 2 e − e   3 

2 3

1 − t 2

 3  cos  t  2 

(B)

(D)

∼

–

[2008]

2 3 2 3

1 − t 2

 3  sin  t  2  [2008]

(A) 1. 4 ∠0° A (C) 2. 8 ∠ 0° A

(B) 2. 0 ∠ 0° A (D) 3. 2 ∠ 0° A

21. In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t = 0. The current i (t) for all t is [2012]

1 − t 1  − 23 t −e 2  e 3  1  3t    3t  1 − t  2 − (B) e cos  sin    3  2     2 

1  3t  − t e 2 sin   3  2  1  3t  − t e 2 cos   3  2 

18. The following circuit is driven by a sinusoidal input = Vi = Vpcos(t/RC). The steady state output Vo is [2011] + Vi ∼ −

(A) (Vp/3)cos(t/RC) (C) (Vp/2)cos(t/RC)

10Ω

(A) i (t) = 15exp(-2 × 103t)A (B) i (t) = 5exp(-2 × 103t)A

Unit 01.indd 49 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 51

V1(s)

V2(s) 100 µF

−

(B) (Vp/3) sin(t/RC) (D) (Vp/2) sin(t/RC)

–

+ 10 kΩ

+ Vo

i(t )

100 V

100 µF +

19. In the following circuit, the initial charge on the capacitor is 2. 5 mC in which the voltage polarity as indicated. The switch is closed at time t = 0. The current i (t) at time t after the switch is closed is [2011]

+ –

i(t )

(A) Zero (B) A step function (C) An exponentially decaying function (D) An impulse function V (s ) 22. The transfer function 2 of the following circuit is V 1 (s ) [2013]

C R

t =0

6Ω

1 − t

(A)

(D)

6Ω

te 2

17. For t > 0, the voltage across the resistor is

(C)

−j 4Ω

6Ω 14∠0°V

16. For t > 0, the output voltage Vc(t) is

j 4Ω

Vc(t )

−

(A)

0.5s + 1 s +1

(B)

3s + 6 s+2

(C)

s+2 s +1

(D)

s +1 s+2

Common Data for Questions 23 and 24: Consider the following figure:

50µF

+ 5Ω 10 V

1Ω 2Ω

− 2A

4/25/2019 10:20:41 AM 29/04/2017 14:24:56

1.50 || Part Networks 3.52 III • Unit 1 • Networks 23. The current Is in A in the voltage source and voltage Vs in Volts across the current source, respectively, are [2013] (A) 13, -20 (B) 8, -10 (C) -8, 20 (D) -13, 20 24. The current in the 1 W resistor in A is [2013] (A) 2 (B) 3.33 (C) 10 (D) 12 25. A 230 V RMS source supplies power to two loads connected in parallel. The first load draws 10 kW at 0. 8 leading power factor and the second one draws 10 kVA at 0. 8 lagging power factor. The complex power delivered by the source is [2014] (A) (18 + j 1. 5) kVA (B) (18 - j 1. 5) kVA (C) (20 + j 1. 5) kVA (D) (20 - j 1. 5) kVA 26. In the following circuit, the value of capacitor C (in mF) needed to have critically damped response i (t) is ____________. [2014] 40 Ω

C +

i(t )

i 10µF

(D) i(t )

t=0

30. Consider the building block called ‘Network N’ shown in the figure. Let C = 100 µF and R = 10 kW. Network N + V1(s)

1kΩ

+ −

2kΩ

(

)

(

)

(

)

(

)

C 1µF

5 2 1 − e − t /t t = m sec 3 3 5 2 (B) I (t ) = 1 − e − t /t t = m sec 2 3 5 (C) I (t ) = 1 − e −t /t ,t = 3m sec 3 5 (D) I (t ) = 1 − e −t /t ,t = 3m sec 2 29. A series RC circuit is connected to a DC voltage source at time t = 0. The relationship between the source voltage Vs, resistance R, capacitance C, and current i (t) is

Unit 01.indd 50 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 52

(A) I (t) =

1mH

28. In the figure shown, the capacitor is initially uncharged. Which one of the following expressions describes the current I (t) (in mA) for t > 0? [2014]

1kΩ

4kΩ

10V + −

(B) i(t )

− Vo

27. In the following figure, the ideal switch has been open for a long time. If it is closed at t = 0, then the magnitude of the current (in mA) through the 4 kW resistor at t = 0+ is ____________. [2014] 5kΩ

1 i (u )du c ∫0 Which one of the following represents the current i (t) ? [2014] (A) i(t ) Vs = R i (t) +

C R

−

V2(s) −

Two such blocks are connected in cascade, as shown in the figure. + V1(s)

+ Network N

−

The transfer function

+ Network N

−

V2(s) −

V 3 (s ) of the cascaded network is V 1 (s ) [2014]

4/25/2019 10:20:44 AM 29/04/2017 14:24:59

M02_

4:24:59

Hints/Solutions Chapter 3 • Transient Analysis (AC and DC) || 1.51 3.53 (A)

s 1+ s

 s  (C)   1 + s 

(B)

s2 1 + 3s + s 2

(D)

s 2+s

+ −

t=0

C y(t )

10 V

3Ω

2Ω

sin(ωt )

1 3RC

80 V

1 2 (D) RC RC 32. In the circuit shown in the figure, the value of v0(t) (in Volts) for t → ∞ is _________. [2014] ix

40 V

100 V, 50 Hz

38. In the circuit shown, the average value of the voltage Vab (in volts) in steady state condition is _______. [2015] 1 kΩ

2H + −

Vc(t)

3RC

(C)

10 u(t ) A

−

37. The voltage (VC) across the capacitor (in volts) in the network shown is ______. [2015]

(B)

5 F 6

C C

(A)

R C 2 L (D) 2 L R C 36. In the circuit shown, switch SW is closed at t = 0. Assuming zero initial conditions, the value of VC(t) (in volts) at t = 1 sec is _____. [2015] (C)

31. The steady state output of the circuit shown in the figure is given by y (t) = A (w) sin(wt + F (w) ). If the amplitude |A (w) | = 0. 25, then the frequency w is [2014] R

35. The damping ratio of a series RLC circuit can be expressed as [2015] R 2C 2L (A) (B) 2 2L R C

2ix

5Ω

b −

5p sin(5000t)

1µF Vab

2 kΩ

1mH

Vo(t ) −

5Ω

33. In the circuit shown, at resonance, the amplitude of the sinusoidal voltage (in volts) across the capacitor is ______. [2015] 4Ω 10 cos ωt (volts)

0.1mH

t=0

+ 1µF

−

+ −

34. In the circuit shown, the switch SW is thrown from position A to position B at time t = 0. The energy (in µJ) taken from the 3 V source to charge the 0.1 µF capacitor from 0 V to 3 V is [2015] 3V

39. In the circuit shown, the initial voltages across the capacitors C1 and C2 are 1 V and 3 V, respectively. The switch is closed at time t = 0. The total energy dissipated (in Joules) in the resistor R until steady state is reached, is _____. [2015]

120 Ω

0.1 µF

(A) 0.3 (C) 0.9

Unit 01.indd 51 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 53

(B) 0.45 (D) 3

C1 = 3F

−

C2 = 1F

40. At very high frequencies, the peak output voltage V0 (in volts) is _______. [2015] 100 µF

B SW A t=0

R = 10Ω

1kΩ

1kΩ 100 µF

1kΩ

1.0sin(w t) V

100 µF

4/25/2019 10:20:46 AM 29/04/2017 14:25:02

R3 d

c R2

1.52 || Part Networks 3.54 •2 Networks R1 III • Unit 1 R R2

The steady state magnitude of the capacitor voltage Vc (in volts) is _______. [2016] 44. In the RLC circuit shown in the figure, the input voltage is given by Vi(t) = 2cos (200t) + 4sin (500t). The output voltage v0(t) is [2016]

41. An ACavoltage source V = 10 sin (t) volts is applied to b Rnetwork 1 the following assume that R1 = 3 k W R2 = 6kW AC and R3 = 9kW, and that the diode is ideal. V =10 sin (t)

10 V

RMS current Irms (in mA) through the diode is R2 R2 __________ . R3 [2016] 3 42. The switch has been inRposition 1 for a long time and abruptly changes to position 2 atf t = R 0. g

3Ω

: t=0 + R2

b 0.1F

VC −

2Ω R1

4Ω

2Ω

+ 10V −

2Ω

V =10 sin (t)

If time t is in seconds, the capacitor voltage VC (in volts) for t > 0 is given by [2016] RMS current Irms (in mA) through the diode is (A) 4(1 - exp (-t/0.5) (B) 10 - 6exp (-t/0.5) __________ . [2016] (C) 4(1 - exp(-t/0.6) (D) 10 - 6exp(-t/0.6) 42. The switch has been in position 1 for a long time and 43. The switch S in the circuit shown has been closed for a abruptly changes to position 2 at t = 0. long time. it is opened at time t = 0 and remains open 4Ω has zero 2Ω 3Ω Assume that 1 the 2 diode after that. reverse current and zero forward voltage drop : t=0 +

+ 10V −

2Ω

VC −

0.1F

2Ω

If time t is in seconds, the capacitor voltage VC (in volts) for t > 0 is given by [2016] (A) 4(1 - exp (-t/0.5) (B) 10 - 6exp (-t/0.5) (C) 4(1 - exp(-t/0.6) (D) 10 - 6exp(-t/0.6) 43. The switch S in the circuit shown has been closed for a long time. it is opened at time t = 0 and remains open after that. Assume that the diode has zero reverse current and zero forward voltage drop

) volts is applied to R1 = 3 k W R2 = 6kW ideal. R3

10 V

t=0

1Ω

1 mH

10 µF

+ −

ugh the diode is [2016]

M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 54

+ + −

2Ω

+ +

10 µF 10

1 mH 0.4H

Vi (t)

−

Vo (t) 2Ω

The steady state magnitude of the capacitor voltage Vc – [2016] – is _______. (in volts) 44. In RLC circuit shown in the figure, the input volt(A)thecos (200t) + 2sin (500t) age is given by (B) 2cos (200t) + 4sin (500t) V (t) =sin2cos (200t) + 4sin (500t). i (C) (200t) + 2cos (500t) The output voltage v (t) is [2016] 0 (D) 2sin (200t) + 4cos (500t)

100 µF 0.25H 45. Assume that the circuit in the figure has reached the steady state before time t = 0 when the 3 W resistor + + suddenly burns out, resulting in an open circuit. The 2Ω current i(t) (in ampere) at t = 0 is __________ . [2016] 10

0.4H

Vi (t)

2ΩVo (t)

2F 1Ω

–

2Ω i (t )

12 V ±

–

2Ω

2Ω (A) cos (200t) + 2sin (500t) 2F (B) 2cos (200t) + 4sin (500t) (C) sin (200t) + 2cos (500t) 46. (D) In the2sin circuit shown, positive angular frequency w (200t) + 4costhe(500t) (in radians per second) at the magnitude of the the 45. Assume that the circuit inwhich the figure has reached phase difference between the voltages V and V equals steady state before time t = 0 when the 1 3 W 2resistor p/4 radians, is ____________. [2017] suddenly burns out, resulting in an open circuit. The 47. current The figure an RLC by the .sinusoii(t) shows (in ampere) at tcircuit = 0 is exited __________ [2016]

dal voltage 100 cos (3t) volts, where t is in 2Ωseconds. The 2F amplitude of V2 1Ω ratio is ________. [2017] amplitude of V1 i (t )

12 V ± 2F

4Ω 100 cos 3t

+ −

t=0 µF 100 10.25H Ω

2Ω 2Ω 1H 5Ω 1 36 F

48. In the circuit shown, the voltage Vin(t) is described by: for t < 0 ⎧ 0, Vin (t ) = ⎨ ⎩15 volts, for ≤ 0 where t is in seconds. The time (in seconds) at which the 29/04/2017 14:25:03 current I in the circuit will reach the value 2 amperes is ___________. [2017]

0.25H 100 µF +

+ 2Ω

0.4H

Vi (t)

Vo (t) 2Ω

or a longUnit time and52 01.indd M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 54 –

–

4/25/2019 10:20:50 AM 29/04/2017 14:25:03

Hints/Solutions | 1.53 1Ω

100 kΩ

+ Vin(t)

5 sin (5t) V

1 μF

–

49. The switch in the circuit, shown in the figure, was open for long time and is closed at t = 0 i(t) 10 A

5Ω

t=0 2.5 H

100 kΩ

(A) (B) (C) (D)

1.25 2 sin (5t – 0.25p) 1.25 2 sin 5t – 0.125p 2.5 2 sin (5t – 0.25p) 2.5 2 sin (5t – 0.125p)

51. For the circuit given in the figure, the magnitude of the loop current (in amperes, correct to three decimal places) 0.5 second after closing the switch is ________. [2018] 1V – +

The current i(t) (in ampere) at t = 0.5 s is __________. [2017] 50. For the circuit given in the figure, the voltage Vc (in volts) across the capacitor is [2018]

Unit 01.indd 53

1Ω

1 kΩ

4/25/2019 10:20:52 AM

Chapter 3 • Transient Analysis (AC and DC) | 3.55

hints/solutions Practice Problems 1 1. At t = 0, the switch is closed. The inductor acts as open circuit at t = 0+, so i(0+) = 0 R + RS

IsRs −

For t = 0+, ISRS = (R + RS) i(0+) + L

di (o + ) dt

di (o + ) I S R S = dt L Hence, the correct option is (B).

i.e.,

2. A1 = A + A 2 2

Z(s )

1 −

10 6 s + –

10 s

1K I(s)

10 s

10 10 = 3 3 10 + 10 s 10 (s + 103 ) 10 −2 s + 1000 i (t) = 0. 01 e-1000t Hence, the correct option is (B). 4. i (t) = i1 (t) + i2 (t) I (s) =

10 2 cos(t + 10 ) 10 5 cos( 2t + 10 ) + R + jw 1L R + jw 2 L

w1 = 1, w2 = 2

1   R×  Cs  = LS ||  1  R+   Cs   R  = LS ||   RCs + 1 =

106 + 103 s

i (t) =

10 2 cos(t + 10 ) 10 5 cos ( 2t + 10 ) + 1+ j 1+ 2 j

10 2 cos(t + 10 ) 10 5 cos ( 2t + 10 ) + 5 ∠ tan −1 2 2 ∠ tan −1 1

= 10 cos (t - 35°) + 10 cos (2t + 10° - tan-12) Hence, the correct option is (C).

Unit 01.indd 54 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 55

1 CS

 1  || R  Z (s) = LS ||   Cs 

i (t) =

2 3

= 32 + 4 2 = 5A Hence, the correct option is (B).

I (s) =

5. Ls = 0. 002s L = 0. 002 LI0 = 1 mV 1 mV I0 = = 0. 5 A 0.002 Hence, the correct option is (A). 6. After 2 s, the capacitor is fully charged, and then, the total supply voltage appears across the capacitor only. So, the output voltage is zero. Hence, the correct option is (C). 7.

Ls × R RLs = RCs + 1 RCs + 1 R RLCs 2 + Ls + R Ls + RCs + 1 RCs + 1

RLs RLCs 2 + Ls + R RL = 0. 2 L = 0. 1 H 1 RLC = 1 C = =5F 0.2 R=2W Hence, the correct option is (D). 8. Steady state established with switch in position 1. For t > 0, in s-domain =

40 Ω

10 + S −

−3

2 × 10 s I(s )

20 × 10 −3 ×

50 40

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3.56 | Part III • Unit 1 • Networks

Hints/Solutions | 1.55

10 + 25 × 10 −3 s I(s) = 40 + 20 × 10 −3 s

⇒

10 + 25 × 10 −3 s A B = = + −3 20 × 10 s (s + 2000) s s + 2000

di (0 + ) d 2 i (0 + ) , are 0, 10, and -50 dt dt 2 Hence, the correct option is (D). 13. The given circuit is a second-order circuit. The response will have oscillations and the oscillations will die down with time due to the presence of R. Hence, the correct option is (C). 14. For t = 0+, the circuit can be drawn as shown below.

= 62. 5 s- 1

VC (0+) = VC (0-) = 100 V VR = VC = 100 e- 62. 5t V i = R = 0.25 e −62.5t R Hence, the correct option is (C). 10. R = 5 W, XL = wL = (5,000) (2 × 10-3) = 10 W Z = 5 + j10 = 11. 8 ∠63. 4° V = 150 ∠- 90° V V 150 ∠ − 90 I= = = 13. 4 ∠- 153. 4° A Z 11.8∠63.4 i = 13. 4 cos(5,000t - 153. 4°) i = 13. 4 sin(5,000t - 63. 4)° A Hence, the correct option is (D). 11. s

+ V1 −

Z11 =

+ 5V −

5 = 250 W 20 × 10 −3 Hence, the correct option is (B). 32 + 16 48 15. From the circuit, i(0-) = = = 2A and 16 + 8 24 i(∞) = 4A. The option satisfying both the conditions

−

V1 I1

s2 +1 1 1  s + || =   s  s s ( s 2 + 2) s2 +1 s 4 + 3s 2 + 1 +s = 2 s (s + 2) s 3 + 2s Hence, the correct option is (A). 12. At t = 0+, the capacitor acts as an open circuit i (0+) = 0 On applying KVL Ri (0+) + L L

di (0 + ) 1 + ∫ i (0 + )dt = 20 dt C

di (0 + ) = 20 dt

⇒

di (0 + ) 20 = = 10 dt 2

Unit 01.indd 55 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 56

RΩ

i(0+)

100 Ω

d 2 i (0 + ) R di (0 + ) =− = −50 2 L dt dt

i (0+),

i (t) = 0. 25 + e- 2000t Hence, the correct option is (B). 9. VR = VC for t > 0 RC

di (0 + ) d 2 i (0 + ) i (0 + ) +L + =0 dt C dt 2

is (B). Hence, the correct option is (B). 16. V (t) is unit step, i.e., u (t). The response is i(t). Transfer function  1  I (s )  (s + 5)  s = = 1 V (s ) s+5 s V (s ) s + 5 5 = =1+ Impedance of the given network Ι (s ) s s 1 z (s) = 1 + 1 s 5 From z (s), it can be inferred that the circuit contains a resistance of 1 W and a capacitance of 1/5 F in series. Hence, the correct option is (B). 17. When S1 is closed at t = 0, the current flows in the circuit and it decays exponentially starting from 5 A. When S2 is closed at t = 4 s, the current increases instantaneously and then decreases exponentially. Hence, the correct option is (D). 18. Let the current through the inductor be IL. The current through the resistor be IR. I = IR + IL Voltage V = V 1 + V 2

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Chapter 3 • Transient Analysis (AC and DC) | 3.57

1.56 | Networks V1

 2s + 3  I (s)  =4  s  V1 + V 2

I (s) =

I IL

4s 2s + 3

I (s) = L−1

The phase angle of I with respect to the voltage V2 is 0°. Hence, the correct option is (A). 19. iL (0+) = iL(0-) From the figure 20 Ω

4s 2s = 2s + 3 s + 3

−3 d −2.3 − 3 2 t e = e = −3e 2 t dt 2 Hence, the correct option is (B). 22. The transform network will be,

I (t) = 2

1 2s a

•

200V t=0

+ 3µF –

20 Ω

iL (0-) =

between a and b. 1 Zab = + 0.5s 2s Zab =

(1 + s ) 0.5s / 2s 2

1+ s2 + 0.5s 2s

(1 + s ) 0.5s + 1 = (1 + s ) s + 1 Z= 2s 1 + 2s (1 + 2s ) 2s 2

2Ω

= + −1F

1+ s2 2s

zab || 0. 5s =

b 5∠0

0.5s V2

•

The direction of the current is as indicated in the figure. The capacitor is charged to a voltage of 100 V with the polarity indicated in the figure. After switch is closed the voltage across the capacitance is applied to inductor. As the inductor is in parallel with capacitor. VL (0+) = 100 V Hence, the correct option is (C). 20. At steady state, t → ∞, i (t) = 0 Hence, the correct option is (C). 21. a

1 2s

0.5s

200 = 5A 20 + 20

−3 t 2

s + s +1 4

(1 + 2s ) 2s 2

Hence, the correct option is (D). 23. 4W

−

Vc (0 )= 2V

6Ω

S B

The transformed circuit is

3Ω

1 4

4  1 2 = 2 + +  I (s ) s  s s

Unit 01.indd 56 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 57

−

V(0) s

= =

10V

1 s Vc

+ −

(0−) s

10 = 1A (when switch at A) 4+6 t = 0+, switch is at B i1 = 0, i2 = i in loop 2, I(0-) =

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3.58 | Part III • Unit 1 • Networks 6i + 1

Hints/Solutions | 1.57 Vc (1 ms) = 20. 16 × e- 2. 48 = 1. 69 volts Vc (t) = 1. 69. e-(t-1m)/t; for t > 1 ms t = 25 × 103 × 0. 5 × 10-6 = 12. 5 ms Vc (5 ms) = 1. 69. e-4/ 12. 5 = 1. 23 volts Hence, the correct option is (A). 26. At t = 0Vc(0-) = VC(0+) = 0 V t > 0 :-

di + 3i = 5i dt

di + 4i = 0 dt i = ce-4t,, c = i0 = 1 A \i = e-4t at t = 4s, i = 0. 1 mA Hence, the correct option is (A). 24. 25 mF

10 Ω

V c V c − e −t 1 dV c + + . =0 20 10 20 dt

2 mH

10 A = I (0 − ) = 1 A 10 when ‘s’ is closed V(c) = 0, capacitor shorted When switch is opened, current through inductor will not change instantaneously. Also, V c (0 − ) = 0 The voltage around the capacitor also cannot change instantaneously, \ Vc(0+)= 0 Hence, the correct option is (A). 25. At t < 0 :- Switch is closed I=

1kΩ

25V

–

Vc(0 )

5kΩ

25kΩ

dV c + 3 v c = 2.e −t dt dV c (0 + ) = 2 V/s dt Differentiating (i) w.r.t t and t = 0+,

d 2V c dV + 3. c = − 2.e −t dt dt 2 at t = 0+ d 2V c (0 + ) = − 2 − 3× 2 dt 2 = - 8 V/s2 Hence, the correct option is (C). 27. 10 ∠30° = 4. I1 - 0. 5ix + (-j2). ix (-j2). ix = (I - ix) j3 ⇒ I = ix/3

−

Vc (0 ) − 25 Vc (0 ) Vc (0 ) + + =0 1 kΩ 5 kΩ 25 kΩ Vc (0-) [25 + 5 + 1] = 25 × 25 = 20. 16 V Req = 1 kW || 5 kW || 25 kW 1 1 1 1 = + + Re q 1 kΩ 5 kΩ 25 kΩ 1 25 + 5 + 1 = Re q 25 kΩ

⇒ ix =

25 kW 31 25 t = RC = × 0.5 ms = 403. 225 ms 31

Vc (t) = 20. 16. e

-2480t

Unit 01.indd 57 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 58

V, for 0 < t < 1 ms

(ii)

10 ∠ 30 2.17 ∠ − 67.38

= 4. 62 ∠97. 38° Hence, the correct option is (B). 28. From the given data Q0 = 2 .5 mc Q=CV V0 =

Req =

(i)

4  10 < 30° =  − 0.5 − j 2 . I x 3 

– −

(i)

2.5 × 10 −3 = -50 V = Vc (0+) 50 × 10 −6

at t → ∞ Vc (∞) = 100 V ⇒ V (t) = V(∞) + (V(0+) - V(∞)). e-t/t t = RC = 0. 5 ms \ Vc (t) = 100 - 150. e-t/t Volts

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Chapter 3 • Transient Analysis (AC and DC) | 3.59

1.58 | Networks

for t > 0: Now, on opening the switch S, the circuit in s-domain will be as shown below.

3 dv c = 15. e −2 ×10 t amps dt Hence, the correct option is (A). 29. At t < 0: S →opened and L → S. C C → O.C At t = 0-, the equivalent circuit is shown below

ic = c.

2Ω

2Ω I(s)

4 Ω s

SΩ

2Ω

± Li 0 ± 1V

12V

1Ω

4. I (s) + s.I (s) +

12 = 4A 3 Vc(0-) = 1 × 4 = 4 V t > o: At t = 0+. The equivalent ckt is shown below i

1Ω

2.5 Ω

31. 4A

12 − 4 8 = = 3. 2 A 2.5 2.5 ic = 4 - i = 0. 8 A Hence, the correct option is (A). 30. For t < 0 :Switch is closed at t = 0- ⇒ L → S. C C → O.C

But VA = 5∠0° - V1

V 1 3V 1 5 − V1 + = 4 4 1 − j1.5 20 − 4V 1 1 − j1.5

4V1(1 - j 1. 5) = 20 - 4V1 8V1 = 20 + j6V1 2V1(4 - j3) = 20 V1 =

10 = 2 ∠36. 86° 4 − j3

V1 = 0. 5 ∠36. 86° 4 Complex power S = V. I I1 =

2Ω I L(0–)

V1 3 VA + V1 = 4 4 1 − j1.5

4V1 =

2V ±

1 = 0. 5 s 2 Hence, the correct option is (B).

\t=

12V

4 2 I (s ) = + 1 s s

4 2  I (s) s + 4 +  = 1 + 3 s  1 ⇒ i (t) = e-2t A I (s) = s+2

⇒ i(0-) =

1Ω

V c(0–)

i (0–)

2Ω

V c(0–)

= 5 × 0. 5 ∠36. 87° = 2. 5 ∠36. 9°

2 =1A 2 Vc(0-) = 2 V

iL(0-) =

Unit 01.indd 58 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 59

Power factor = cos 36. 9 = 0. 8 leading (R. C) Hence, the correct option is (B).

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3.60 | Part III • Unit 1 • Networks

Hints/Solutions | 1.59 For w = w1 = 1

32. From the given data R = 1Ω

I(t)

V(t) ~

H ( j1 ) ∠f1 =

H ( j2 ) ∠f2 =

1+ w

1 1 Y (s ) sC = = 2 1 U (s ) LCs + RCs + 1 + R + Ls sC 1 LC = 1 R 2 s + s+ L LC 1 2 wn = LC 1 wn = LC R 2xw n = L

1 5

R C 2 L For no oscillation, x ≥ 1 L R≥2 C Hence, the correct option is (C). 5. At t = 0, capacitor acts as short circuit. So the voltage across capacitor is zero. Hence, the correct option is (A). (5 − j 3) (5 + j 3) 6. VAB = 5∠30° × 5 − j3 + 5 + j3 ( 25 + 9) 10 = 17∠30° V Hence, the correct option is (D). = 5∠30° ×

Unit 01.indd 59 M02_GATE-ECE-GUIDE-00_SE_XXXX_CH03.indd 60

∠ − 45°

∠ − tan −1 2

(

)

i (t) = 10cos(t -35°) + 10cos(2t +10° − tan-12)A Hence, the correct option is (C). 7. Apply Thevenin’s theorem at the capacitor side 20 k = 5V VTH = 10 × 20 k + 20 k RTH = 20k || 20k = 10k 10 k Ω

4 µF

5V − i(t )

− t 5 3 −6 e 40 ×10 ×10 3 10 × 10 = 0. 5 e- 25t mA Hence, the correct option is (A).

i (t) =

8. 10 Ω

R C LC = R 2x = L L x =

cos 2t + 10° − tan −1 2

< − tan −1w

Practice Problems 2

\ I = H(jw). V \ Steady-state current I (t) = 1 1 10 2 . .cos (t + 10° − 45°) + 10 5 . . 2 5

Given V (j) = 10 2 cos (t + 10°) + 10 5 . cos ( 2t + 10°)V w1 = 1 rad/s and w2 = 2rad/s 1

∠ − tan −11 =

For w2 = 2 rad/s

L = 1H

For RL series circuit I 1 1 = W + ( jw ) = = V R + jwL 1 + jw

H ( jw ) ∠f =

1H i(t )

100 V

I (s) =

100 100 A B = = + s s (s + 10) s s + 10 10 + s

100 = 10 s + 10 S = 0

100 = −10 s S = −10

I (s) =

10 10 − s s + 10

i (t) = 10 -10 e- 10t i (t) = 10 (1 - e- 10t) Hence, the correct option is (B).

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