CHAPTER – 39
ALTERNATING CURRENT 1.
= 50 Hz = 0 Sin Wt Peak value =
0
1
2.
2
= 0 Sin Wt
2
0
2
= Sin Wt = Sin
4
= Wt. 4
or, t =
1 1 = = = = 0.0025 s = 2.5 ms 400 4 2 8 8 50
Erms = 220 V Frequency = 50 Hz E (a) Erms = 0 2 E0 = Erms 2 = 2 × 220 = 1.414 × 220 = 311.08 V = 311 V (b) Time taken for the current to reach the peak value = Time taken to reach the 0 value from r.m.s = 0 0 = 0 Sin t 2 2
4 1 t= = = = = 2.5 ms 4 4 2f 850 400 P = 60 W V = 220 V = E t =
3.
R=
v2 220 220 = = 806.67 P 60
0 =
4.
2 E = 1.414 × 220 = 311.08 0 806.67 0 = = = 0.385 ≈ 0.39 A R 311.08 E = 12 volts i2 Rt = i2rms RT
E2
E 2 rms
2
E0 2 2 2 R R 2 2 2 2 E0 = 2E E0 = 2 × 12 = 2 × 144
5.
6.
=
2
E =
E0 = 2 144 = 16.97 ≈ 17 V P0 = 80 W (given) P Prms = 0 = 40 W 2 Energy consumed = P × t = 40 × 100 = 4000 J = 4.0 KJ 6 2 E = 3 × 10 V/m, A = 20 cm , d = 0.1 mm Potential diff. across the capacitor = Ed = 3 × 106 × 0.1 × 10–3 = 300 V V 300 = = 212 V Max. rms Voltage = 2 2 39.1
Alternating Current 7.
i = i0e
–ur
i2 =
i 2 1 i0 2 e 2 t / dt = 0
0
i2 = 8.
e 2 t / dt =
0
i i0 2 1 2 1 = 0 e 2 e –6
C = 10 F = 10 × 10 E = (10 V) Sin t E0 E = a) = 0 = Xc 1 C
2
2
i0 i e 2t / = 0 e 2 1 2 2 0
e 2 1 2 –5
F = 10 F
10 –3 = 1 × 10 A 1 10 10 5
b) = 100 s–1 E0 10 –2 = = = 1 × 10 A = 0.01 A 1 1 C 100 10 5 c) = 500 s–1 E0 10 –2 = = = 5 × 10 A = 0.05 A 1 1 C 500 10 5
9.
d) = 1000 s–1 E0 10 –1 = = 1 × 10 A = 0.1 A 1 1 C 1000 10 5 Inductance = 5.0 mH = 0.005 H –1 a) = 100 s 5 XL = L = 100 × = 0.5 Ω 1000 10 i= 0 = = 20 A XL 0 .5 –1
b) = 500 s
XL = L = 500 × i=
5 = 2.5 Ω 1000
0 10 = =4A XL 2 .5
c) = 1000 s
–1
XL = L = 1000 × i=
0 10 = =2A XL 5
10. R = 10 Ω, E = 6.5 V, Z=
5 =5Ω 1000
R 2 XL 2 =
L = 0.4 Henry 30 = Hz
R 2 (2L)2
Power = Vrms rms cos 6 .5 R 6.5 6.5 10 5 6.5 6.5 10 6.5 6.5 10 = 6.5 × = = = = 0.625 = 2 2 Z Z 100 576 8 30 R 2 (2L )2 0 .4 10 2 2 39.2
Alternating Current
V2 T, R
11. H =
H
H= =
dH =
0
= 250 ,
E0 = 12 V,
2
E 0 Sin 2 t 144 dt = sin 2 t dt = 1.44 100 R
R = 100 Ω
1 cos 2t dt 2
3 10 3 1.44 10 Sin2t 103 dt Cos2t dt = 0.72 10 3 0 2 0 2 0
1 ( 2) 1 –4 = 0.72 = 0.72 = 0.0002614 = 2.61 × 10 J 1000 1000 500 –6 12. R = 300Ω, C = 25 F = 25 × 10 F, 0 = 50 V, = 50 Hz Xc =
10 4 1 1 = = 50 c 25 2 25 10 6 2
R Xc
Z=
2
10 4 (300) 25 2
=
2
(300)2 ( 400 )2 = 500
=
E0 50 = = 0.1 A Z 500 (b) Average Power dissipitated, = Erms rms Cos
(a) Peak current =
=
E0 2
E0 2Z
E 2 R 50 50 300 3 = 02 = = = 1.5 . Z 2 500 500 2 2Z
13. Power = 55 W,
Voltage = 110 V,
frequency () = 50 Hz,
V Current in the circuit = = Z
220 220 (220 )2 (100 L )2
220 × 2 =
L
R
R 2 ( L ) 2
110 V
VR
–
R 2 ( L ) 2
220 V
= 110
(220)2 (100 L)2 4 2
V2 110 110 = = 220 Ω P 55
= 2 = 2 × 50 = 100 V
Voltage drop across the resistor = ir = =
Resistance =
2
2
2
(220) + (100L) = (440) 4 2
2
2
48400 + 10 L = 193600 10 L = 193600 – 48400 142500 2 L = 2 = 1.4726 L = 1.2135 ≈ 1.2 Hz 10 4 –6 14. R = 300 Ω, C = 20 F = 20 × 10 F 50 Hz L = 1 Henry, E = 50 V V= E (a) 0 = 0 , Z Z=
R 2 ( X c XL ) 2 =
1 2L (300)2 2 C
1 50 = (300 )2 2 1 50 20 10 6 2 E0 50 0 = = = 0.1 A Z 500
2
2
=
39.3
10 4 (300 ) 100 20 2
2
= 500
Alternating Current (b) Potential across the capacitor = i0 × Xc = 0.1 × 500 = 50 V Potential difference across the resistor = i0 × R = 0.1 × 300 = 30 V Potential difference across the inductor = i0 × XL = 0.1 × 100 = 10 V Rms. potential = 50 V Net sum of all potential drops = 50 V + 30 V + 10 V = 90 V Sum or potential drops > R.M.S potential applied. 15. R = 300 Ω –6 C = 20 F = 20 × 10 F L = 1H, Z = 500 (from 14) E 50 0 = 50 V, 0 = 0 = = 0.1 A Z 500 2 –6 –3 Electric Energy stored in Capacitor = (1/2) CV = (1/2) × 20 × 10 × 50 × 50 = 25 × 10 J = 25 mJ 2 2 –3 Magnetic field energy stored in the coil = (1/2) L 0 = (1/2) × 1 × (0.1) = 5 × 10 J = 5 mJ 16. (a)For current to be maximum in a circuit (Resonant Condition) Xl = Xc WL = 2
W =
1 WC
10 6 1 1 = = LC 36 2 18 10 6
10 3 10 3 2 = 6 6 1000 = = 26.537 Hz ≈ 27 Hz 6 2 E (b) Maximum Current = (in resonance and) R 20 2 = = A = 2 mA 3 10 10 10 3 17. Erms = 24 V r = 4 Ω, rms = 6 A W=
E 24 = =4Ω 6 Internal Resistance = 4 Ω Hence net resistance = 4 + 4 = 8 Ω 12 Current = = 1.5 A 8 –3 18. V1 = 10 × 10 V 3 R = 1 × 10 Ω –9 C = 10 × 10 F R=
(a) Xc =
Z= 0 =
10 Ω V1
1 10 4 5000 1 1 1 = = = = = 3 9 4 WC 2 C 2 2 10 10 10 10 2 10 R 2 Xc 2 =
E0 V = 1 = Z Z
1 10
3 2
2
5000 =
5000 10 6
10 10 3 5000 10 6
2
39.4
2
10 nF
V0
Alternating Current (b) Xc = Z= 0 =
1 10 3 500 1 1 1 = = = = = 5 9 3 WC 2 C 2 2 10 10 10 2 10 R 2 Xc 2 =
2
500 =
3 2
500 10 6
2
10 10 3
E0 V = 1 = Z Z
V0 = 0 Xc =
10
500 10 6
10 10 3 500 10 6
2
2
500 = 1.6124 V ≈ 1.6 mV
6
(c) = 1 MHz = 10 Hz Xc =
1 10 2 1 1 1 50 = = = = = 6 9 2 WC 2 C 2 2 10 10 10 2 10
Z=
R 2 Xc 2 =
0 =
2
50 =
3 2
50 10 6
2
10 10 3
E0 V = 1 = Z Z
V0 = 0 Xc =
10
50 10 6
10 10 3 50 10 6
2
2
50 ≈ 0.16 mV
7
(d) = 10 MHz = 10 Hz 1 1 1 1 10 5 Xc = = = = = = 7 9 1 WC 2 C 2 2 10 10 10 2 10 Z= 0 =
R 2 Xc 2 =
E0 V = 1 = Z Z
V0 = 0 Xc =
10
3 2
2
5 =
5 10 6
2
10 10 3 5 10 6
10 10 3 5 10 6
2
2
5 ≈ 16 V
19. Transformer works upon the principle of induction which is only possible in case of AC. Hence when DC is supplied to it, the primary coil blocks the Current supplied to it and hence induced current supplied to it and hence induced Current in the secondary coil is zero.
39.5
P1
Sec