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  2f 850 400 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  (2L)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  (2L )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 2t   dt 2 

 

3 10 3   1.44  10  Sin2t  103  dt Cos2t 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) + (100L) = (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   2L  (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