CHAPTER – 39

ALTERNATING CURRENT

1.  = 50 Hz
 = 0 Sin Wt
0
Peak value  =
2
0
= 0 Sin Wt
2
1 
 = Sin Wt = Sin 
2 4
   1 1
 = Wt. or, t = =

= = = 0.0025 s = 2.5 ms
4 400 4  2 8 8  50
2. 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

 t = 
4
   1
t= = = = = 2.5 ms
4 4  2f 850 400
3. P = 60 W V = 220 V = E

= 220  220 = 806.67

v2
R=
P 60
0 = 2 E = 1.414 × 220 = 311.08
 806.67
0 = 0
= = 0.385 ≈ 0.39 A
R 311.08
4. E = 12 volts
i2 Rt = i2 rmsRT
E2 E2rms E02
 =  E2
R2 R2 = 2
 E02 = 2E2  E 20 = 2 × 122 = 2 × 144
 E0 = 2  144 = 16.97 ≈ 17 V
5. P0 = 80 W (given)
P0
Prms = = 40 W
2
Energy consumed = P × t = 40 × 100 = 4000 J = 4.0 KJ
6. E = 3 × 106 V/m, A = 20 cm 2, d = 0.1 mm
Potential diff. across the capacitor = Ed = 3 × 10 6 × 0.1 × 10–3 = 300 V
V 300
Max. rms Voltage = = = 212 V
2 2

39.1
 
7. i = i0e–ur
2 1  2  2t / 
i0
2   2t /  i02
   2t /   
i 02

 2

i e   e 1

i = e dt = dt =  0 =   2



0
  e
0 0  2
i0 2  1  i0  e2 
 1
i2 =  2 1 = e  2 
2 e   
8. C = 10 F = 10 × 10 F = 10 F
–6 –5

E = (10 V) Sin t
10
a)  = E0 = E0
= = 1 × 10–3 A
Xc  1   1 5 
  10  10 
 C   
b)  = 100 s–1
E0 10
= = = 1 × 10–2 A = 0.01 A
 1   1 5 
  100  10 
 C   
c)  = 500 s–1
E0 10
= = = 5 × 10–2 A = 0.05 A
 1   1 5 
  500  10 
 C   
d)  = 1000 s–1
E0 10
 = = 1 × 10–1 A = 0.1 A
 1   1 

  1000  105 
 C   
9. Inductance = 5.0 mH = 0.005 H
a)  = 100 s–1
5
XL = L = 100 × = 0.5 Ω
1000
 0 = 10 = 20 A
i=
XL 0.5
b)  = 500 s–1
5
XL = L = 500 × = 2.5 Ω
1000
0 10
i= = =4A
XL 2.5
c)  = 1000 s–1
5
XL = L = 1000 × =5Ω
1000
0 10
i= = =2A
XL 5
10. R = 10 Ω, L = 0.4 Henry
30
E = 6.5 V, = Hz

Z= R2  XL
2
= R2  (2L)2
Power = Vrms  rms cos 
= 6.5 × 6.5  R = 6.5  6.5  10 6.5  6.5  10 6.5  6.5  10 = 0.625 = 5 
= = 

Z Z  R2  (2L) 2 
2
 30 2 100  576 8
102  2   0.4
    

39.2
 V2
11. H = T, E0 = 12 V,  = 250 , R = 100 Ω
R 2
H E0 Sin t 2
144  1 cos 2t 
2

  

H=
0
dH =
R 100
dt = sin
 2
 dt
t dt = 1.44
 
 3  Sin2t  103 
1.44  10 3 103 
 
 
=  
dt  Cos2t dt  = 0.7210    
2  0 0
   2 0 
 1 1  (  2)
= 0.72  =  0.72 = 0.0002614 = 2.61 × 10–4 J
 1000 500 
 1000
12. R = 300Ω, C = 25 F = 25 × 10–6 F, 0 = 50 V,  = 50 Hz
1 1 104
Xc = =
c 50  = 25
 2  25  10 6

  104 2
Z= R2  Xc (300)2   
2
= = (300)2  (400)2 = 500
 25 
50
E = 0.1 A
(a) Peak current = 0 =
500
Z
(b) Average Power dissipitated, = E rms rms Cos 
E0 R
E 2 50  50  300 3
= 0  2Z  Z = 0 = 2  500  500 = 2 = 1.5 .
E
2 2Z2
2 110  110
13. Power = 55 W, Voltage = 110 V, Resistance = V = = 220 Ω
P 55
frequency () = 50 Hz, = 2 = 2 × 50 = 100 
V R L
V
Current in the circuit = =
Z R2  (L)2
110 V
VR –
Voltage drop across the resistor = ir = 220 V
R2  (L)2
220  220
= = 110
(220) 2  (100L) 2
(220)2  (100L)2
 220 × 2 =  (220)2 + (100L) 2 = (440)2
 48400 + 10  L = 193600 4 2 2
 1042 L2 = 193600 – 48400
142500
 L2 = 1.4726  L = 1.2135 ≈ 1.2 Hz
= 2  104
14. R = 300 Ω, C = 20 F = 20 × 10 –6 F
50
L = 1 Henry, E = 50 V V= Hz

E0
(a) 0 = ,
Z
2
 1 
Z= R2  (Xc  XL )2 = (300)2    2L
 2C 
2
 
 1 50   104
2

= (300) 2     2   1 = (300) 2      100  = 500
 50 6    20 
 2   20  10   
  
E0 50
= = = 0.1 A
0
Z 500

39.3
 (b) Potential across the capacitor = i 0 × Xc = 0.1 × 500 = 50 V
Potential difference across the resistor = i 0 × R = 0.1 × 300 = 30 V
Potential difference across the inductor = i 0 × 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
C = 20 F = 20 × 10 –6 F
L = 1H, Z = 500 (from 14)
E0 50
0 = 50 V, 0 = = = 0.1 A
Z 500
Electric Energy stored in Capacitor = (1/2) CV 2 = (1/2) × 20 × 10–6 × 50 × 50 = 25 × 10 –3 J = 25 mJ
Magnetic field energy stored in the coil = (1/2) L 02 = (1/2) × 1 × (0.1)2 = 5 × 10–3 J = 5 mJ
16. (a)For current to be maximum in a circuit
Xl = Xc (Resonant Condition)
1
 WL =
WC
1 1
= = 106
W 2
LC 2  18  106 =
3 3 36
10 10

W=  2 =
6 6
1000
= 26.537 Hz ≈ 27 Hz
  = 6  2
 E
(in resonance and)
(b) Maximum Current = R
2
20 = A = 2 mA
=
10  103 103
17. Erms = 24 V
r = 4 Ω,  rms = 6 A
E 24
R= = =4Ω
 6
Internal Resistance = 4 Ω
Hence net resistance = 4 + 4 = 8 Ω
12
 Current = = 1.5 A
8 10 Ω
18. V1 = 10 × 10–3 V
R = 1 × 103 Ω V1 10 nF V0
C = 10 × 10–9 F
1 1 1 1 104 5000
(a) Xc = = = = = =
WC 2C 2  10  103
 10  109 2  104 2 

1 10 
2
 5000   5000 2
Z= R  Xc
2 2
= 3 2   = 10 6   
      
E0 V1 10  103
0 = = =
Z Z  5000 2
10 6   
  

39.4
 1 1 1 1 103 500
(b) Xc = = = = = =
WC 2C 2  105  10  109 2  103 2 

c 10    = 10   
      
E0 V1 10  103
0 = = =
Z Z
 500 2
10 6  
 
10  103 500

V0 = 0 Xc =  = 1.6124 V ≈ 1.6 mV

 500 2
10 6  
 
(c)  = 1 MHz = 106 Hz
1 1 1 1 102 50
Xc = = = = = =
WC 2C 2  10  10 
6
2  10 2 2 
109

10 
2 2
  50   50 
Z= R2  X c2 = 3 2    = 106   
      
 E0 V1 10  103
 0 = = =
Z Z 2
  50 
106   
   
 10  10
3
50
 V0 = 0 Xc =  ≈ 0.16 mV

  50 2
10 6  
  
(d)  = 10 MHz = 107 Hz
1 1 1 1 10 5
Xc = = = = = =
WC 2C 2  10  10  10
7 9
2  101 2 

10 
2 2
 5   5 
Z= R2  X c2 = 3 2    = 10 6  

   
V1 10  103
0 = E0 = =
Z Z  5 2
10 6  

10  103 5
V0 = 0 Xc =  ≈ 16 V

 5 2
10 6  


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 P1 Sec
secondary coil is zero.

    
39.5