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 Page 1


39.1
CHAPTER – 39
ALTERNATING CURRENT
1. ? = 50 Hz
? = ?
0
Sin Wt
Peak value ? = 
2
0
?
2
0
?
= ?
0
Sin Wt
?
2
1
= Sin Wt = Sin 
4
?
?
4
?
= Wt. or, t = 
400
?
= 
? ? ?
?
2 4
= 
? 8
1
= 
50 8
1
?
= 0.0025 s = 2.5 ms 
2. E
rms
= 220 V
Frequency = 50 Hz
(a) E
rms
= 
2
E
0
? E
0
= E
rms 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
? = 
2
0
?
?
2
0
?
= ?
0
Sin ?t
? ?t = 
4
?
? t = 
?
?
4
= 
f 2 4 ? ?
?
= 
50 8 ?
?
= 
400
1
= 2.5 ms ?
3. P = 60 W V = 220 V = E
R = 
P
v
2
= 
60
220 220 ?
= 806.67
?
0
= 2 E = 1.414 × 220 = 311.08
?
0
= 
R
0
?
= 
08 . 311
67 . 806
= 0.385 ˜ 0.39 A
4. E = 12 volts
i
2
Rt = i
2
rms
RT
?
2
2
R
E
= 
2
rms
2
R
E
? E
2
= 
2
E
2
0
? E
0
2
= 2E
2
  ? E
0
2
= 2 × 12
2
= 2 × 144 
? E
0
= 144 2 ? = 16.97 ˜ 17 V 
5. P
0
= 80 W (given)
P
rms
=
2
P
0
= 40 W
Energy consumed = P × t = 40 × 100 = 4000 J = 4.0 KJ
6. E = 3 × 10
6
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
Max. rms Voltage = 
2
V
= 
2
300
= 212 V
Page 2


39.1
CHAPTER – 39
ALTERNATING CURRENT
1. ? = 50 Hz
? = ?
0
Sin Wt
Peak value ? = 
2
0
?
2
0
?
= ?
0
Sin Wt
?
2
1
= Sin Wt = Sin 
4
?
?
4
?
= Wt. or, t = 
400
?
= 
? ? ?
?
2 4
= 
? 8
1
= 
50 8
1
?
= 0.0025 s = 2.5 ms 
2. E
rms
= 220 V
Frequency = 50 Hz
(a) E
rms
= 
2
E
0
? E
0
= E
rms 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
? = 
2
0
?
?
2
0
?
= ?
0
Sin ?t
? ?t = 
4
?
? t = 
?
?
4
= 
f 2 4 ? ?
?
= 
50 8 ?
?
= 
400
1
= 2.5 ms ?
3. P = 60 W V = 220 V = E
R = 
P
v
2
= 
60
220 220 ?
= 806.67
?
0
= 2 E = 1.414 × 220 = 311.08
?
0
= 
R
0
?
= 
08 . 311
67 . 806
= 0.385 ˜ 0.39 A
4. E = 12 volts
i
2
Rt = i
2
rms
RT
?
2
2
R
E
= 
2
rms
2
R
E
? E
2
= 
2
E
2
0
? E
0
2
= 2E
2
  ? E
0
2
= 2 × 12
2
= 2 × 144 
? E
0
= 144 2 ? = 16.97 ˜ 17 V 
5. P
0
= 80 W (given)
P
rms
=
2
P
0
= 40 W
Energy consumed = P × t = 40 × 100 = 4000 J = 4.0 KJ
6. E = 3 × 10
6
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
Max. rms Voltage = 
2
V
= 
2
300
= 212 V
Alternating Current
39.2
7. i = i
0
e
–ur
2
i = 
?
?
? ?
?
0
/ t 2 2
0
dt e i
1
= 
?
?
? ?
?
0
/ t 2
2
0
dt e
i
= 
? ? ?
?
?
?
?
?
? ?
?
?
0
/ t 2
2
0
e
2
i
= ? ? 1 e
2
i
2
2
0
? ?
?
?
?
?
?
2
i = ?
?
?
?
?
?
? ? 1
e
1
2
i
2
2
0
= 
?
?
?
?
?
?
?
?
?
2
1 e
e
i
2
0
8. C = 10 ?F = 10 × 10
–6
F = 10
–5
F
E = (10 V) Sin ?t
a) ? = 
Xc
E
0
= 
?
?
?
?
?
?
?C
1
E
0
= 
?
?
?
?
?
?
?
?5
10 10
1
10
= 1 × 10
–3
A
b) ? = 100 s
–1
? = 
?
?
?
?
?
?
?C
1
E
0
= 
?
?
?
?
?
?
?
?5
10 100
1
10
= 1 × 10
–2
A = 0.01 A
c) ? = 500 s
–1
? = 
?
?
?
?
?
?
?C
1
E
0
= 
?
?
?
?
?
?
?
?5
10 500
1
10
= 5 × 10
–2
A = 0.05 A
d) ? = 1000 s
–1
???
?
?
?
?
?
?
?C
1
E
0
= 
?
?
?
?
?
?
?
?5
10 1000
1
10
= 1 × 10
–1
A = 0.1 A ?
9. Inductance = 5.0 mH  = 0.005 H
a) ? = 100 s
–1
X
L
= ?L = 100 × 
1000
5
= 0.5 O
i = 
L
0
X
?
= 
5 . 0
10
= 20 A
b) ? = 500 s
–1
X
L
= ?L = 500 × 
1000
5
= 2.5 O
i = 
L
0
X
?
= 
5 . 2
10
= 4 A
c) ? = 1000 s
–1
X
L
= ?L = 1000 × 
1000
5
= 5 O
i = 
L
0
X
?
= 
5
10
= 2 A
10. R = 10 O, L = 0.4 Henry
E = 6.5 V, ? = 
?
30
Hz
Z = 
2
L
2
X R ? = 
2 2
) L 2 ( R ? ? ?
Power = V
rms
?
rms
cos ?
= 6.5 × 
Z
R
Z
5 . 6
? = 
2
2 2
) L 2 ( R
10 5 . 6 5 . 6
?
?
?
?
?
?
? ? ?
? ?
= 
2
2
4 . 0
30
2 10
10 5 . 6 5 . 6
?
?
?
?
?
?
?
?
? ? ?
? ?
= 
576 100
10 5 . 6 5 . 6
?
? ?
= 0.625 = 
8
5
??
Page 3


39.1
CHAPTER – 39
ALTERNATING CURRENT
1. ? = 50 Hz
? = ?
0
Sin Wt
Peak value ? = 
2
0
?
2
0
?
= ?
0
Sin Wt
?
2
1
= Sin Wt = Sin 
4
?
?
4
?
= Wt. or, t = 
400
?
= 
? ? ?
?
2 4
= 
? 8
1
= 
50 8
1
?
= 0.0025 s = 2.5 ms 
2. E
rms
= 220 V
Frequency = 50 Hz
(a) E
rms
= 
2
E
0
? E
0
= E
rms 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
? = 
2
0
?
?
2
0
?
= ?
0
Sin ?t
? ?t = 
4
?
? t = 
?
?
4
= 
f 2 4 ? ?
?
= 
50 8 ?
?
= 
400
1
= 2.5 ms ?
3. P = 60 W V = 220 V = E
R = 
P
v
2
= 
60
220 220 ?
= 806.67
?
0
= 2 E = 1.414 × 220 = 311.08
?
0
= 
R
0
?
= 
08 . 311
67 . 806
= 0.385 ˜ 0.39 A
4. E = 12 volts
i
2
Rt = i
2
rms
RT
?
2
2
R
E
= 
2
rms
2
R
E
? E
2
= 
2
E
2
0
? E
0
2
= 2E
2
  ? E
0
2
= 2 × 12
2
= 2 × 144 
? E
0
= 144 2 ? = 16.97 ˜ 17 V 
5. P
0
= 80 W (given)
P
rms
=
2
P
0
= 40 W
Energy consumed = P × t = 40 × 100 = 4000 J = 4.0 KJ
6. E = 3 × 10
6
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
Max. rms Voltage = 
2
V
= 
2
300
= 212 V
Alternating Current
39.2
7. i = i
0
e
–ur
2
i = 
?
?
? ?
?
0
/ t 2 2
0
dt e i
1
= 
?
?
? ?
?
0
/ t 2
2
0
dt e
i
= 
? ? ?
?
?
?
?
?
? ?
?
?
0
/ t 2
2
0
e
2
i
= ? ? 1 e
2
i
2
2
0
? ?
?
?
?
?
?
2
i = ?
?
?
?
?
?
? ? 1
e
1
2
i
2
2
0
= 
?
?
?
?
?
?
?
?
?
2
1 e
e
i
2
0
8. C = 10 ?F = 10 × 10
–6
F = 10
–5
F
E = (10 V) Sin ?t
a) ? = 
Xc
E
0
= 
?
?
?
?
?
?
?C
1
E
0
= 
?
?
?
?
?
?
?
?5
10 10
1
10
= 1 × 10
–3
A
b) ? = 100 s
–1
? = 
?
?
?
?
?
?
?C
1
E
0
= 
?
?
?
?
?
?
?
?5
10 100
1
10
= 1 × 10
–2
A = 0.01 A
c) ? = 500 s
–1
? = 
?
?
?
?
?
?
?C
1
E
0
= 
?
?
?
?
?
?
?
?5
10 500
1
10
= 5 × 10
–2
A = 0.05 A
d) ? = 1000 s
–1
???
?
?
?
?
?
?
?C
1
E
0
= 
?
?
?
?
?
?
?
?5
10 1000
1
10
= 1 × 10
–1
A = 0.1 A ?
9. Inductance = 5.0 mH  = 0.005 H
a) ? = 100 s
–1
X
L
= ?L = 100 × 
1000
5
= 0.5 O
i = 
L
0
X
?
= 
5 . 0
10
= 20 A
b) ? = 500 s
–1
X
L
= ?L = 500 × 
1000
5
= 2.5 O
i = 
L
0
X
?
= 
5 . 2
10
= 4 A
c) ? = 1000 s
–1
X
L
= ?L = 1000 × 
1000
5
= 5 O
i = 
L
0
X
?
= 
5
10
= 2 A
10. R = 10 O, L = 0.4 Henry
E = 6.5 V, ? = 
?
30
Hz
Z = 
2
L
2
X R ? = 
2 2
) L 2 ( R ? ? ?
Power = V
rms
?
rms
cos ?
= 6.5 × 
Z
R
Z
5 . 6
? = 
2
2 2
) L 2 ( R
10 5 . 6 5 . 6
?
?
?
?
?
?
? ? ?
? ?
= 
2
2
4 . 0
30
2 10
10 5 . 6 5 . 6
?
?
?
?
?
?
?
?
? ? ?
? ?
= 
576 100
10 5 . 6 5 . 6
?
? ?
= 0.625 = 
8
5
??
Alternating Current
39.3
11. H = T
R
V
2
, E
0
= 12 V, ? = 250 ?, R = 100 O
H = 
?
H
0
dH = 
?
?
dt
R
t Sin E
2 2
0
= 
?
? dt t sin
100
144
2
= 1.44 
?
?
?
?
?
?
? ? ?
2
t 2 cos 1
dt
= 
?
?
?
?
?
?
? ?
? ?
? ? 3 3
10
0
10
0
dt t 2 Cos dt
2
44 . 1
= 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
3
10
0
3
2
t 2 Sin
10 72 . 0
= 
?
?
?
?
?
?
?
?
500
1
1000
1
72 . 0 = 72 . 0
1000
) 2 (
?
?
? ?
= 0.0002614 = 2.61 × 10
–4
J
12. R = 300O, C = 25 ?F = 25 × 10
–6
F, ?
0
= 50 V, ? = 50 Hz
X
c
= 
c
1
?
= 
6
10 25 2
50
1
?
? ? ? ?
?
= 
25
10
4
Z = 
2
c
2
X R ? = 
2
4
2
25
10
) 300 (
?
?
?
?
?
?
?
?
? = 
2 2
) 400 ( ) 300 ( ? = 500
(a) Peak current = 
Z
E
0
= 
500
50
= 0.1 A
(b) Average Power dissipitated, = E
rms 
?
rms
Cos ?
= 
Z
R
Z 2
E
2
E
0 0
? ? = 
2
2
0
Z 2
E
= 
500 500 2
300 50 50
? ?
? ?
= 
2
3
= 1.5 ?.
13. Power = 55 W, Voltage = 110 V, Resistance = 
P
V
2
= 
55
110 110 ?
= 220 O
frequency ( ?) = 50 Hz, ?= 2 ?? = 2 ? × 50 = 100 ?
Current in the circuit = 
Z
V
= 
2 2
) L ( R
V
? ?
Voltage drop across the resistor = ir = 
2 2
) L ( R
VR
? ?
= 
2 2
) L 100 ( ) 220 (
220 220
? ?
?
= 110
? 220 × 2 = 
2 2
) L 100 ( ) 220 ( ? ? ? (220)
2
+ (100 ?L)
2
= (440)
2
? 48400 + 10
4
?
2
L
2
= 193600 ? 10
4
?
2
L
2
= 193600 – 48400
? L
2
= 
4 2
10
142500
? ?
= 1.4726 ? L = 1.2135 ˜ 1.2 Hz  ?
14. R = 300 O, C = 20 ?F = 20 × 10
–6
F
L = 1 Henry, E = 50 V V = 
?
50
Hz
(a) ?
0
= 
Z
E
0
, 
Z = 
2
L c
2
) X X ( R ? ? = 
2
2
L 2
C 2
1
) 300 (
?
?
?
?
?
?
?
?
? ? ?
? ?
?
= 
2
6
2
1
50
2
10 20
50
2
1
) 300 (
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
? ?
?
? ?
?
?
= 
2
4
2
100
20
10
) 300 (
?
?
?
?
?
?
?
?
? ? = 500
?
0
= 
Z
E
0
= 
500
50
= 0.1 A 
R
–
110 V ?
220 V ?
L
Page 4


39.1
CHAPTER – 39
ALTERNATING CURRENT
1. ? = 50 Hz
? = ?
0
Sin Wt
Peak value ? = 
2
0
?
2
0
?
= ?
0
Sin Wt
?
2
1
= Sin Wt = Sin 
4
?
?
4
?
= Wt. or, t = 
400
?
= 
? ? ?
?
2 4
= 
? 8
1
= 
50 8
1
?
= 0.0025 s = 2.5 ms 
2. E
rms
= 220 V
Frequency = 50 Hz
(a) E
rms
= 
2
E
0
? E
0
= E
rms 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
? = 
2
0
?
?
2
0
?
= ?
0
Sin ?t
? ?t = 
4
?
? t = 
?
?
4
= 
f 2 4 ? ?
?
= 
50 8 ?
?
= 
400
1
= 2.5 ms ?
3. P = 60 W V = 220 V = E
R = 
P
v
2
= 
60
220 220 ?
= 806.67
?
0
= 2 E = 1.414 × 220 = 311.08
?
0
= 
R
0
?
= 
08 . 311
67 . 806
= 0.385 ˜ 0.39 A
4. E = 12 volts
i
2
Rt = i
2
rms
RT
?
2
2
R
E
= 
2
rms
2
R
E
? E
2
= 
2
E
2
0
? E
0
2
= 2E
2
  ? E
0
2
= 2 × 12
2
= 2 × 144 
? E
0
= 144 2 ? = 16.97 ˜ 17 V 
5. P
0
= 80 W (given)
P
rms
=
2
P
0
= 40 W
Energy consumed = P × t = 40 × 100 = 4000 J = 4.0 KJ
6. E = 3 × 10
6
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
Max. rms Voltage = 
2
V
= 
2
300
= 212 V
Alternating Current
39.2
7. i = i
0
e
–ur
2
i = 
?
?
? ?
?
0
/ t 2 2
0
dt e i
1
= 
?
?
? ?
?
0
/ t 2
2
0
dt e
i
= 
? ? ?
?
?
?
?
?
? ?
?
?
0
/ t 2
2
0
e
2
i
= ? ? 1 e
2
i
2
2
0
? ?
?
?
?
?
?
2
i = ?
?
?
?
?
?
? ? 1
e
1
2
i
2
2
0
= 
?
?
?
?
?
?
?
?
?
2
1 e
e
i
2
0
8. C = 10 ?F = 10 × 10
–6
F = 10
–5
F
E = (10 V) Sin ?t
a) ? = 
Xc
E
0
= 
?
?
?
?
?
?
?C
1
E
0
= 
?
?
?
?
?
?
?
?5
10 10
1
10
= 1 × 10
–3
A
b) ? = 100 s
–1
? = 
?
?
?
?
?
?
?C
1
E
0
= 
?
?
?
?
?
?
?
?5
10 100
1
10
= 1 × 10
–2
A = 0.01 A
c) ? = 500 s
–1
? = 
?
?
?
?
?
?
?C
1
E
0
= 
?
?
?
?
?
?
?
?5
10 500
1
10
= 5 × 10
–2
A = 0.05 A
d) ? = 1000 s
–1
???
?
?
?
?
?
?
?C
1
E
0
= 
?
?
?
?
?
?
?
?5
10 1000
1
10
= 1 × 10
–1
A = 0.1 A ?
9. Inductance = 5.0 mH  = 0.005 H
a) ? = 100 s
–1
X
L
= ?L = 100 × 
1000
5
= 0.5 O
i = 
L
0
X
?
= 
5 . 0
10
= 20 A
b) ? = 500 s
–1
X
L
= ?L = 500 × 
1000
5
= 2.5 O
i = 
L
0
X
?
= 
5 . 2
10
= 4 A
c) ? = 1000 s
–1
X
L
= ?L = 1000 × 
1000
5
= 5 O
i = 
L
0
X
?
= 
5
10
= 2 A
10. R = 10 O, L = 0.4 Henry
E = 6.5 V, ? = 
?
30
Hz
Z = 
2
L
2
X R ? = 
2 2
) L 2 ( R ? ? ?
Power = V
rms
?
rms
cos ?
= 6.5 × 
Z
R
Z
5 . 6
? = 
2
2 2
) L 2 ( R
10 5 . 6 5 . 6
?
?
?
?
?
?
? ? ?
? ?
= 
2
2
4 . 0
30
2 10
10 5 . 6 5 . 6
?
?
?
?
?
?
?
?
? ? ?
? ?
= 
576 100
10 5 . 6 5 . 6
?
? ?
= 0.625 = 
8
5
??
Alternating Current
39.3
11. H = T
R
V
2
, E
0
= 12 V, ? = 250 ?, R = 100 O
H = 
?
H
0
dH = 
?
?
dt
R
t Sin E
2 2
0
= 
?
? dt t sin
100
144
2
= 1.44 
?
?
?
?
?
?
? ? ?
2
t 2 cos 1
dt
= 
?
?
?
?
?
?
? ?
? ?
? ? 3 3
10
0
10
0
dt t 2 Cos dt
2
44 . 1
= 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
3
10
0
3
2
t 2 Sin
10 72 . 0
= 
?
?
?
?
?
?
?
?
500
1
1000
1
72 . 0 = 72 . 0
1000
) 2 (
?
?
? ?
= 0.0002614 = 2.61 × 10
–4
J
12. R = 300O, C = 25 ?F = 25 × 10
–6
F, ?
0
= 50 V, ? = 50 Hz
X
c
= 
c
1
?
= 
6
10 25 2
50
1
?
? ? ? ?
?
= 
25
10
4
Z = 
2
c
2
X R ? = 
2
4
2
25
10
) 300 (
?
?
?
?
?
?
?
?
? = 
2 2
) 400 ( ) 300 ( ? = 500
(a) Peak current = 
Z
E
0
= 
500
50
= 0.1 A
(b) Average Power dissipitated, = E
rms 
?
rms
Cos ?
= 
Z
R
Z 2
E
2
E
0 0
? ? = 
2
2
0
Z 2
E
= 
500 500 2
300 50 50
? ?
? ?
= 
2
3
= 1.5 ?.
13. Power = 55 W, Voltage = 110 V, Resistance = 
P
V
2
= 
55
110 110 ?
= 220 O
frequency ( ?) = 50 Hz, ?= 2 ?? = 2 ? × 50 = 100 ?
Current in the circuit = 
Z
V
= 
2 2
) L ( R
V
? ?
Voltage drop across the resistor = ir = 
2 2
) L ( R
VR
? ?
= 
2 2
) L 100 ( ) 220 (
220 220
? ?
?
= 110
? 220 × 2 = 
2 2
) L 100 ( ) 220 ( ? ? ? (220)
2
+ (100 ?L)
2
= (440)
2
? 48400 + 10
4
?
2
L
2
= 193600 ? 10
4
?
2
L
2
= 193600 – 48400
? L
2
= 
4 2
10
142500
? ?
= 1.4726 ? L = 1.2135 ˜ 1.2 Hz  ?
14. R = 300 O, C = 20 ?F = 20 × 10
–6
F
L = 1 Henry, E = 50 V V = 
?
50
Hz
(a) ?
0
= 
Z
E
0
, 
Z = 
2
L c
2
) X X ( R ? ? = 
2
2
L 2
C 2
1
) 300 (
?
?
?
?
?
?
?
?
? ? ?
? ?
?
= 
2
6
2
1
50
2
10 20
50
2
1
) 300 (
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
? ?
?
? ?
?
?
= 
2
4
2
100
20
10
) 300 (
?
?
?
?
?
?
?
?
? ? = 500
?
0
= 
Z
E
0
= 
500
50
= 0.1 A 
R
–
110 V ?
220 V ?
L
Alternating Current
39.4
(b) Potential across the capacitor = i
0
× X
c
= 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
× X
L
= 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 O
C = 20 ?F = 20 × 10
–6
F
L = 1H, Z = 500 (from 14)
?
0
= 50 V, ?
0
= 
Z
E
0
= 
500
50
= 0.1 A
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 ?
0
2
= (1/2) × 1 × (0.1)
2
= 5 × 10
–3
J = 5 mJ
16. (a)For current to be maximum in a circuit
X
l
= X
c
(Resonant Condition)
? WL = 
WC
1
? W
2
= 
LC
1
= 
6
10 18 2
1
?
? ?
= 
36
10
6
? W = 
6
10
3
? 2 ?? = 
6
10
3
? ? = 
? ? 2 6
1000
= 26.537 Hz ˜ 27 Hz ?
(b) Maximum Current = 
R
E
(in resonance and)
= 
3
10 10
20
?
= 
3
10
2
A = 2 mA
17. E
rms
= 24 V
r = 4 O, ?
rms
= 6 A
R = 
?
E
= 
6
24
= 4 O
Internal Resistance = 4 O
Hence net resistance = 4 + 4 = 8 O
? Current = 
8
12
= 1.5 A 
18. V
1
= 10 × 10
–3
V
R = 1 × 10
3
O
C = 10 × 10
–9
F
(a) X
c
= 
WC
1
= 
C 2
1
? ?
=
9 3
10 10 10 10 2
1
?
? ? ? ? ?
= 
4
10 2
1
?
? ?
= 
? 2
10
4
= 
?
5000
Z = 
2
c
2
X R ? = ? ?
2
2
3
5000
10 1 ?
?
?
?
?
?
?
? ? = 
2
6
5000
10 ?
?
?
?
?
?
?
?
?
0
= 
Z
E
0
= 
Z
V
1
= 
2
6
3
5000
10
10 10
?
?
?
?
?
?
?
?
?
?
10 nF
10 O ?
V 0
V 1
Page 5


39.1
CHAPTER – 39
ALTERNATING CURRENT
1. ? = 50 Hz
? = ?
0
Sin Wt
Peak value ? = 
2
0
?
2
0
?
= ?
0
Sin Wt
?
2
1
= Sin Wt = Sin 
4
?
?
4
?
= Wt. or, t = 
400
?
= 
? ? ?
?
2 4
= 
? 8
1
= 
50 8
1
?
= 0.0025 s = 2.5 ms 
2. E
rms
= 220 V
Frequency = 50 Hz
(a) E
rms
= 
2
E
0
? E
0
= E
rms 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
? = 
2
0
?
?
2
0
?
= ?
0
Sin ?t
? ?t = 
4
?
? t = 
?
?
4
= 
f 2 4 ? ?
?
= 
50 8 ?
?
= 
400
1
= 2.5 ms ?
3. P = 60 W V = 220 V = E
R = 
P
v
2
= 
60
220 220 ?
= 806.67
?
0
= 2 E = 1.414 × 220 = 311.08
?
0
= 
R
0
?
= 
08 . 311
67 . 806
= 0.385 ˜ 0.39 A
4. E = 12 volts
i
2
Rt = i
2
rms
RT
?
2
2
R
E
= 
2
rms
2
R
E
? E
2
= 
2
E
2
0
? E
0
2
= 2E
2
  ? E
0
2
= 2 × 12
2
= 2 × 144 
? E
0
= 144 2 ? = 16.97 ˜ 17 V 
5. P
0
= 80 W (given)
P
rms
=
2
P
0
= 40 W
Energy consumed = P × t = 40 × 100 = 4000 J = 4.0 KJ
6. E = 3 × 10
6
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
Max. rms Voltage = 
2
V
= 
2
300
= 212 V
Alternating Current
39.2
7. i = i
0
e
–ur
2
i = 
?
?
? ?
?
0
/ t 2 2
0
dt e i
1
= 
?
?
? ?
?
0
/ t 2
2
0
dt e
i
= 
? ? ?
?
?
?
?
?
? ?
?
?
0
/ t 2
2
0
e
2
i
= ? ? 1 e
2
i
2
2
0
? ?
?
?
?
?
?
2
i = ?
?
?
?
?
?
? ? 1
e
1
2
i
2
2
0
= 
?
?
?
?
?
?
?
?
?
2
1 e
e
i
2
0
8. C = 10 ?F = 10 × 10
–6
F = 10
–5
F
E = (10 V) Sin ?t
a) ? = 
Xc
E
0
= 
?
?
?
?
?
?
?C
1
E
0
= 
?
?
?
?
?
?
?
?5
10 10
1
10
= 1 × 10
–3
A
b) ? = 100 s
–1
? = 
?
?
?
?
?
?
?C
1
E
0
= 
?
?
?
?
?
?
?
?5
10 100
1
10
= 1 × 10
–2
A = 0.01 A
c) ? = 500 s
–1
? = 
?
?
?
?
?
?
?C
1
E
0
= 
?
?
?
?
?
?
?
?5
10 500
1
10
= 5 × 10
–2
A = 0.05 A
d) ? = 1000 s
–1
???
?
?
?
?
?
?
?C
1
E
0
= 
?
?
?
?
?
?
?
?5
10 1000
1
10
= 1 × 10
–1
A = 0.1 A ?
9. Inductance = 5.0 mH  = 0.005 H
a) ? = 100 s
–1
X
L
= ?L = 100 × 
1000
5
= 0.5 O
i = 
L
0
X
?
= 
5 . 0
10
= 20 A
b) ? = 500 s
–1
X
L
= ?L = 500 × 
1000
5
= 2.5 O
i = 
L
0
X
?
= 
5 . 2
10
= 4 A
c) ? = 1000 s
–1
X
L
= ?L = 1000 × 
1000
5
= 5 O
i = 
L
0
X
?
= 
5
10
= 2 A
10. R = 10 O, L = 0.4 Henry
E = 6.5 V, ? = 
?
30
Hz
Z = 
2
L
2
X R ? = 
2 2
) L 2 ( R ? ? ?
Power = V
rms
?
rms
cos ?
= 6.5 × 
Z
R
Z
5 . 6
? = 
2
2 2
) L 2 ( R
10 5 . 6 5 . 6
?
?
?
?
?
?
? ? ?
? ?
= 
2
2
4 . 0
30
2 10
10 5 . 6 5 . 6
?
?
?
?
?
?
?
?
? ? ?
? ?
= 
576 100
10 5 . 6 5 . 6
?
? ?
= 0.625 = 
8
5
??
Alternating Current
39.3
11. H = T
R
V
2
, E
0
= 12 V, ? = 250 ?, R = 100 O
H = 
?
H
0
dH = 
?
?
dt
R
t Sin E
2 2
0
= 
?
? dt t sin
100
144
2
= 1.44 
?
?
?
?
?
?
? ? ?
2
t 2 cos 1
dt
= 
?
?
?
?
?
?
? ?
? ?
? ? 3 3
10
0
10
0
dt t 2 Cos dt
2
44 . 1
= 
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
3
10
0
3
2
t 2 Sin
10 72 . 0
= 
?
?
?
?
?
?
?
?
500
1
1000
1
72 . 0 = 72 . 0
1000
) 2 (
?
?
? ?
= 0.0002614 = 2.61 × 10
–4
J
12. R = 300O, C = 25 ?F = 25 × 10
–6
F, ?
0
= 50 V, ? = 50 Hz
X
c
= 
c
1
?
= 
6
10 25 2
50
1
?
? ? ? ?
?
= 
25
10
4
Z = 
2
c
2
X R ? = 
2
4
2
25
10
) 300 (
?
?
?
?
?
?
?
?
? = 
2 2
) 400 ( ) 300 ( ? = 500
(a) Peak current = 
Z
E
0
= 
500
50
= 0.1 A
(b) Average Power dissipitated, = E
rms 
?
rms
Cos ?
= 
Z
R
Z 2
E
2
E
0 0
? ? = 
2
2
0
Z 2
E
= 
500 500 2
300 50 50
? ?
? ?
= 
2
3
= 1.5 ?.
13. Power = 55 W, Voltage = 110 V, Resistance = 
P
V
2
= 
55
110 110 ?
= 220 O
frequency ( ?) = 50 Hz, ?= 2 ?? = 2 ? × 50 = 100 ?
Current in the circuit = 
Z
V
= 
2 2
) L ( R
V
? ?
Voltage drop across the resistor = ir = 
2 2
) L ( R
VR
? ?
= 
2 2
) L 100 ( ) 220 (
220 220
? ?
?
= 110
? 220 × 2 = 
2 2
) L 100 ( ) 220 ( ? ? ? (220)
2
+ (100 ?L)
2
= (440)
2
? 48400 + 10
4
?
2
L
2
= 193600 ? 10
4
?
2
L
2
= 193600 – 48400
? L
2
= 
4 2
10
142500
? ?
= 1.4726 ? L = 1.2135 ˜ 1.2 Hz  ?
14. R = 300 O, C = 20 ?F = 20 × 10
–6
F
L = 1 Henry, E = 50 V V = 
?
50
Hz
(a) ?
0
= 
Z
E
0
, 
Z = 
2
L c
2
) X X ( R ? ? = 
2
2
L 2
C 2
1
) 300 (
?
?
?
?
?
?
?
?
? ? ?
? ?
?
= 
2
6
2
1
50
2
10 20
50
2
1
) 300 (
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
? ?
?
? ?
?
?
= 
2
4
2
100
20
10
) 300 (
?
?
?
?
?
?
?
?
? ? = 500
?
0
= 
Z
E
0
= 
500
50
= 0.1 A 
R
–
110 V ?
220 V ?
L
Alternating Current
39.4
(b) Potential across the capacitor = i
0
× X
c
= 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
× X
L
= 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 O
C = 20 ?F = 20 × 10
–6
F
L = 1H, Z = 500 (from 14)
?
0
= 50 V, ?
0
= 
Z
E
0
= 
500
50
= 0.1 A
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 ?
0
2
= (1/2) × 1 × (0.1)
2
= 5 × 10
–3
J = 5 mJ
16. (a)For current to be maximum in a circuit
X
l
= X
c
(Resonant Condition)
? WL = 
WC
1
? W
2
= 
LC
1
= 
6
10 18 2
1
?
? ?
= 
36
10
6
? W = 
6
10
3
? 2 ?? = 
6
10
3
? ? = 
? ? 2 6
1000
= 26.537 Hz ˜ 27 Hz ?
(b) Maximum Current = 
R
E
(in resonance and)
= 
3
10 10
20
?
= 
3
10
2
A = 2 mA
17. E
rms
= 24 V
r = 4 O, ?
rms
= 6 A
R = 
?
E
= 
6
24
= 4 O
Internal Resistance = 4 O
Hence net resistance = 4 + 4 = 8 O
? Current = 
8
12
= 1.5 A 
18. V
1
= 10 × 10
–3
V
R = 1 × 10
3
O
C = 10 × 10
–9
F
(a) X
c
= 
WC
1
= 
C 2
1
? ?
=
9 3
10 10 10 10 2
1
?
? ? ? ? ?
= 
4
10 2
1
?
? ?
= 
? 2
10
4
= 
?
5000
Z = 
2
c
2
X R ? = ? ?
2
2
3
5000
10 1 ?
?
?
?
?
?
?
? ? = 
2
6
5000
10 ?
?
?
?
?
?
?
?
?
0
= 
Z
E
0
= 
Z
V
1
= 
2
6
3
5000
10
10 10
?
?
?
?
?
?
?
?
?
?
10 nF
10 O ?
V 0
V 1
Alternating Current
39.5
(b) X
c
= 
WC
1
= 
C 2
1
? ?
= 
9 5
10 10 10 2
1
?
? ? ? ?
= 
3
10 2
1
?
? ?
= 
? 2
10
3
= 
?
500
Z=  
2
c
2
X R ? = ? ?
2
2
3
500
10 ?
?
?
?
?
?
?
? = 
2
6
500
10 ?
?
?
?
?
?
?
?
?
0
= 
Z
E
0
= 
Z
V
1
= 
2
6
3
500
10
10 10
?
?
?
?
?
?
?
?
?
?
V
0
= ?
0
X
c
= 
?
?
?
?
?
?
?
?
?
?
?
?
500
500
10
10 10
2
6
3
= 1.6124 V ˜ 1.6 mV
(c) ? = 1 MHz = 10
6
Hz
X
c
= 
WC
1
= 
C 2
1
? ?
= 
9 6
10 10 10 2
1
?
? ? ? ?
= 
2
10 2
1
?
? ?
= 
? 2
10
2
= 
?
50
Z=  
2
c
2
X R ? = ? ?
2
2
3
50
10 ?
?
?
?
?
?
?
? = 
2
6
50
10 ?
?
?
?
?
?
?
?
?
0
= 
Z
E
0
= 
Z
V
1
= 
2
6
3
50
10
10 10
?
?
?
?
?
?
?
?
?
?
V
0
= ?
0
X
c
= 
?
?
?
?
?
?
?
?
?
?
?
?
50
50
10
10 10
2
6
3
˜ 0.16 mV
(d) ? = 10 MHz = 10
7
Hz
X
c
= 
WC
1
= 
C 2
1
? ?
= 
9 7
10 10 10 2
1
?
? ? ? ?
= 
1
10 2
1
?
? ?
= 
? 2
10
= 
?
5
Z=  
2
c
2
X R ? = ? ?
2
2
3
5
10 ?
?
?
?
?
?
?
? = 
2
6
5
10 ?
?
?
?
?
?
?
?
?
0
= 
Z
E
0
= 
Z
V
1
= 
2
6
3
5
10
10 10
?
?
?
?
?
?
?
?
?
?
V
0
= ?
0
X
c
= 
?
?
?
?
?
?
?
?
?
?
?
?
5
5
10
10 10
2
6
3
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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.
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FAQs on HC Verma Solutions: Chapter 39 - Alternating Current - Physics Class 11 - NEET

1. How is alternating current (AC) different from direct current (DC)?
Ans. Alternating current (AC) and direct current (DC) are two different types of electric current. The main difference between AC and DC is the direction of flow of electrons. In AC, the flow of electrons periodically changes direction, while in DC, the flow of electrons is unidirectional. AC is commonly used in power grids and household appliances, while DC is commonly used in batteries and electronic devices.
2. What is the frequency of alternating current?
Ans. The frequency of alternating current is the number of complete cycles (oscillations) it completes in one second. It is measured in Hertz (Hz). In most countries, the frequency of the AC mains supply is 50 Hz or 60 Hz. This means that the current changes its direction 50 or 60 times per second, depending on the country.
3. How is the RMS value of alternating current calculated?
Ans. The RMS (Root Mean Square) value of an alternating current is the equivalent value of a direct current that produces the same amount of heat in a given resistor. It is calculated using the formula RMS = (Imax / √2), where Imax is the maximum value of the current. For example, if the maximum value of the current is 10 A, then the RMS value would be (10 / √2) ≈ 7.07 A.
4. What is the phase difference in alternating current?
Ans. The phase difference in alternating current refers to the difference in timing or phase between two alternating quantities, such as voltage or current. It is measured in degrees or radians. In AC circuits, the phase difference between voltage and current determines whether the load is resistive, inductive, or capacitive. For example, in a purely resistive circuit, the phase difference between voltage and current is zero degrees.
5. What is the significance of reactance in alternating current?
Ans. Reactance is the opposition offered by a circuit element (such as an inductor or a capacitor) to the flow of alternating current. It is measured in ohms. Reactance plays a crucial role in AC circuits as it determines the overall impedance of the circuit and affects the flow of current. In combination with resistance, reactance forms the complex impedance, which determines the behavior of the circuit in terms of phase shifts and power consumption.
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Document Description: HC Verma Solutions: Chapter 39 - Alternating Current for NEET 2024 is part of Physics Class 11 preparation. The notes and questions for HC Verma Solutions: Chapter 39 - Alternating Current have been prepared according to the NEET exam syllabus. Information about HC Verma Solutions: Chapter 39 - Alternating Current covers topics like and HC Verma Solutions: Chapter 39 - Alternating Current Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for HC Verma Solutions: Chapter 39 - Alternating Current.
Introduction of HC Verma Solutions: Chapter 39 - Alternating Current in English is available as part of our Physics Class 11 for NEET & HC Verma Solutions: Chapter 39 - Alternating Current in Hindi for Physics Class 11 course. Download more important topics related with notes, lectures and mock test series for NEET Exam by signing up for free. NEET: HC Verma Solutions: Chapter 39 - Alternating Current | Physics Class 11 - NEET
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