Chapter 39 : Alternating Current - HC Verma Solution, Physics Class 11 Notes | EduRev

HC Verma and Irodov Solutions

JEE : Chapter 39 : Alternating Current - HC Verma Solution, Physics Class 11 Notes | EduRev

 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
˜ 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.
? ? ? ? ?
Sec ? P 1
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