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 Page 1


32.1
ELECTRIC CURRENT IN CONDUCTORS
CHAPTER - 32
1. Q(t) = At
2
+ Bt + c
a) At
2
= Q
? A = 
1 1
2 2
Q A'T'
A T
t T
?
?
? ?
b) Bt = Q
? B = 
Q A 'T'
A
t T
? ?
c) C = [Q]
? C = A ?T ?
d) Current t = 
? ?
2
dQ d
At Bt C
dt dt
? ? ?
= 2At + B = 2 ? 5 ? 5 + 3 = 53 A.
2. No. of electrons per second = 2 ? 10
16
electrons / sec.
Charge passing per second = 2 ? 10
16
? 1.6 ? 10
–9
coulomb
sec
= 3.2 ? 10
–9
Coulomb/sec
Current = 3.2 ? 10
–3
A.
3. i ? = 2 ?A, t = 5 min = 5 ? 60 sec.
q = i t = 2 ? 10
–6
? 5 ? 60
= 10 ? 60 ? 10
–6
c = 6 ? 10
–4
c
4. i = i
0
+ ?t, t = 10 sec, i
0
= 10 A, ? = 4 A/sec.
q = 
t t t t
0 0
0 0 0 0
idt (i t)dt i dt tdt ? ? ? ? ? ?
? ? ? ?
= i
0
t + 
2
t 10 10
10 10 4
2 2
?
? ? ? ? ?
= 100 + 200 = 300 C.
5. i = 1 A, A = 1 mm
2
= 1 ? 10
–6
m
2
f ? cu = 9000 kg/m
3
Molecular mass has N
0
atoms 
= m Kg has (N
0
/M ? m) atoms = 
0
3
N AI9000
63.5 10
?
?
No.of atoms = No.of electrons
n = 
0 0
N Af N f No.of electrons
Unit volume mAI M
? ?
= 
23
3
6 10 9000
63.5 10
?
? ?
?
i = V
d
n A e.
? V
d
= 
23
6 19
3
i 1
nAe 6 10 9000
10 1.6 10
63.5 10
? ?
?
?
? ?
? ? ?
?
= 
3
23 6 19
63.5 10
6 10 9000 10 1.6 10
?
? ?
?
? ? ? ? ?
= 
3
26 19 6
63.5 10
6 9 1.6 10 10 10
?
? ?
?
? ? ? ? ?
Page 2


32.1
ELECTRIC CURRENT IN CONDUCTORS
CHAPTER - 32
1. Q(t) = At
2
+ Bt + c
a) At
2
= Q
? A = 
1 1
2 2
Q A'T'
A T
t T
?
?
? ?
b) Bt = Q
? B = 
Q A 'T'
A
t T
? ?
c) C = [Q]
? C = A ?T ?
d) Current t = 
? ?
2
dQ d
At Bt C
dt dt
? ? ?
= 2At + B = 2 ? 5 ? 5 + 3 = 53 A.
2. No. of electrons per second = 2 ? 10
16
electrons / sec.
Charge passing per second = 2 ? 10
16
? 1.6 ? 10
–9
coulomb
sec
= 3.2 ? 10
–9
Coulomb/sec
Current = 3.2 ? 10
–3
A.
3. i ? = 2 ?A, t = 5 min = 5 ? 60 sec.
q = i t = 2 ? 10
–6
? 5 ? 60
= 10 ? 60 ? 10
–6
c = 6 ? 10
–4
c
4. i = i
0
+ ?t, t = 10 sec, i
0
= 10 A, ? = 4 A/sec.
q = 
t t t t
0 0
0 0 0 0
idt (i t)dt i dt tdt ? ? ? ? ? ?
? ? ? ?
= i
0
t + 
2
t 10 10
10 10 4
2 2
?
? ? ? ? ?
= 100 + 200 = 300 C.
5. i = 1 A, A = 1 mm
2
= 1 ? 10
–6
m
2
f ? cu = 9000 kg/m
3
Molecular mass has N
0
atoms 
= m Kg has (N
0
/M ? m) atoms = 
0
3
N AI9000
63.5 10
?
?
No.of atoms = No.of electrons
n = 
0 0
N Af N f No.of electrons
Unit volume mAI M
? ?
= 
23
3
6 10 9000
63.5 10
?
? ?
?
i = V
d
n A e.
? V
d
= 
23
6 19
3
i 1
nAe 6 10 9000
10 1.6 10
63.5 10
? ?
?
?
? ?
? ? ?
?
= 
3
23 6 19
63.5 10
6 10 9000 10 1.6 10
?
? ?
?
? ? ? ? ?
= 
3
26 19 6
63.5 10
6 9 1.6 10 10 10
?
? ?
?
? ? ? ? ?
Electric Current in Conductors
32.2
= 
3 3
63.5 10 63.5 10
6 9 1.6 10 6 9 16
? ?
? ?
?
? ? ? ? ?
= 0.074 ? 10
–3
m/s = 0.074 mm/s.
6. ? = 1 m, r = 0.1 mm = 0.1 ? 10
–3
m
R = 100 ?, f = ?
? R = f ? / a
? f = 
6
Ra 100 3.14 0.1 0.1 10
1
?
? ? ? ?
?
?
= 3.14 ? 10
–6
= ? ? 10
–6
?-m.
?
7. ? ? = 2 ?
volume of the wire remains constant.
A ? = A ? ? ?
? A ? = A ? ? 2 ?
? A ? = A/2
f = Specific resistance
R = 
f
A
?
; R ? = 
f '
A '
?
100 ? = 
f2 4f
A / 2 A
?
? ?
= 4R
? 4 ? 100 ? = 400 ? ?
8. ? = 4 m, A = 1 mm
2
= 1 ? 10
–6
m
2
I = 2 A, n/V = 10
29
, t = ?
i = n A V
d
e
? e = 10
29
? 1 ? 10
–6
? V
d
? 1.6 ? 10
–19
?
d
29 6 19
2
V
10 10 1.6 10
? ?
?
? ? ?
= 
4
1 1
8000 0.8 10
?
?
t = 
d
4
4 8000
V 1/ 8000
? ? ?
?
= 32000 = 3.2 ? 10
4 
sec.
9. f
cu
= 1.7 ? 10
–8
?-m
A = 0.01 mm
2
= 0.01 ? 10
–6
m
2
R = 1 K ? = 10
3
?
R = 
f
a
?
? 10
3
= 
8
6
1.7 10
10
?
?
? ? ?
?
3
10
1.7
? ? = 0.58 ? 10
3
m = 0.6 km. ?
10. dR, due to the small strip dx at a distanc x d = R = 
2
fdx
y ?
…(1)
tan ? = 
y a b a
x L
? ?
?
?
y a b a
x L
? ?
?
? L(y – a) = x(b – a)
x ?
b–a 
dx 
a ?
y ?
? ? Y–a 
b ?
Page 3


32.1
ELECTRIC CURRENT IN CONDUCTORS
CHAPTER - 32
1. Q(t) = At
2
+ Bt + c
a) At
2
= Q
? A = 
1 1
2 2
Q A'T'
A T
t T
?
?
? ?
b) Bt = Q
? B = 
Q A 'T'
A
t T
? ?
c) C = [Q]
? C = A ?T ?
d) Current t = 
? ?
2
dQ d
At Bt C
dt dt
? ? ?
= 2At + B = 2 ? 5 ? 5 + 3 = 53 A.
2. No. of electrons per second = 2 ? 10
16
electrons / sec.
Charge passing per second = 2 ? 10
16
? 1.6 ? 10
–9
coulomb
sec
= 3.2 ? 10
–9
Coulomb/sec
Current = 3.2 ? 10
–3
A.
3. i ? = 2 ?A, t = 5 min = 5 ? 60 sec.
q = i t = 2 ? 10
–6
? 5 ? 60
= 10 ? 60 ? 10
–6
c = 6 ? 10
–4
c
4. i = i
0
+ ?t, t = 10 sec, i
0
= 10 A, ? = 4 A/sec.
q = 
t t t t
0 0
0 0 0 0
idt (i t)dt i dt tdt ? ? ? ? ? ?
? ? ? ?
= i
0
t + 
2
t 10 10
10 10 4
2 2
?
? ? ? ? ?
= 100 + 200 = 300 C.
5. i = 1 A, A = 1 mm
2
= 1 ? 10
–6
m
2
f ? cu = 9000 kg/m
3
Molecular mass has N
0
atoms 
= m Kg has (N
0
/M ? m) atoms = 
0
3
N AI9000
63.5 10
?
?
No.of atoms = No.of electrons
n = 
0 0
N Af N f No.of electrons
Unit volume mAI M
? ?
= 
23
3
6 10 9000
63.5 10
?
? ?
?
i = V
d
n A e.
? V
d
= 
23
6 19
3
i 1
nAe 6 10 9000
10 1.6 10
63.5 10
? ?
?
?
? ?
? ? ?
?
= 
3
23 6 19
63.5 10
6 10 9000 10 1.6 10
?
? ?
?
? ? ? ? ?
= 
3
26 19 6
63.5 10
6 9 1.6 10 10 10
?
? ?
?
? ? ? ? ?
Electric Current in Conductors
32.2
= 
3 3
63.5 10 63.5 10
6 9 1.6 10 6 9 16
? ?
? ?
?
? ? ? ? ?
= 0.074 ? 10
–3
m/s = 0.074 mm/s.
6. ? = 1 m, r = 0.1 mm = 0.1 ? 10
–3
m
R = 100 ?, f = ?
? R = f ? / a
? f = 
6
Ra 100 3.14 0.1 0.1 10
1
?
? ? ? ?
?
?
= 3.14 ? 10
–6
= ? ? 10
–6
?-m.
?
7. ? ? = 2 ?
volume of the wire remains constant.
A ? = A ? ? ?
? A ? = A ? ? 2 ?
? A ? = A/2
f = Specific resistance
R = 
f
A
?
; R ? = 
f '
A '
?
100 ? = 
f2 4f
A / 2 A
?
? ?
= 4R
? 4 ? 100 ? = 400 ? ?
8. ? = 4 m, A = 1 mm
2
= 1 ? 10
–6
m
2
I = 2 A, n/V = 10
29
, t = ?
i = n A V
d
e
? e = 10
29
? 1 ? 10
–6
? V
d
? 1.6 ? 10
–19
?
d
29 6 19
2
V
10 10 1.6 10
? ?
?
? ? ?
= 
4
1 1
8000 0.8 10
?
?
t = 
d
4
4 8000
V 1/ 8000
? ? ?
?
= 32000 = 3.2 ? 10
4 
sec.
9. f
cu
= 1.7 ? 10
–8
?-m
A = 0.01 mm
2
= 0.01 ? 10
–6
m
2
R = 1 K ? = 10
3
?
R = 
f
a
?
? 10
3
= 
8
6
1.7 10
10
?
?
? ? ?
?
3
10
1.7
? ? = 0.58 ? 10
3
m = 0.6 km. ?
10. dR, due to the small strip dx at a distanc x d = R = 
2
fdx
y ?
…(1)
tan ? = 
y a b a
x L
? ?
?
?
y a b a
x L
? ?
?
? L(y – a) = x(b – a)
x ?
b–a 
dx 
a ?
y ?
? ? Y–a 
b ?
Electric Current in Conductors
32.3
? Ly – La = xb – xa 
?
dy
L 0 b a
dx
? ? ? (diff. w.r.t. x)
?
dy
L b a
dx
? ?
? dx = 
Ldy
b a ?
…(2)
Putting the value of dx in equation (1)
dR = 
2
fLdy
y (b a) ? ?
? dR = 
2
fI dy
(b a) y ? ?
?
R b
2
0 a
fI dy
dR
(b a) y
?
? ?
? ?
? R = 
fI (b a) fl
(b a) ab ab
?
?
? ? ?
.
11. r = 0.1 mm = 10
–4
m
R = 1 K ? = 10
3
?, V = 20 V
a) No.of electrons transferred 
i = 
3
V 20
R 10
? = 20 ? 10
–3
= 2 ? 10
–2
A
q = i t = 2 ? 10
–2
? 1 = 2 ? 10
–2
C.
No. of electrons transferred = 
2 17
19
2 10 2 10
1.6 1.6 10
? ?
?
? ?
?
?
= 1.25 ? 10
17
.
b) Current density of wire
= 
2
6
8
i 2 10 2
10
A 3.14 10
?
?
?
?
? ? ?
? ?
= 0.6369 ? 10
+6
= 6.37 ? 10
5
A/m
2
.
12. A = 2 ? 10
–6
m
2
, I = 1 A
f = 1.7 ? 10
–8
?-m
E = ?
R = 
8
6
f 1.7 10
A 2 10
?
?
? ?
?
?
? ?
V = IR = 
8
6
1 1.7 10
2 10
?
?
? ? ?
?
?
E = 
8
2
6
dV V 1.7 10 1.7
10 V /m
dL I 2 2 10
?
?
?
? ?
? ? ? ?
?
?
?
= 8.5 mV/m. ?
13. I = 2 m, R = 5 ?, i = 10 A, E = ?
V = iR = 10 ? 5 = 50 V
E = 
V 50
I 2
? = 25 V/m.
14. R ?
Fe
= R
Fe
(1 + ?
Fe
??), R ?
Cu
= R
Cu
(1 + ?
Cu
???)
R ?
Fe
= R ?
Cu
? R
Fe
(1 + ?
Fe
??), = R
Cu
(1 + ?
Cu
???)
Page 4


32.1
ELECTRIC CURRENT IN CONDUCTORS
CHAPTER - 32
1. Q(t) = At
2
+ Bt + c
a) At
2
= Q
? A = 
1 1
2 2
Q A'T'
A T
t T
?
?
? ?
b) Bt = Q
? B = 
Q A 'T'
A
t T
? ?
c) C = [Q]
? C = A ?T ?
d) Current t = 
? ?
2
dQ d
At Bt C
dt dt
? ? ?
= 2At + B = 2 ? 5 ? 5 + 3 = 53 A.
2. No. of electrons per second = 2 ? 10
16
electrons / sec.
Charge passing per second = 2 ? 10
16
? 1.6 ? 10
–9
coulomb
sec
= 3.2 ? 10
–9
Coulomb/sec
Current = 3.2 ? 10
–3
A.
3. i ? = 2 ?A, t = 5 min = 5 ? 60 sec.
q = i t = 2 ? 10
–6
? 5 ? 60
= 10 ? 60 ? 10
–6
c = 6 ? 10
–4
c
4. i = i
0
+ ?t, t = 10 sec, i
0
= 10 A, ? = 4 A/sec.
q = 
t t t t
0 0
0 0 0 0
idt (i t)dt i dt tdt ? ? ? ? ? ?
? ? ? ?
= i
0
t + 
2
t 10 10
10 10 4
2 2
?
? ? ? ? ?
= 100 + 200 = 300 C.
5. i = 1 A, A = 1 mm
2
= 1 ? 10
–6
m
2
f ? cu = 9000 kg/m
3
Molecular mass has N
0
atoms 
= m Kg has (N
0
/M ? m) atoms = 
0
3
N AI9000
63.5 10
?
?
No.of atoms = No.of electrons
n = 
0 0
N Af N f No.of electrons
Unit volume mAI M
? ?
= 
23
3
6 10 9000
63.5 10
?
? ?
?
i = V
d
n A e.
? V
d
= 
23
6 19
3
i 1
nAe 6 10 9000
10 1.6 10
63.5 10
? ?
?
?
? ?
? ? ?
?
= 
3
23 6 19
63.5 10
6 10 9000 10 1.6 10
?
? ?
?
? ? ? ? ?
= 
3
26 19 6
63.5 10
6 9 1.6 10 10 10
?
? ?
?
? ? ? ? ?
Electric Current in Conductors
32.2
= 
3 3
63.5 10 63.5 10
6 9 1.6 10 6 9 16
? ?
? ?
?
? ? ? ? ?
= 0.074 ? 10
–3
m/s = 0.074 mm/s.
6. ? = 1 m, r = 0.1 mm = 0.1 ? 10
–3
m
R = 100 ?, f = ?
? R = f ? / a
? f = 
6
Ra 100 3.14 0.1 0.1 10
1
?
? ? ? ?
?
?
= 3.14 ? 10
–6
= ? ? 10
–6
?-m.
?
7. ? ? = 2 ?
volume of the wire remains constant.
A ? = A ? ? ?
? A ? = A ? ? 2 ?
? A ? = A/2
f = Specific resistance
R = 
f
A
?
; R ? = 
f '
A '
?
100 ? = 
f2 4f
A / 2 A
?
? ?
= 4R
? 4 ? 100 ? = 400 ? ?
8. ? = 4 m, A = 1 mm
2
= 1 ? 10
–6
m
2
I = 2 A, n/V = 10
29
, t = ?
i = n A V
d
e
? e = 10
29
? 1 ? 10
–6
? V
d
? 1.6 ? 10
–19
?
d
29 6 19
2
V
10 10 1.6 10
? ?
?
? ? ?
= 
4
1 1
8000 0.8 10
?
?
t = 
d
4
4 8000
V 1/ 8000
? ? ?
?
= 32000 = 3.2 ? 10
4 
sec.
9. f
cu
= 1.7 ? 10
–8
?-m
A = 0.01 mm
2
= 0.01 ? 10
–6
m
2
R = 1 K ? = 10
3
?
R = 
f
a
?
? 10
3
= 
8
6
1.7 10
10
?
?
? ? ?
?
3
10
1.7
? ? = 0.58 ? 10
3
m = 0.6 km. ?
10. dR, due to the small strip dx at a distanc x d = R = 
2
fdx
y ?
…(1)
tan ? = 
y a b a
x L
? ?
?
?
y a b a
x L
? ?
?
? L(y – a) = x(b – a)
x ?
b–a 
dx 
a ?
y ?
? ? Y–a 
b ?
Electric Current in Conductors
32.3
? Ly – La = xb – xa 
?
dy
L 0 b a
dx
? ? ? (diff. w.r.t. x)
?
dy
L b a
dx
? ?
? dx = 
Ldy
b a ?
…(2)
Putting the value of dx in equation (1)
dR = 
2
fLdy
y (b a) ? ?
? dR = 
2
fI dy
(b a) y ? ?
?
R b
2
0 a
fI dy
dR
(b a) y
?
? ?
? ?
? R = 
fI (b a) fl
(b a) ab ab
?
?
? ? ?
.
11. r = 0.1 mm = 10
–4
m
R = 1 K ? = 10
3
?, V = 20 V
a) No.of electrons transferred 
i = 
3
V 20
R 10
? = 20 ? 10
–3
= 2 ? 10
–2
A
q = i t = 2 ? 10
–2
? 1 = 2 ? 10
–2
C.
No. of electrons transferred = 
2 17
19
2 10 2 10
1.6 1.6 10
? ?
?
? ?
?
?
= 1.25 ? 10
17
.
b) Current density of wire
= 
2
6
8
i 2 10 2
10
A 3.14 10
?
?
?
?
? ? ?
? ?
= 0.6369 ? 10
+6
= 6.37 ? 10
5
A/m
2
.
12. A = 2 ? 10
–6
m
2
, I = 1 A
f = 1.7 ? 10
–8
?-m
E = ?
R = 
8
6
f 1.7 10
A 2 10
?
?
? ?
?
?
? ?
V = IR = 
8
6
1 1.7 10
2 10
?
?
? ? ?
?
?
E = 
8
2
6
dV V 1.7 10 1.7
10 V /m
dL I 2 2 10
?
?
?
? ?
? ? ? ?
?
?
?
= 8.5 mV/m. ?
13. I = 2 m, R = 5 ?, i = 10 A, E = ?
V = iR = 10 ? 5 = 50 V
E = 
V 50
I 2
? = 25 V/m.
14. R ?
Fe
= R
Fe
(1 + ?
Fe
??), R ?
Cu
= R
Cu
(1 + ?
Cu
???)
R ?
Fe
= R ?
Cu
? R
Fe
(1 + ?
Fe
??), = R
Cu
(1 + ?
Cu
???)
Electric Current in Conductors
32.4
? 3.9 [ 1 + 5 ? 10
–3
(20 – ?)] = 4.1 [1 + 4 x 10
–3
(20 – ?)]
? 3.9 + 3.9 ? 5 ? 10
–3
(20 – ?) = 4.1 + 4.1 ? 4 ? 10
–3
(20 – ?)
?? 4.1 ? 4 ? 10
–3
(20 – ?) – 3.9 ? 5 ? 10
–3
(20 – ?) = 3.9 – 4.1
? 16.4(20 – ?) – 19.5(20 – ?) = 0.2 ? 10
3
? (20 – ?) (–3.1) = 0.2 ? 10
3
? ? – 20 = 200 ?
? ? = 220°C. ?
15. Let the voltmeter reading when, the voltage is 0 be X.
1 1
2 2
I R V
I R V
?
?
1.75 14.4 V 0.35 14.4 V
2.75 22.4 V 0.55 22.4 V
? ?
? ? ?
? ?
?
0.07 14.4 V 7 14.4 V
0.11 22.4 V 11 22.4 V
? ?
? ? ?
? ?
? 7(22.4 – V) = 11(14.4 – V) ? 156.8 – 7V = 158.4 – 11V
? (7 – 11)V = 156.8 – 158.4 ? –4V = –1.6
? V = 0.4 V.
16. a) When switch is open, no current passes through the ammeter. In the upper part of 
the circuit the Voltmenter has ? resistance. Thus current in it is 0.
? Voltmeter read the emf. (There is not Pot. Drop across the resistor).
b) When switch is closed current passes through the circuit and if its value of i.
The voltmeter reads
? – ir = 1.45
? 1.52 – ir = 1.45
? ir = 0.07
? 1 r = 0.07 ? r = 0.07 ?. ?
17. E = 6 V, r = 1 ?, V = 5.8 V, R = ?
I = 
E 6
R r R 1
?
? ?
, V = E – Ir
? 5.8 = 
6
6 1
R 1
? ?
?
?
6
R 1 ?
= 0.2
? R + 1 = 30 ? R = 29 ?. ?
18. V = ? + ir
? 7.2 = 6 + 2 ? r
? 1.2 = 2r ? r = 0.6 ?. ?
19. a) net emf while charging
9 – 6 = 3V
Current = 3/10 = 0.3 A
b) When completely charged.
Internal resistance ‘r’ = 1 ?
Current = 3/1 = 3 A ?
20. a) 0.1i
1
+ 1 i
1
– 6 + 1i
1
– 6 = 0
? 0.1 i
1
+ 1i
1
+ 1i
1
= 12
? i
1
= 
12
2.1
ABCDA
? 0.1i
2
+ 1i – 6 = 0
? 0.1i
2
+ 1i
?? ?
V 
A 
r ?
r ?
2 ?
6 ?
1 ?
0.1 
6 ?
1 ?
i 1 ?
6 ?
Page 5


32.1
ELECTRIC CURRENT IN CONDUCTORS
CHAPTER - 32
1. Q(t) = At
2
+ Bt + c
a) At
2
= Q
? A = 
1 1
2 2
Q A'T'
A T
t T
?
?
? ?
b) Bt = Q
? B = 
Q A 'T'
A
t T
? ?
c) C = [Q]
? C = A ?T ?
d) Current t = 
? ?
2
dQ d
At Bt C
dt dt
? ? ?
= 2At + B = 2 ? 5 ? 5 + 3 = 53 A.
2. No. of electrons per second = 2 ? 10
16
electrons / sec.
Charge passing per second = 2 ? 10
16
? 1.6 ? 10
–9
coulomb
sec
= 3.2 ? 10
–9
Coulomb/sec
Current = 3.2 ? 10
–3
A.
3. i ? = 2 ?A, t = 5 min = 5 ? 60 sec.
q = i t = 2 ? 10
–6
? 5 ? 60
= 10 ? 60 ? 10
–6
c = 6 ? 10
–4
c
4. i = i
0
+ ?t, t = 10 sec, i
0
= 10 A, ? = 4 A/sec.
q = 
t t t t
0 0
0 0 0 0
idt (i t)dt i dt tdt ? ? ? ? ? ?
? ? ? ?
= i
0
t + 
2
t 10 10
10 10 4
2 2
?
? ? ? ? ?
= 100 + 200 = 300 C.
5. i = 1 A, A = 1 mm
2
= 1 ? 10
–6
m
2
f ? cu = 9000 kg/m
3
Molecular mass has N
0
atoms 
= m Kg has (N
0
/M ? m) atoms = 
0
3
N AI9000
63.5 10
?
?
No.of atoms = No.of electrons
n = 
0 0
N Af N f No.of electrons
Unit volume mAI M
? ?
= 
23
3
6 10 9000
63.5 10
?
? ?
?
i = V
d
n A e.
? V
d
= 
23
6 19
3
i 1
nAe 6 10 9000
10 1.6 10
63.5 10
? ?
?
?
? ?
? ? ?
?
= 
3
23 6 19
63.5 10
6 10 9000 10 1.6 10
?
? ?
?
? ? ? ? ?
= 
3
26 19 6
63.5 10
6 9 1.6 10 10 10
?
? ?
?
? ? ? ? ?
Electric Current in Conductors
32.2
= 
3 3
63.5 10 63.5 10
6 9 1.6 10 6 9 16
? ?
? ?
?
? ? ? ? ?
= 0.074 ? 10
–3
m/s = 0.074 mm/s.
6. ? = 1 m, r = 0.1 mm = 0.1 ? 10
–3
m
R = 100 ?, f = ?
? R = f ? / a
? f = 
6
Ra 100 3.14 0.1 0.1 10
1
?
? ? ? ?
?
?
= 3.14 ? 10
–6
= ? ? 10
–6
?-m.
?
7. ? ? = 2 ?
volume of the wire remains constant.
A ? = A ? ? ?
? A ? = A ? ? 2 ?
? A ? = A/2
f = Specific resistance
R = 
f
A
?
; R ? = 
f '
A '
?
100 ? = 
f2 4f
A / 2 A
?
? ?
= 4R
? 4 ? 100 ? = 400 ? ?
8. ? = 4 m, A = 1 mm
2
= 1 ? 10
–6
m
2
I = 2 A, n/V = 10
29
, t = ?
i = n A V
d
e
? e = 10
29
? 1 ? 10
–6
? V
d
? 1.6 ? 10
–19
?
d
29 6 19
2
V
10 10 1.6 10
? ?
?
? ? ?
= 
4
1 1
8000 0.8 10
?
?
t = 
d
4
4 8000
V 1/ 8000
? ? ?
?
= 32000 = 3.2 ? 10
4 
sec.
9. f
cu
= 1.7 ? 10
–8
?-m
A = 0.01 mm
2
= 0.01 ? 10
–6
m
2
R = 1 K ? = 10
3
?
R = 
f
a
?
? 10
3
= 
8
6
1.7 10
10
?
?
? ? ?
?
3
10
1.7
? ? = 0.58 ? 10
3
m = 0.6 km. ?
10. dR, due to the small strip dx at a distanc x d = R = 
2
fdx
y ?
…(1)
tan ? = 
y a b a
x L
? ?
?
?
y a b a
x L
? ?
?
? L(y – a) = x(b – a)
x ?
b–a 
dx 
a ?
y ?
? ? Y–a 
b ?
Electric Current in Conductors
32.3
? Ly – La = xb – xa 
?
dy
L 0 b a
dx
? ? ? (diff. w.r.t. x)
?
dy
L b a
dx
? ?
? dx = 
Ldy
b a ?
…(2)
Putting the value of dx in equation (1)
dR = 
2
fLdy
y (b a) ? ?
? dR = 
2
fI dy
(b a) y ? ?
?
R b
2
0 a
fI dy
dR
(b a) y
?
? ?
? ?
? R = 
fI (b a) fl
(b a) ab ab
?
?
? ? ?
.
11. r = 0.1 mm = 10
–4
m
R = 1 K ? = 10
3
?, V = 20 V
a) No.of electrons transferred 
i = 
3
V 20
R 10
? = 20 ? 10
–3
= 2 ? 10
–2
A
q = i t = 2 ? 10
–2
? 1 = 2 ? 10
–2
C.
No. of electrons transferred = 
2 17
19
2 10 2 10
1.6 1.6 10
? ?
?
? ?
?
?
= 1.25 ? 10
17
.
b) Current density of wire
= 
2
6
8
i 2 10 2
10
A 3.14 10
?
?
?
?
? ? ?
? ?
= 0.6369 ? 10
+6
= 6.37 ? 10
5
A/m
2
.
12. A = 2 ? 10
–6
m
2
, I = 1 A
f = 1.7 ? 10
–8
?-m
E = ?
R = 
8
6
f 1.7 10
A 2 10
?
?
? ?
?
?
? ?
V = IR = 
8
6
1 1.7 10
2 10
?
?
? ? ?
?
?
E = 
8
2
6
dV V 1.7 10 1.7
10 V /m
dL I 2 2 10
?
?
?
? ?
? ? ? ?
?
?
?
= 8.5 mV/m. ?
13. I = 2 m, R = 5 ?, i = 10 A, E = ?
V = iR = 10 ? 5 = 50 V
E = 
V 50
I 2
? = 25 V/m.
14. R ?
Fe
= R
Fe
(1 + ?
Fe
??), R ?
Cu
= R
Cu
(1 + ?
Cu
???)
R ?
Fe
= R ?
Cu
? R
Fe
(1 + ?
Fe
??), = R
Cu
(1 + ?
Cu
???)
Electric Current in Conductors
32.4
? 3.9 [ 1 + 5 ? 10
–3
(20 – ?)] = 4.1 [1 + 4 x 10
–3
(20 – ?)]
? 3.9 + 3.9 ? 5 ? 10
–3
(20 – ?) = 4.1 + 4.1 ? 4 ? 10
–3
(20 – ?)
?? 4.1 ? 4 ? 10
–3
(20 – ?) – 3.9 ? 5 ? 10
–3
(20 – ?) = 3.9 – 4.1
? 16.4(20 – ?) – 19.5(20 – ?) = 0.2 ? 10
3
? (20 – ?) (–3.1) = 0.2 ? 10
3
? ? – 20 = 200 ?
? ? = 220°C. ?
15. Let the voltmeter reading when, the voltage is 0 be X.
1 1
2 2
I R V
I R V
?
?
1.75 14.4 V 0.35 14.4 V
2.75 22.4 V 0.55 22.4 V
? ?
? ? ?
? ?
?
0.07 14.4 V 7 14.4 V
0.11 22.4 V 11 22.4 V
? ?
? ? ?
? ?
? 7(22.4 – V) = 11(14.4 – V) ? 156.8 – 7V = 158.4 – 11V
? (7 – 11)V = 156.8 – 158.4 ? –4V = –1.6
? V = 0.4 V.
16. a) When switch is open, no current passes through the ammeter. In the upper part of 
the circuit the Voltmenter has ? resistance. Thus current in it is 0.
? Voltmeter read the emf. (There is not Pot. Drop across the resistor).
b) When switch is closed current passes through the circuit and if its value of i.
The voltmeter reads
? – ir = 1.45
? 1.52 – ir = 1.45
? ir = 0.07
? 1 r = 0.07 ? r = 0.07 ?. ?
17. E = 6 V, r = 1 ?, V = 5.8 V, R = ?
I = 
E 6
R r R 1
?
? ?
, V = E – Ir
? 5.8 = 
6
6 1
R 1
? ?
?
?
6
R 1 ?
= 0.2
? R + 1 = 30 ? R = 29 ?. ?
18. V = ? + ir
? 7.2 = 6 + 2 ? r
? 1.2 = 2r ? r = 0.6 ?. ?
19. a) net emf while charging
9 – 6 = 3V
Current = 3/10 = 0.3 A
b) When completely charged.
Internal resistance ‘r’ = 1 ?
Current = 3/1 = 3 A ?
20. a) 0.1i
1
+ 1 i
1
– 6 + 1i
1
– 6 = 0
? 0.1 i
1
+ 1i
1
+ 1i
1
= 12
? i
1
= 
12
2.1
ABCDA
? 0.1i
2
+ 1i – 6 = 0
? 0.1i
2
+ 1i
?? ?
V 
A 
r ?
r ?
2 ?
6 ?
1 ?
0.1 
6 ?
1 ?
i 1 ?
6 ?
Electric Current in Conductors
32.5
ADEFA, 
? i – 6 + 6 – (i
2
– i)1 = 0
? i – i
2
+ i = 0
? 2i – i
2
= 0 ? –2i ± 0.2i = 0
? i
2
= 0.
b) 1i
1
+ 1 i
1
– 6 + 1i
1
= 0
? 3i
1
= 12 ? i
1
= 4
DCFED
? i
2
+ i – 6 = 0 ? i
2
+ i = 6
ABCDA,
i
2
+ (i
2
– i) – 6  = 0
? i
2
+ i
2
– i = 6 ? 2i
2
– i = 6
? –2i
2
± 2i = 6 ? i = –2
i
2
+ i = 6
? i
2
– 2 = 6  ? i
2
= 8
1
2
i 4 1
i 8 2
? ? .
c) 10i
1
+ 1i
1
– 6 + 1i
1
– 6 = 0
? 12i
1
= 12 ? i
1
= 1
10i
2
– i
1
– 6 = 0
? 10i
2
– i
1
= 6
? 10i
2
+ (i
2
– i)1 – 6 = 0
? 11i
2
= 6 
? –i
2
= 0
21. a) Total emf = n
1
E
in 1 row
Total emf in all news = n
1
E 
Total resistance in one row = n
1
r
Total resistance in all rows = 
1
2
n r
n
Net resistance = 
1
2
n r
n
+ R
Current = 
1 1 2
1 2 1 2
n E n n E
n /n r R n r n R
?
? ?
b) I = 
1 2
1 2
n n E
n r n R ?
for I = max,
n
1
r + n
2
R = min
?
? ?
2
1 2 1 2
n r n R 2 n rn R ? ? = min
it is min, when 
1 2
n r n R ?
? n
1
r = n
2
R
I is max when n
1
r = n
2
R.
i ?
i 2 ?
A ?
0.1 ? ?
1 ? ?
i 2 –i ?
6 ?
B ? C ?
1 ? ?
6 ?
E ?
D 
F ?
1 ?
1 ?
6 ?
1 ?
i 1 ?
6 ?
i ?
i 2 ?
E ?
1 ? ?
1 ? ?
i 2 –i ?
6 ?
D C ?
1 ? ?
6 ?
B ?
F ?
A ?
1 ?
1 ?
6 ?
1 ?
i 1 ?
6 ?
i ?
i 2 ?
E ?
10 ? ?
1 ? ?
i 2 –i ?
6 ?
D ? C ?
1 ? ?
6 ?
B ?
F ?
A ?
n 1
r ? r ? r ?
r ? r ? r ?
R  
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FAQs on HC Verma Solutions: Chapter 32 - Electric Current in Conductors - Physics Class 11 - NEET

1. What is electric current and how is it defined?
Ans. Electric current is the flow of electric charge in a conductor. It is defined as the rate of flow of charge through a cross-sectional area of a conductor. Mathematically, it is given by the equation I = Q/t, where I is the current, Q is the charge, and t is the time.
2. What factors affect the electric current in a conductor?
Ans. There are several factors that affect the electric current in a conductor. Some of these factors include the voltage applied across the conductor, the resistance of the conductor, the length and cross-sectional area of the conductor, and the temperature of the conductor.
3. What is Ohm's Law and how is it related to electric current?
Ans. Ohm's Law states that the current flowing through a conductor is directly proportional to the voltage applied across it and inversely proportional to the resistance of the conductor. Mathematically, it is represented by the equation I = V/R, where I is the current, V is the voltage, and R is the resistance. This law describes the relationship between current, voltage, and resistance in a conductor.
4. What is the difference between direct current (DC) and alternating current (AC)?
Ans. The main difference between direct current (DC) and alternating current (AC) lies in the direction of the flow of electric charge. In DC, the flow of electric charge is unidirectional, meaning it only flows in one direction. In AC, the flow of electric charge constantly changes direction, usually in a sinusoidal waveform. DC is commonly used in batteries and electronic devices, while AC is used in household electricity supply.
5. How does the resistance of a conductor affect the electric current?
Ans. The resistance of a conductor affects the electric current flowing through it. According to Ohm's Law, as the resistance increases, the current decreases, and vice versa. This means that a higher resistance in a conductor will impede the flow of electric charge, resulting in a lower current. Conversely, a lower resistance will allow for a higher current flow. Resistance is dependent on factors such as the material of the conductor, its length, its cross-sectional area, and its temperature.
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