Chapter 32 : Electric Current in Conductors - HC Verma Solution, Physics Class 11 Notes | EduRev

Physics Class 12

Class 11 : Chapter 32 : Electric Current in Conductors - HC Verma Solution, Physics Class 11 Notes | EduRev

 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  
Read More
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