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 Page 1


Introductory Exercise 20.1
1. i
q
t
= , here q e = , t
r
v
=
2 p
\ i
ev
r
=
2p
 
         =
´ ´ ´
´ ´ ´
-
-
1.6 10 2.2 10
2 3.14 5 10
19 6
11
         = ´
-
1.12 10
3
 A 
         =1.12mA
2. No. of at oms in 63.45 g of Cu = ´ 6.023 10
23
 
\ No. of atoms in 1 cm
3
 (8.89 g) of Cu
         =
´
´
6.023 10
63.54
8.89
23
         = ´ 8.43 10
22
As one conduction electron is present per
atoms,
n = ´ 8.43 10
22
 cm
-3
 or 8.43 m ´
-
10
28 3
As i neAv
d
=
Þ v
i
neA
d
=
    =
´ ´ ´ ´
´ ´
-
-
2.0
8.43 10 1.6 10 3.14
(0.5 10 )
28 19
32
    = ´
-
1.88 10
6
 ms
-1
3. Yes. 
As cur rent al ways flows in the di rec tion of
elec tr ic field.
4. False. 
In the ab sence of po ten tial dif fer ence,
elec trons passes ran dom mo tion.
5. Cur rent due to both pos i tive and neg a tive
ions is from left to right, hence, there is a net 
cur rent from left to right.
6. i t = + 10 4 Þ 
dq
dt
t = + 10 4
Þ             dq t dt
q
0 0
10
10 4
ò ò
= + ( )
Þ               q t t = + = [ ] 10 2 300
2
0
10
C
Introductory Exercise 20.2
1. R
L
A
=
r
  = ´ ´
´ ´
æ
è
ç
ö
ø
÷
-
-
1.72
3.14
2.05
10
35
2
10
8
3
2
  =0.57W
2. (a)     J
E
=
r
   Þ   i JA
EA
= =
r
=
´ ´ ´
´
-
-
0.49
2.75
3 14 042 10
10
3 2
8
. ( . )
       =9.87 A
(b) V EL = = ´ = 0.49 5.88 12 V
(c) R
V
i
= = =
5.88
9.87
0.6 W
3. Let us con sider the con duc tor to be made up
of a num ber of el e men tary discs. The
con duc tor is sup posed to be ex tended to form 
a com plete cone and the ver tex O of the cone
Current Electricity
20
Page 2


Introductory Exercise 20.1
1. i
q
t
= , here q e = , t
r
v
=
2 p
\ i
ev
r
=
2p
 
         =
´ ´ ´
´ ´ ´
-
-
1.6 10 2.2 10
2 3.14 5 10
19 6
11
         = ´
-
1.12 10
3
 A 
         =1.12mA
2. No. of at oms in 63.45 g of Cu = ´ 6.023 10
23
 
\ No. of atoms in 1 cm
3
 (8.89 g) of Cu
         =
´
´
6.023 10
63.54
8.89
23
         = ´ 8.43 10
22
As one conduction electron is present per
atoms,
n = ´ 8.43 10
22
 cm
-3
 or 8.43 m ´
-
10
28 3
As i neAv
d
=
Þ v
i
neA
d
=
    =
´ ´ ´ ´
´ ´
-
-
2.0
8.43 10 1.6 10 3.14
(0.5 10 )
28 19
32
    = ´
-
1.88 10
6
 ms
-1
3. Yes. 
As cur rent al ways flows in the di rec tion of
elec tr ic field.
4. False. 
In the ab sence of po ten tial dif fer ence,
elec trons passes ran dom mo tion.
5. Cur rent due to both pos i tive and neg a tive
ions is from left to right, hence, there is a net 
cur rent from left to right.
6. i t = + 10 4 Þ 
dq
dt
t = + 10 4
Þ             dq t dt
q
0 0
10
10 4
ò ò
= + ( )
Þ               q t t = + = [ ] 10 2 300
2
0
10
C
Introductory Exercise 20.2
1. R
L
A
=
r
  = ´ ´
´ ´
æ
è
ç
ö
ø
÷
-
-
1.72
3.14
2.05
10
35
2
10
8
3
2
  =0.57W
2. (a)     J
E
=
r
   Þ   i JA
EA
= =
r
=
´ ´ ´
´
-
-
0.49
2.75
3 14 042 10
10
3 2
8
. ( . )
       =9.87 A
(b) V EL = = ´ = 0.49 5.88 12 V
(c) R
V
i
= = =
5.88
9.87
0.6 W
3. Let us con sider the con duc tor to be made up
of a num ber of el e men tary discs. The
con duc tor is sup posed to be ex tended to form 
a com plete cone and the ver tex O of the cone
Current Electricity
20
is taken as or i gin with the con duc tor placed
along x-axis with its two ends at x r = and 
x l r = + . Let q be the semi-ver ti cal an gle of
the cone.
Consider an elementary disc of thickness dx
at a distance x from origin.
Resistance of this disc,
dR
dx
A
= r
If y be the radius of this disc, then
A y = p
2
But y x = tan q
dR
dx
x
= r
p q
2 2
tan
\ Resistance of conductor
R dR
dx
x
r
l r
= =
ò ò
+ r
p q
2 2
tan
        R
x
r
l r
= -
é
ë
ê
ù
û
ú
+
r
p q tan
2
1
           = -
+
é
ë
ê
ù
û
ú
r
p q tan
2
1 1
r l r
             R
l
r l r
=
+
r
p q ( )tan
2
But,     r a tanq=
     ( )tan r l b + = q
            R
l
ab
=
r
p
4. True.         r
s
=
1
\ r s
s
s ´ = ´ =
1
1
5. R R
Cu Fe
=
4.1 3.9
Cu Fe
( ) ( ) 1 1 + = + a a D D T T
4.1 4.0 [ ( )] 1 10 20
3
+ ´ -
-
T
= + ´ -
-
3.9 5.0 [ ( )] 1 10 20
3
T
4.1 16.4 + ´ -
-
10 20
3
( ) T
= + ´ -
-
3.9 19.5 10 20
3
( ) T
 3.1 0.2 ´ - =
-
10 20
3
( ) T
T - =
´
-
20
10
3
0.2
3.1
  
   = ° 64.5 C
T = ° 84.5 C
Introductory Exercise 20.3
1. Potential difference across both the re sis tors 
is 10 V.
Hence, i
1
10
2
5 = = A
and i
2
10
4
= = 2.5 A
2. As A is grounded, V
A
= 0
V V
B A
= + = 2 2 V
V V
C A
= + = 5 5 V
   V V
D C
= + = 10 15 V
\      i
V V
C B
1
1
3 =
-
= A
and          i
V V
D A
2
2
15
2
=
-
= =7.5 A
3. Cur rent in the given loop is
i
E
=
+ 15
8
 V E i E
E
AB
= - = - ×
+ æ
è
ç
ö
ø
÷ = 2 2
15
8
0
Þ E = 5 V       
4. Ef fec tive emf,
E = ´ - ´ = 8 1 2 1 6 V
Effective resistance of circuit
3 
i
1 i
2
4W 2W 10 V
i
2
i
1
1W
B
2W
A
2 V 10 V
5 V
C
D
i
1
R
1
R
2
R
3 10 V
i
3
i
2
10 V
Page 3


Introductory Exercise 20.1
1. i
q
t
= , here q e = , t
r
v
=
2 p
\ i
ev
r
=
2p
 
         =
´ ´ ´
´ ´ ´
-
-
1.6 10 2.2 10
2 3.14 5 10
19 6
11
         = ´
-
1.12 10
3
 A 
         =1.12mA
2. No. of at oms in 63.45 g of Cu = ´ 6.023 10
23
 
\ No. of atoms in 1 cm
3
 (8.89 g) of Cu
         =
´
´
6.023 10
63.54
8.89
23
         = ´ 8.43 10
22
As one conduction electron is present per
atoms,
n = ´ 8.43 10
22
 cm
-3
 or 8.43 m ´
-
10
28 3
As i neAv
d
=
Þ v
i
neA
d
=
    =
´ ´ ´ ´
´ ´
-
-
2.0
8.43 10 1.6 10 3.14
(0.5 10 )
28 19
32
    = ´
-
1.88 10
6
 ms
-1
3. Yes. 
As cur rent al ways flows in the di rec tion of
elec tr ic field.
4. False. 
In the ab sence of po ten tial dif fer ence,
elec trons passes ran dom mo tion.
5. Cur rent due to both pos i tive and neg a tive
ions is from left to right, hence, there is a net 
cur rent from left to right.
6. i t = + 10 4 Þ 
dq
dt
t = + 10 4
Þ             dq t dt
q
0 0
10
10 4
ò ò
= + ( )
Þ               q t t = + = [ ] 10 2 300
2
0
10
C
Introductory Exercise 20.2
1. R
L
A
=
r
  = ´ ´
´ ´
æ
è
ç
ö
ø
÷
-
-
1.72
3.14
2.05
10
35
2
10
8
3
2
  =0.57W
2. (a)     J
E
=
r
   Þ   i JA
EA
= =
r
=
´ ´ ´
´
-
-
0.49
2.75
3 14 042 10
10
3 2
8
. ( . )
       =9.87 A
(b) V EL = = ´ = 0.49 5.88 12 V
(c) R
V
i
= = =
5.88
9.87
0.6 W
3. Let us con sider the con duc tor to be made up
of a num ber of el e men tary discs. The
con duc tor is sup posed to be ex tended to form 
a com plete cone and the ver tex O of the cone
Current Electricity
20
is taken as or i gin with the con duc tor placed
along x-axis with its two ends at x r = and 
x l r = + . Let q be the semi-ver ti cal an gle of
the cone.
Consider an elementary disc of thickness dx
at a distance x from origin.
Resistance of this disc,
dR
dx
A
= r
If y be the radius of this disc, then
A y = p
2
But y x = tan q
dR
dx
x
= r
p q
2 2
tan
\ Resistance of conductor
R dR
dx
x
r
l r
= =
ò ò
+ r
p q
2 2
tan
        R
x
r
l r
= -
é
ë
ê
ù
û
ú
+
r
p q tan
2
1
           = -
+
é
ë
ê
ù
û
ú
r
p q tan
2
1 1
r l r
             R
l
r l r
=
+
r
p q ( )tan
2
But,     r a tanq=
     ( )tan r l b + = q
            R
l
ab
=
r
p
4. True.         r
s
=
1
\ r s
s
s ´ = ´ =
1
1
5. R R
Cu Fe
=
4.1 3.9
Cu Fe
( ) ( ) 1 1 + = + a a D D T T
4.1 4.0 [ ( )] 1 10 20
3
+ ´ -
-
T
= + ´ -
-
3.9 5.0 [ ( )] 1 10 20
3
T
4.1 16.4 + ´ -
-
10 20
3
( ) T
= + ´ -
-
3.9 19.5 10 20
3
( ) T
 3.1 0.2 ´ - =
-
10 20
3
( ) T
T - =
´
-
20
10
3
0.2
3.1
  
   = ° 64.5 C
T = ° 84.5 C
Introductory Exercise 20.3
1. Potential difference across both the re sis tors 
is 10 V.
Hence, i
1
10
2
5 = = A
and i
2
10
4
= = 2.5 A
2. As A is grounded, V
A
= 0
V V
B A
= + = 2 2 V
V V
C A
= + = 5 5 V
   V V
D C
= + = 10 15 V
\      i
V V
C B
1
1
3 =
-
= A
and          i
V V
D A
2
2
15
2
=
-
= =7.5 A
3. Cur rent in the given loop is
i
E
=
+ 15
8
 V E i E
E
AB
= - = - ×
+ æ
è
ç
ö
ø
÷ = 2 2
15
8
0
Þ E = 5 V       
4. Ef fec tive emf,
E = ´ - ´ = 8 1 2 1 6 V
Effective resistance of circuit
3 
i
1 i
2
4W 2W 10 V
i
2
i
1
1W
B
2W
A
2 V 10 V
5 V
C
D
i
1
R
1
R
2
R
3 10 V
i
3
i
2
10 V
   R R r = +
external
10 = + ´ = 2 10 1 12W
\        i
E
R
= = =
6
12
0.5 A
5. As R R
2 3
= and V V
1 2
=
Potential difference across R
1
 is zero.
Hence, current through R i
1 1
0 Þ =
and current through R
2
Þ i
V
R
2
1
2
=
      = =
10
10
1 A
6. i
E
R r
=
+
Also, V E ir = -
i
E V
r
=
-
Introductory Exercise 20.4
1.
Applying KCL at junction R
i = + = 1 2 3 A
V V V
ST RU QP
= =
Taking V V
ST RU
=
6 1 = - E
E = - 5 V
And from
V V
ST QP
=
6 12 = - + ir
r
i
=
-
= =
12 6 6
3
2 W
2. Power de liv ered by the 12 V power sup ply,
P Vi
1
= = ´ = 12 3 36 W
and power dissipated in 3 W resister,
P i R
3 3
2
3
2
2 3 12 = = ´ = W
Introductory Exercise 20.5
1. E
E
r
E
r
E
r
r r r
=
+ +
+ +
=
+ +
+ +
1
1
2
2
3
3
1 2 3
1 1 1
10
1
4
2
6
2
1
1
1
2
1
2
         =
+ + 10 2 3
2
               = 7.5 V
 and
1 1 1 1
1 2 3
r r r r
= + +
     = + + =
1
1
1
2
1
2
2
Þ                     r =
1
2
              =0.5W
2. i
E
R r
=
+
Rate of dissipation of energy
P i R =
2
 =
+
E R
R r
2
2
( )
For maximum or minimum power
dP
dR
= 0
Þ E
R r R R r
R r
2
2
4
2
0
( ) ( )
( )
+ - +
+
é
ë
ê
ù
û
ú
=
 4
Q
i
r
12 V
P
U
T
2A
1A
R
S
E
1W
3W
E r
R
I
E
r
O R
I
E
r
O V E
(a) (b)
Page 4


Introductory Exercise 20.1
1. i
q
t
= , here q e = , t
r
v
=
2 p
\ i
ev
r
=
2p
 
         =
´ ´ ´
´ ´ ´
-
-
1.6 10 2.2 10
2 3.14 5 10
19 6
11
         = ´
-
1.12 10
3
 A 
         =1.12mA
2. No. of at oms in 63.45 g of Cu = ´ 6.023 10
23
 
\ No. of atoms in 1 cm
3
 (8.89 g) of Cu
         =
´
´
6.023 10
63.54
8.89
23
         = ´ 8.43 10
22
As one conduction electron is present per
atoms,
n = ´ 8.43 10
22
 cm
-3
 or 8.43 m ´
-
10
28 3
As i neAv
d
=
Þ v
i
neA
d
=
    =
´ ´ ´ ´
´ ´
-
-
2.0
8.43 10 1.6 10 3.14
(0.5 10 )
28 19
32
    = ´
-
1.88 10
6
 ms
-1
3. Yes. 
As cur rent al ways flows in the di rec tion of
elec tr ic field.
4. False. 
In the ab sence of po ten tial dif fer ence,
elec trons passes ran dom mo tion.
5. Cur rent due to both pos i tive and neg a tive
ions is from left to right, hence, there is a net 
cur rent from left to right.
6. i t = + 10 4 Þ 
dq
dt
t = + 10 4
Þ             dq t dt
q
0 0
10
10 4
ò ò
= + ( )
Þ               q t t = + = [ ] 10 2 300
2
0
10
C
Introductory Exercise 20.2
1. R
L
A
=
r
  = ´ ´
´ ´
æ
è
ç
ö
ø
÷
-
-
1.72
3.14
2.05
10
35
2
10
8
3
2
  =0.57W
2. (a)     J
E
=
r
   Þ   i JA
EA
= =
r
=
´ ´ ´
´
-
-
0.49
2.75
3 14 042 10
10
3 2
8
. ( . )
       =9.87 A
(b) V EL = = ´ = 0.49 5.88 12 V
(c) R
V
i
= = =
5.88
9.87
0.6 W
3. Let us con sider the con duc tor to be made up
of a num ber of el e men tary discs. The
con duc tor is sup posed to be ex tended to form 
a com plete cone and the ver tex O of the cone
Current Electricity
20
is taken as or i gin with the con duc tor placed
along x-axis with its two ends at x r = and 
x l r = + . Let q be the semi-ver ti cal an gle of
the cone.
Consider an elementary disc of thickness dx
at a distance x from origin.
Resistance of this disc,
dR
dx
A
= r
If y be the radius of this disc, then
A y = p
2
But y x = tan q
dR
dx
x
= r
p q
2 2
tan
\ Resistance of conductor
R dR
dx
x
r
l r
= =
ò ò
+ r
p q
2 2
tan
        R
x
r
l r
= -
é
ë
ê
ù
û
ú
+
r
p q tan
2
1
           = -
+
é
ë
ê
ù
û
ú
r
p q tan
2
1 1
r l r
             R
l
r l r
=
+
r
p q ( )tan
2
But,     r a tanq=
     ( )tan r l b + = q
            R
l
ab
=
r
p
4. True.         r
s
=
1
\ r s
s
s ´ = ´ =
1
1
5. R R
Cu Fe
=
4.1 3.9
Cu Fe
( ) ( ) 1 1 + = + a a D D T T
4.1 4.0 [ ( )] 1 10 20
3
+ ´ -
-
T
= + ´ -
-
3.9 5.0 [ ( )] 1 10 20
3
T
4.1 16.4 + ´ -
-
10 20
3
( ) T
= + ´ -
-
3.9 19.5 10 20
3
( ) T
 3.1 0.2 ´ - =
-
10 20
3
( ) T
T - =
´
-
20
10
3
0.2
3.1
  
   = ° 64.5 C
T = ° 84.5 C
Introductory Exercise 20.3
1. Potential difference across both the re sis tors 
is 10 V.
Hence, i
1
10
2
5 = = A
and i
2
10
4
= = 2.5 A
2. As A is grounded, V
A
= 0
V V
B A
= + = 2 2 V
V V
C A
= + = 5 5 V
   V V
D C
= + = 10 15 V
\      i
V V
C B
1
1
3 =
-
= A
and          i
V V
D A
2
2
15
2
=
-
= =7.5 A
3. Cur rent in the given loop is
i
E
=
+ 15
8
 V E i E
E
AB
= - = - ×
+ æ
è
ç
ö
ø
÷ = 2 2
15
8
0
Þ E = 5 V       
4. Ef fec tive emf,
E = ´ - ´ = 8 1 2 1 6 V
Effective resistance of circuit
3 
i
1 i
2
4W 2W 10 V
i
2
i
1
1W
B
2W
A
2 V 10 V
5 V
C
D
i
1
R
1
R
2
R
3 10 V
i
3
i
2
10 V
   R R r = +
external
10 = + ´ = 2 10 1 12W
\        i
E
R
= = =
6
12
0.5 A
5. As R R
2 3
= and V V
1 2
=
Potential difference across R
1
 is zero.
Hence, current through R i
1 1
0 Þ =
and current through R
2
Þ i
V
R
2
1
2
=
      = =
10
10
1 A
6. i
E
R r
=
+
Also, V E ir = -
i
E V
r
=
-
Introductory Exercise 20.4
1.
Applying KCL at junction R
i = + = 1 2 3 A
V V V
ST RU QP
= =
Taking V V
ST RU
=
6 1 = - E
E = - 5 V
And from
V V
ST QP
=
6 12 = - + ir
r
i
=
-
= =
12 6 6
3
2 W
2. Power de liv ered by the 12 V power sup ply,
P Vi
1
= = ´ = 12 3 36 W
and power dissipated in 3 W resister,
P i R
3 3
2
3
2
2 3 12 = = ´ = W
Introductory Exercise 20.5
1. E
E
r
E
r
E
r
r r r
=
+ +
+ +
=
+ +
+ +
1
1
2
2
3
3
1 2 3
1 1 1
10
1
4
2
6
2
1
1
1
2
1
2
         =
+ + 10 2 3
2
               = 7.5 V
 and
1 1 1 1
1 2 3
r r r r
= + +
     = + + =
1
1
1
2
1
2
2
Þ                     r =
1
2
              =0.5W
2. i
E
R r
=
+
Rate of dissipation of energy
P i R =
2
 =
+
E R
R r
2
2
( )
For maximum or minimum power
dP
dR
= 0
Þ E
R r R R r
R r
2
2
4
2
0
( ) ( )
( )
+ - +
+
é
ë
ê
ù
û
ú
=
 4
Q
i
r
12 V
P
U
T
2A
1A
R
S
E
1W
3W
E r
R
I
E
r
O R
I
E
r
O V E
(a) (b)
Þ     E
R r r R
R r
2
4
0
( )( )
( )
+ -
+
=
Þ
E r R
R r
2
3
0
( )
( )
-
+
=
Þ      R r =
d P
dR
E
R r r R R r
R r
2
2
2
3 2
6
1 3
=
+ - - - +
+
é
ë
ê
ù
û
ú
( ) ( ) ( )( )
( )
        =
- -
+
E r R
R r
2
4
4 2 ( )
( )
Clearly
d P
dR
2
2
  is negative at R r = .
Hence, P is maximum at R r =
and P
E r
r r
E
r
max
( )
=
+
=
2
2
2
4
3. When the bat ter ies are con nected in se ries
E E
eff
= = 2 4V, r r
eff
= = 2 2 W 
For maximum power
R r = =
eff
2W
and P
E
r
max
( )
= =
´
=
eff
eff
W
2 2
4
4
4 2
2
4. I
g
= 5 mA, G = 1 W, V = 5 V
R
V
I
G
g
= - =
´
-
-
5
5 10
1
3
                            = 999 W
A 999 W resistance must be connected in
series with the galvanometer.
5. G = 100 W, i
g
= 50 mA, i = 5 mA
\ S
i G
i i
g
g
=
-
=
´ ´
´ - ´
-
- -
50 10 100
5 10 50 10
6
3 6
               =
-
=
1
1
1
0.01 0.99
               =
100
99
W
By connecting a shunt resistance of 
100
99
W.
6. i
V
G
g
=
and R
nV
i
G n G
g
= - = - ( ) 1
7. V E
AB
=
15
16
Potential gradient
k
V
L
E
AB
= =
´
15
16 600
=
E
640
 V/cm    
(a) 
E
kL
2
= Þ L
E
k
= =
2
320 cm
(b) V kl = = ´ =
E E
640
560
7
8
Also, V E ir = - 
\ E ir
E
- =
7
8
    i
E
r
=
8
AIEEE Corner
Subjective Questions (Level 1)
1. i
q
t
ne
t
= =
Given, 
    i=0.7 , t = 1 s, e = ´
-
1.6 C 10
19
\ n
it
e
= =
´
´
-
0.7
1.6
1
10
19
      = ´ 4.375 10
8
 
2. q it = = ´ ´ 3.6 3 3600
=38880 C
3. (a) q it = = ´ = 7.5 337.5 C 45
(b) q ne = Þ n
q
e
=
=
´
= ´
-
337.5
1.6
2.11
21
10
10
19
4. T
r
v
=
2 p
 Þ f
T
v
r
= =
1
2 p
           =
´
´ ´ ´
-
2.2
3.14 5.3
10
2 10
6
11
           = ´ 6.6 10
19
 s
-1
          I
q
T
ef = =
           = ´ ´ ´
-
1.6 6.6 10 10
19 19
           =10.56 A
5 
Page 5


Introductory Exercise 20.1
1. i
q
t
= , here q e = , t
r
v
=
2 p
\ i
ev
r
=
2p
 
         =
´ ´ ´
´ ´ ´
-
-
1.6 10 2.2 10
2 3.14 5 10
19 6
11
         = ´
-
1.12 10
3
 A 
         =1.12mA
2. No. of at oms in 63.45 g of Cu = ´ 6.023 10
23
 
\ No. of atoms in 1 cm
3
 (8.89 g) of Cu
         =
´
´
6.023 10
63.54
8.89
23
         = ´ 8.43 10
22
As one conduction electron is present per
atoms,
n = ´ 8.43 10
22
 cm
-3
 or 8.43 m ´
-
10
28 3
As i neAv
d
=
Þ v
i
neA
d
=
    =
´ ´ ´ ´
´ ´
-
-
2.0
8.43 10 1.6 10 3.14
(0.5 10 )
28 19
32
    = ´
-
1.88 10
6
 ms
-1
3. Yes. 
As cur rent al ways flows in the di rec tion of
elec tr ic field.
4. False. 
In the ab sence of po ten tial dif fer ence,
elec trons passes ran dom mo tion.
5. Cur rent due to both pos i tive and neg a tive
ions is from left to right, hence, there is a net 
cur rent from left to right.
6. i t = + 10 4 Þ 
dq
dt
t = + 10 4
Þ             dq t dt
q
0 0
10
10 4
ò ò
= + ( )
Þ               q t t = + = [ ] 10 2 300
2
0
10
C
Introductory Exercise 20.2
1. R
L
A
=
r
  = ´ ´
´ ´
æ
è
ç
ö
ø
÷
-
-
1.72
3.14
2.05
10
35
2
10
8
3
2
  =0.57W
2. (a)     J
E
=
r
   Þ   i JA
EA
= =
r
=
´ ´ ´
´
-
-
0.49
2.75
3 14 042 10
10
3 2
8
. ( . )
       =9.87 A
(b) V EL = = ´ = 0.49 5.88 12 V
(c) R
V
i
= = =
5.88
9.87
0.6 W
3. Let us con sider the con duc tor to be made up
of a num ber of el e men tary discs. The
con duc tor is sup posed to be ex tended to form 
a com plete cone and the ver tex O of the cone
Current Electricity
20
is taken as or i gin with the con duc tor placed
along x-axis with its two ends at x r = and 
x l r = + . Let q be the semi-ver ti cal an gle of
the cone.
Consider an elementary disc of thickness dx
at a distance x from origin.
Resistance of this disc,
dR
dx
A
= r
If y be the radius of this disc, then
A y = p
2
But y x = tan q
dR
dx
x
= r
p q
2 2
tan
\ Resistance of conductor
R dR
dx
x
r
l r
= =
ò ò
+ r
p q
2 2
tan
        R
x
r
l r
= -
é
ë
ê
ù
û
ú
+
r
p q tan
2
1
           = -
+
é
ë
ê
ù
û
ú
r
p q tan
2
1 1
r l r
             R
l
r l r
=
+
r
p q ( )tan
2
But,     r a tanq=
     ( )tan r l b + = q
            R
l
ab
=
r
p
4. True.         r
s
=
1
\ r s
s
s ´ = ´ =
1
1
5. R R
Cu Fe
=
4.1 3.9
Cu Fe
( ) ( ) 1 1 + = + a a D D T T
4.1 4.0 [ ( )] 1 10 20
3
+ ´ -
-
T
= + ´ -
-
3.9 5.0 [ ( )] 1 10 20
3
T
4.1 16.4 + ´ -
-
10 20
3
( ) T
= + ´ -
-
3.9 19.5 10 20
3
( ) T
 3.1 0.2 ´ - =
-
10 20
3
( ) T
T - =
´
-
20
10
3
0.2
3.1
  
   = ° 64.5 C
T = ° 84.5 C
Introductory Exercise 20.3
1. Potential difference across both the re sis tors 
is 10 V.
Hence, i
1
10
2
5 = = A
and i
2
10
4
= = 2.5 A
2. As A is grounded, V
A
= 0
V V
B A
= + = 2 2 V
V V
C A
= + = 5 5 V
   V V
D C
= + = 10 15 V
\      i
V V
C B
1
1
3 =
-
= A
and          i
V V
D A
2
2
15
2
=
-
= =7.5 A
3. Cur rent in the given loop is
i
E
=
+ 15
8
 V E i E
E
AB
= - = - ×
+ æ
è
ç
ö
ø
÷ = 2 2
15
8
0
Þ E = 5 V       
4. Ef fec tive emf,
E = ´ - ´ = 8 1 2 1 6 V
Effective resistance of circuit
3 
i
1 i
2
4W 2W 10 V
i
2
i
1
1W
B
2W
A
2 V 10 V
5 V
C
D
i
1
R
1
R
2
R
3 10 V
i
3
i
2
10 V
   R R r = +
external
10 = + ´ = 2 10 1 12W
\        i
E
R
= = =
6
12
0.5 A
5. As R R
2 3
= and V V
1 2
=
Potential difference across R
1
 is zero.
Hence, current through R i
1 1
0 Þ =
and current through R
2
Þ i
V
R
2
1
2
=
      = =
10
10
1 A
6. i
E
R r
=
+
Also, V E ir = -
i
E V
r
=
-
Introductory Exercise 20.4
1.
Applying KCL at junction R
i = + = 1 2 3 A
V V V
ST RU QP
= =
Taking V V
ST RU
=
6 1 = - E
E = - 5 V
And from
V V
ST QP
=
6 12 = - + ir
r
i
=
-
= =
12 6 6
3
2 W
2. Power de liv ered by the 12 V power sup ply,
P Vi
1
= = ´ = 12 3 36 W
and power dissipated in 3 W resister,
P i R
3 3
2
3
2
2 3 12 = = ´ = W
Introductory Exercise 20.5
1. E
E
r
E
r
E
r
r r r
=
+ +
+ +
=
+ +
+ +
1
1
2
2
3
3
1 2 3
1 1 1
10
1
4
2
6
2
1
1
1
2
1
2
         =
+ + 10 2 3
2
               = 7.5 V
 and
1 1 1 1
1 2 3
r r r r
= + +
     = + + =
1
1
1
2
1
2
2
Þ                     r =
1
2
              =0.5W
2. i
E
R r
=
+
Rate of dissipation of energy
P i R =
2
 =
+
E R
R r
2
2
( )
For maximum or minimum power
dP
dR
= 0
Þ E
R r R R r
R r
2
2
4
2
0
( ) ( )
( )
+ - +
+
é
ë
ê
ù
û
ú
=
 4
Q
i
r
12 V
P
U
T
2A
1A
R
S
E
1W
3W
E r
R
I
E
r
O R
I
E
r
O V E
(a) (b)
Þ     E
R r r R
R r
2
4
0
( )( )
( )
+ -
+
=
Þ
E r R
R r
2
3
0
( )
( )
-
+
=
Þ      R r =
d P
dR
E
R r r R R r
R r
2
2
2
3 2
6
1 3
=
+ - - - +
+
é
ë
ê
ù
û
ú
( ) ( ) ( )( )
( )
        =
- -
+
E r R
R r
2
4
4 2 ( )
( )
Clearly
d P
dR
2
2
  is negative at R r = .
Hence, P is maximum at R r =
and P
E r
r r
E
r
max
( )
=
+
=
2
2
2
4
3. When the bat ter ies are con nected in se ries
E E
eff
= = 2 4V, r r
eff
= = 2 2 W 
For maximum power
R r = =
eff
2W
and P
E
r
max
( )
= =
´
=
eff
eff
W
2 2
4
4
4 2
2
4. I
g
= 5 mA, G = 1 W, V = 5 V
R
V
I
G
g
= - =
´
-
-
5
5 10
1
3
                            = 999 W
A 999 W resistance must be connected in
series with the galvanometer.
5. G = 100 W, i
g
= 50 mA, i = 5 mA
\ S
i G
i i
g
g
=
-
=
´ ´
´ - ´
-
- -
50 10 100
5 10 50 10
6
3 6
               =
-
=
1
1
1
0.01 0.99
               =
100
99
W
By connecting a shunt resistance of 
100
99
W.
6. i
V
G
g
=
and R
nV
i
G n G
g
= - = - ( ) 1
7. V E
AB
=
15
16
Potential gradient
k
V
L
E
AB
= =
´
15
16 600
=
E
640
 V/cm    
(a) 
E
kL
2
= Þ L
E
k
= =
2
320 cm
(b) V kl = = ´ =
E E
640
560
7
8
Also, V E ir = - 
\ E ir
E
- =
7
8
    i
E
r
=
8
AIEEE Corner
Subjective Questions (Level 1)
1. i
q
t
ne
t
= =
Given, 
    i=0.7 , t = 1 s, e = ´
-
1.6 C 10
19
\ n
it
e
= =
´
´
-
0.7
1.6
1
10
19
      = ´ 4.375 10
8
 
2. q it = = ´ ´ 3.6 3 3600
=38880 C
3. (a) q it = = ´ = 7.5 337.5 C 45
(b) q ne = Þ n
q
e
=
=
´
= ´
-
337.5
1.6
2.11
21
10
10
19
4. T
r
v
=
2 p
 Þ f
T
v
r
= =
1
2 p
           =
´
´ ´ ´
-
2.2
3.14 5.3
10
2 10
6
11
           = ´ 6.6 10
19
 s
-1
          I
q
T
ef = =
           = ´ ´ ´
-
1.6 6.6 10 10
19 19
           =10.56 A
5 
5. (a) I t = - 55 0.65
2
        I
dq
dt
=
Þ    dq Idt =
Þ       q Idt =
ò
\     q Idt t dt = = -
ò ò
0
8
2
0
8
55 ( ) 0.65
         = -
é
ë
ê
ù
û
ú
55
2
0
8
2
0
8
[ ] t
t
0.65
= - = 440 20.8 419.2C
(b) If current is constant
         I
q
t
=   = =
419.2
52.4 A
8
6. i v
d
µ
\      
v
v
i
i
d
d
2
1
2
1
=
Þ     v
i
i
v
d d
2 1
2
1
4
6 00
1 20
10 = = ´ ´
-
.
.
1.20 
           = ´
-
6.00 10
4
 ms
-1
 
7.  v
i
neA
d
=
=
´ ´ ´ ´ ´
- -
1
10 10 1 10
28 19 4
8.5 1.6
    = ´
-
0.735 10
6
 ms
-1
    =0.735mm/s
   t
l
v
d
= =
´
-
10
10
3
6
0.735
    = ´ 1.36 10
9
 s = 43 yr
8. Dis tance cov ered by one elec tron in 1 s
= ´ = 1 0.05 0.05 cm
Number of electrons in 1 cm of wire
= ´ 2 10
21
\ Number of electrons crossing a given area
per second
= Number of electrons in 0.05 cm of wire
     = ´ ´ = 0.05 2 10 10
21 20
    i
q
t
ne
t
= =
     =
´ ´
= ´
-
10 10
1
10
20 19
1.6
1.6 =16 A
9. R
L
A
=r
Given, 
     r m =0.017 -m W 
           = ´
-
1.7 -m 10
8
W
         l=24.0 m
    A
d
=
æ
è
ç
ö
ø
÷
= ´ ´
æ
è
ç
ö
ø
÷
-
p
2 2
10
2
3
2
3.14
2.05
         = ´
-
3.29 10
6
 m
2
       R= ´ ´
´
-
-
1.7
24.0
3.29
10
10
8
6
         =0.12W
10.     R
L
A
=r
     A
L
R
=
r
If D is density, then
   m DV DAL
D L
R
= = =
r
2
=
´ ´ ´ ´
-
8.9 1.72 3.5
0.125
10 10
3 8 2
( )
     = ´
-
1.5 10
2
 kg = 15 g
11. At 20°C, 
R
1
600 = W, R
2
300 = W 
At 50°C,
R R t
1 1 1
1 ¢ = + ( ) a D
   = + ´ = ´ 600 1 30 600 ( ) 0.001 1.03
   =618W
R R t
2 2 2
1 ¢ = + ( ) a D
   = + ´ = 300 1 30 336 ( ) 0.004
\ R R R ¢= ¢+ ¢= +
1 2
618 336
           =954W
a =
¢-
´
=
-
´
R R
R t D
954 900
900 30
 
         R= + = 600 300 900W
           = °
-
0.002 C
1
 
12. As both the wires are con nected in par al lel,
        V V
Al Cu
=
      i R i R
Al Al Cu Cu
=
   i
L
d
i
L
d
Al Al
Al
Al
Cu Cu
Cu
Cu
r
p
r
p
2 2
=
Þ       d d
i L
i L
Cu Al
Cu Cu Cu
Al Al Al
=
r
r
       = ´
´ ´
´ ´
-
1 10
2 6
3
3
0.017
0.028 7.5
= ´
-
0.569 10
3
 m   
           =0.569 mm.
 6
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FAQs on DC Pandey Solutions: Current Electricity- 1 - DC Pandey Solutions for NEET Physics

1. What is the formula to calculate electric current?
Ans. The formula to calculate electric current is I = Q/t, where I is the electric current, Q is the charge flowing through a conductor, and t is the time taken.
2. How is electric current measured?
Ans. Electric current is measured using an ammeter, which is connected in series with the circuit. The ammeter measures the flow of electric charges and gives the reading in amperes (A).
3. What is the difference between AC and DC current?
Ans. AC (alternating current) and DC (direct current) are two types of electric currents. AC constantly changes direction, while DC flows in only one direction. AC is commonly used for household electrical supply, while DC is commonly used in batteries and electronic devices.
4. What is resistivity and how is it related to current?
Ans. Resistivity is a property of a material that quantifies its ability to resist the flow of electric current. It is denoted by the symbol ρ (rho). The current flowing through a material is inversely proportional to its resistivity. Higher resistivity materials offer more resistance to the flow of current.
5. What is Ohm's Law and how is it used in current electricity?
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 its resistance. Mathematically, it can be expressed as I = V/R, where I is the current, V is the voltage, and R is the resistance. Ohm's Law is widely used in analyzing and calculating current, voltage, and resistance in electrical circuits.
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