DC Pandey Solutions: Current Electricity- 2

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``` Page 1

22. i =
+ +
=
200
5 10 25
5 A
(i) V V
3 0
5 25 - = ´
Þ V
3
125 = V
(ii) V V
0 2
5 10 - = ´
V
2
50 = - V
(iii) V V
2 1
5 5 - = ´
V
1
75 = - V
(iv) V
3 2
5 35 175
-
= ´ = V
(v) V
1 2
5 5 25
-
= - ´ = - V
(vi) V
1 3
200
-
= - V
23. (a)
R R R R r
v a eff
= + + | |
=
´
+
+ +
50 200
50 200
2 1
= 43 W
i
E
R
= = =
eff
4.3
0.1 A
43
\ Reading of ammeter, i = 0.1 A
and reading of voltmeter = i R R
v
( | | )
= ´ = 0.1 V 40 4
(b)
R R R R r
a v eff
= + + ( )| |
=
´
+
+
52 200
52 200
1
= 42.26 W
i
E
R
= =
eff
0.102 W
Reading of voltmeter
V E ir = -
= - ´ 4.3 0.102 1
»4.2W
Reading of Ammeter,
i
V
R R
a
1
42
=
+
= =
4.2
0.08 A
24. Con sider the di rec tions of cur rent as shown
in fig ure.
Applying KVL in loop 1, 2 and 3, we
respectively get,
I I I I
1 1 2 1
6 5 42 + - + = ( )
Þ 12 6 42
1 2
I I - =
Þ 2 7
1 2
I I - = …(i)
4 6 8 10
2 2 1 2 3
I I I I I + - + + = ( ) ( )
Þ 9 3 4 5
2 1 3
I I I - + = …(ii)
8 16 4
2 3 3
( ) I I I + + =
2 6 1
2 3
I I + = …(iii)
On solving, we get,
I
1
= 4.7 A, I
2
= 2.4 A, I
3
= 0.5 A
Resistor 5W 1 W 4W 6W 8W 16W
Current 4.7 A 4.7 A 2.4 A 2.3A 2.9 A 0.5A
25.
9
r
A
V
E
i
i
1
R
i
2
42V 5W
4W
6W
1W
6W
8W
4V
D
B
10V
C
A
I
1
I
2
I
3
r
A
V
E
i
i
1
R
S
i
2
1
0
3
200 V
2
5W
10W
25W
i
1
V
R = 400W
400W
200W 100W
i
2
100W
100W
i
10 V
i
1
200W
100W
i
2
100W
100W
i
10 V
Þ
i
200W
D
B
A
C
Page 2

22. i =
+ +
=
200
5 10 25
5 A
(i) V V
3 0
5 25 - = ´
Þ V
3
125 = V
(ii) V V
0 2
5 10 - = ´
V
2
50 = - V
(iii) V V
2 1
5 5 - = ´
V
1
75 = - V
(iv) V
3 2
5 35 175
-
= ´ = V
(v) V
1 2
5 5 25
-
= - ´ = - V
(vi) V
1 3
200
-
= - V
23. (a)
R R R R r
v a eff
= + + | |
=
´
+
+ +
50 200
50 200
2 1
= 43 W
i
E
R
= = =
eff
4.3
0.1 A
43
\ Reading of ammeter, i = 0.1 A
and reading of voltmeter = i R R
v
( | | )
= ´ = 0.1 V 40 4
(b)
R R R R r
a v eff
= + + ( )| |
=
´
+
+
52 200
52 200
1
= 42.26 W
i
E
R
= =
eff
0.102 W
Reading of voltmeter
V E ir = -
= - ´ 4.3 0.102 1
»4.2W
Reading of Ammeter,
i
V
R R
a
1
42
=
+
= =
4.2
0.08 A
24. Con sider the di rec tions of cur rent as shown
in fig ure.
Applying KVL in loop 1, 2 and 3, we
respectively get,
I I I I
1 1 2 1
6 5 42 + - + = ( )
Þ 12 6 42
1 2
I I - =
Þ 2 7
1 2
I I - = …(i)
4 6 8 10
2 2 1 2 3
I I I I I + - + + = ( ) ( )
Þ 9 3 4 5
2 1 3
I I I - + = …(ii)
8 16 4
2 3 3
( ) I I I + + =
2 6 1
2 3
I I + = …(iii)
On solving, we get,
I
1
= 4.7 A, I
2
= 2.4 A, I
3
= 0.5 A
Resistor 5W 1 W 4W 6W 8W 16W
Current 4.7 A 4.7 A 2.4 A 2.3A 2.9 A 0.5A
25.
9
r
A
V
E
i
i
1
R
i
2
42V 5W
4W
6W
1W
6W
8W
4V
D
B
10V
C
A
I
1
I
2
I
3
r
A
V
E
i
i
1
R
S
i
2
1
0
3
200 V
2
5W
10W
25W
i
1
V
R = 400W
400W
200W 100W
i
2
100W
100W
i
10 V
i
1
200W
100W
i
2
100W
100W
i
10 V
Þ
i
200W
D
B
A
C
As Wheatstone bridge is balanced, 100W
resistance between B and D can be removed,
ie,
\ i i
1 2
10
300
1
30
= = = A
Hence, reading of voltmeter
= Potential difference between B and C
= ´ = 200
20
3
2
i V
=6.67V
26. (a) (i) When S is open.
V
R
R R
E
V V
1
1
1 2
3000
5000
200 =
+
= ´
=120 V
V
R
R R
E
V
V V
2
2
1 2
2000
5000
200 =
+
= ´
= 80 V
(ii) When S is closed,
Now, R
1
and V
1
are in parallel and their
effective resistance
R
R R
R R
V
V
1
1
1
1
1
6000
5
1200 ¢ =
+
= = W
Similarly,
R
2
and V
2
are in parallel with their effective
resistance,
R
R R
R R
V
V
2
2
2
2
2
6000
5
1200 ¢ =
+
= = W
As  R R
1 2
¢ = ¢
Hence,
reading of V
1
= reading of V
2
=
+
´ =
1200
1200 1200
200 100 V
(b) Current distribution is shown in figure
i
E
R R
=
¢ + ¢
1 2
i = =
200
2400
1
12
A
i
R
R R
i
V
V
1
1
1
1
=
+
= ´
3000
5000
1
12
=
1
20
A
i
R
R R
i
V
2
1
1
1
2000
5000
1
12
=
+
= ´
=
1
30
A
\ Current flowing through
S i i = - = -
1 2
1
20
1
30
=
1
60
A
27. Ef fec tive emf of 2 V and 6 V bat ter ies
con nected in par al lel
E
E r E r
r r
¢ =
+
+
=
´ - ´
+
1 2 2 1
1 2
2 1 6 1
1 1
= - 2 V
and r
r r
r r
¢ =
+
=
1 2
1 2
1
2
W
= 0.5W
10
E
'
R
2
'
R
1
R
1
S
R
2
E
V
1
V
2
i
2
i
1
i
2 i
1
i
V
1
V
2
R
1 S
R
2
E
V
1
V
2
R
1 S
R
2
E
i
2
A
100W 200W
100W 200W
i
1
10 V
B
C
D
Page 3

22. i =
+ +
=
200
5 10 25
5 A
(i) V V
3 0
5 25 - = ´
Þ V
3
125 = V
(ii) V V
0 2
5 10 - = ´
V
2
50 = - V
(iii) V V
2 1
5 5 - = ´
V
1
75 = - V
(iv) V
3 2
5 35 175
-
= ´ = V
(v) V
1 2
5 5 25
-
= - ´ = - V
(vi) V
1 3
200
-
= - V
23. (a)
R R R R r
v a eff
= + + | |
=
´
+
+ +
50 200
50 200
2 1
= 43 W
i
E
R
= = =
eff
4.3
0.1 A
43
\ Reading of ammeter, i = 0.1 A
and reading of voltmeter = i R R
v
( | | )
= ´ = 0.1 V 40 4
(b)
R R R R r
a v eff
= + + ( )| |
=
´
+
+
52 200
52 200
1
= 42.26 W
i
E
R
= =
eff
0.102 W
Reading of voltmeter
V E ir = -
= - ´ 4.3 0.102 1
»4.2W
Reading of Ammeter,
i
V
R R
a
1
42
=
+
= =
4.2
0.08 A
24. Con sider the di rec tions of cur rent as shown
in fig ure.
Applying KVL in loop 1, 2 and 3, we
respectively get,
I I I I
1 1 2 1
6 5 42 + - + = ( )
Þ 12 6 42
1 2
I I - =
Þ 2 7
1 2
I I - = …(i)
4 6 8 10
2 2 1 2 3
I I I I I + - + + = ( ) ( )
Þ 9 3 4 5
2 1 3
I I I - + = …(ii)
8 16 4
2 3 3
( ) I I I + + =
2 6 1
2 3
I I + = …(iii)
On solving, we get,
I
1
= 4.7 A, I
2
= 2.4 A, I
3
= 0.5 A
Resistor 5W 1 W 4W 6W 8W 16W
Current 4.7 A 4.7 A 2.4 A 2.3A 2.9 A 0.5A
25.
9
r
A
V
E
i
i
1
R
i
2
42V 5W
4W
6W
1W
6W
8W
4V
D
B
10V
C
A
I
1
I
2
I
3
r
A
V
E
i
i
1
R
S
i
2
1
0
3
200 V
2
5W
10W
25W
i
1
V
R = 400W
400W
200W 100W
i
2
100W
100W
i
10 V
i
1
200W
100W
i
2
100W
100W
i
10 V
Þ
i
200W
D
B
A
C
As Wheatstone bridge is balanced, 100W
resistance between B and D can be removed,
ie,
\ i i
1 2
10
300
1
30
= = = A
Hence, reading of voltmeter
= Potential difference between B and C
= ´ = 200
20
3
2
i V
=6.67V
26. (a) (i) When S is open.
V
R
R R
E
V V
1
1
1 2
3000
5000
200 =
+
= ´
=120 V
V
R
R R
E
V
V V
2
2
1 2
2000
5000
200 =
+
= ´
= 80 V
(ii) When S is closed,
Now, R
1
and V
1
are in parallel and their
effective resistance
R
R R
R R
V
V
1
1
1
1
1
6000
5
1200 ¢ =
+
= = W
Similarly,
R
2
and V
2
are in parallel with their effective
resistance,
R
R R
R R
V
V
2
2
2
2
2
6000
5
1200 ¢ =
+
= = W
As  R R
1 2
¢ = ¢
Hence,
reading of V
1
= reading of V
2
=
+
´ =
1200
1200 1200
200 100 V
(b) Current distribution is shown in figure
i
E
R R
=
¢ + ¢
1 2
i = =
200
2400
1
12
A
i
R
R R
i
V
V
1
1
1
1
=
+
= ´
3000
5000
1
12
=
1
20
A
i
R
R R
i
V
2
1
1
1
2000
5000
1
12
=
+
= ´
=
1
30
A
\ Current flowing through
S i i = - = -
1 2
1
20
1
30
=
1
60
A
27. Ef fec tive emf of 2 V and 6 V bat ter ies
con nected in par al lel
E
E r E r
r r
¢ =
+
+
=
´ - ´
+
1 2 2 1
1 2
2 1 6 1
1 1
= - 2 V
and r
r r
r r
¢ =
+
=
1 2
1 2
1
2
W
= 0.5W
10
E
'
R
2
'
R
1
R
1
S
R
2
E
V
1
V
2
i
2
i
1
i
2 i
1
i
V
1
V
2
R
1 S
R
2
E
V
1
V
2
R
1 S
R
2
E
i
2
A
100W 200W
100W 200W
i
1
10 V
B
C
D
Net emf, E = - = 4 2 2 V
28. (a)
As E E
1 2
>
Current will flow from B to A.
(b) E
1
is doing positive work
(c) As current flows from B to A through
resistor, B is at higher potential.
29. i R
2
2 5 = < W W
Clearly X is doing negative work.
(a) P Vi = Þ V
P
i
= = =
0.5
1.0
5.0 V
(b) E V iR = - = - = 5 2 3 V
(c) It is clear from figure that positive
terminal of X is towards left.
30. i =
-
+
=
150 50
3 2
20 A
V V i
P Q
- = + 50 3.0
V
Q
= - + 100 50 60 ( )
= - 10 V
31. (a) As volt me ter is ideal, it has in fi nite
re sis tance, there fore cur ren t is zero.
(b) V E i r = - Þ = E 5 . 0 V
(c) Reading of voltmeter Þ = V 5.0 V
32. V E i r
1 1
= - Þ E r - = 1.5 8.4          …(i)
V E i r
2 2
= + Þ E r + = 3.5 9.4 …(ii)
On solving, we get
r = 0.2W
E = 8.7 V
33. In case of charg ing
V E i r = + = + ´ = 2 5 0.1 2.5 V
34. Clearly cur rent through each branch is zero.
35. i
E
R G
1
=
+
On shunt i ng the gal va nom e t er with
re sis tan ce S ,
i
E
R R
GS
S G
2
=
+ ¢ +
+
As i i
1 2
=
E
R G
E
R R
GS
G S
+
=
+ ¢ +
+
Þ R R
GS
G S
R G + ¢ +
+
= +
R
G
G S
¢ =
+
2
36.
I
r
R r
I
V
R
2
=
+
=
11
8W
8W 4W
2W
2W
2V 2V
Þ
2V 2V 2V
4W 4W 4W
2V
G
i
2
E
R R'
S
R=2.0W
i A
E
B
+ +
– –
A B
E
1
E
2
+ +
– –
Q
3.0W
2.0W
150 V
50 V
P
i
A
A
V
I
2
B
I r
I
1
G
i
1
E
R
4V
0.5W
2V
6V 1W
1W
ß
0.5W 0.5W
2V
4V
Page 4

22. i =
+ +
=
200
5 10 25
5 A
(i) V V
3 0
5 25 - = ´
Þ V
3
125 = V
(ii) V V
0 2
5 10 - = ´
V
2
50 = - V
(iii) V V
2 1
5 5 - = ´
V
1
75 = - V
(iv) V
3 2
5 35 175
-
= ´ = V
(v) V
1 2
5 5 25
-
= - ´ = - V
(vi) V
1 3
200
-
= - V
23. (a)
R R R R r
v a eff
= + + | |
=
´
+
+ +
50 200
50 200
2 1
= 43 W
i
E
R
= = =
eff
4.3
0.1 A
43
\ Reading of ammeter, i = 0.1 A
and reading of voltmeter = i R R
v
( | | )
= ´ = 0.1 V 40 4
(b)
R R R R r
a v eff
= + + ( )| |
=
´
+
+
52 200
52 200
1
= 42.26 W
i
E
R
= =
eff
0.102 W
Reading of voltmeter
V E ir = -
= - ´ 4.3 0.102 1
»4.2W
Reading of Ammeter,
i
V
R R
a
1
42
=
+
= =
4.2
0.08 A
24. Con sider the di rec tions of cur rent as shown
in fig ure.
Applying KVL in loop 1, 2 and 3, we
respectively get,
I I I I
1 1 2 1
6 5 42 + - + = ( )
Þ 12 6 42
1 2
I I - =
Þ 2 7
1 2
I I - = …(i)
4 6 8 10
2 2 1 2 3
I I I I I + - + + = ( ) ( )
Þ 9 3 4 5
2 1 3
I I I - + = …(ii)
8 16 4
2 3 3
( ) I I I + + =
2 6 1
2 3
I I + = …(iii)
On solving, we get,
I
1
= 4.7 A, I
2
= 2.4 A, I
3
= 0.5 A
Resistor 5W 1 W 4W 6W 8W 16W
Current 4.7 A 4.7 A 2.4 A 2.3A 2.9 A 0.5A
25.
9
r
A
V
E
i
i
1
R
i
2
42V 5W
4W
6W
1W
6W
8W
4V
D
B
10V
C
A
I
1
I
2
I
3
r
A
V
E
i
i
1
R
S
i
2
1
0
3
200 V
2
5W
10W
25W
i
1
V
R = 400W
400W
200W 100W
i
2
100W
100W
i
10 V
i
1
200W
100W
i
2
100W
100W
i
10 V
Þ
i
200W
D
B
A
C
As Wheatstone bridge is balanced, 100W
resistance between B and D can be removed,
ie,
\ i i
1 2
10
300
1
30
= = = A
Hence, reading of voltmeter
= Potential difference between B and C
= ´ = 200
20
3
2
i V
=6.67V
26. (a) (i) When S is open.
V
R
R R
E
V V
1
1
1 2
3000
5000
200 =
+
= ´
=120 V
V
R
R R
E
V
V V
2
2
1 2
2000
5000
200 =
+
= ´
= 80 V
(ii) When S is closed,
Now, R
1
and V
1
are in parallel and their
effective resistance
R
R R
R R
V
V
1
1
1
1
1
6000
5
1200 ¢ =
+
= = W
Similarly,
R
2
and V
2
are in parallel with their effective
resistance,
R
R R
R R
V
V
2
2
2
2
2
6000
5
1200 ¢ =
+
= = W
As  R R
1 2
¢ = ¢
Hence,
reading of V
1
= reading of V
2
=
+
´ =
1200
1200 1200
200 100 V
(b) Current distribution is shown in figure
i
E
R R
=
¢ + ¢
1 2
i = =
200
2400
1
12
A
i
R
R R
i
V
V
1
1
1
1
=
+
= ´
3000
5000
1
12
=
1
20
A
i
R
R R
i
V
2
1
1
1
2000
5000
1
12
=
+
= ´
=
1
30
A
\ Current flowing through
S i i = - = -
1 2
1
20
1
30
=
1
60
A
27. Ef fec tive emf of 2 V and 6 V bat ter ies
con nected in par al lel
E
E r E r
r r
¢ =
+
+
=
´ - ´
+
1 2 2 1
1 2
2 1 6 1
1 1
= - 2 V
and r
r r
r r
¢ =
+
=
1 2
1 2
1
2
W
= 0.5W
10
E
'
R
2
'
R
1
R
1
S
R
2
E
V
1
V
2
i
2
i
1
i
2 i
1
i
V
1
V
2
R
1 S
R
2
E
V
1
V
2
R
1 S
R
2
E
i
2
A
100W 200W
100W 200W
i
1
10 V
B
C
D
Net emf, E = - = 4 2 2 V
28. (a)
As E E
1 2
>
Current will flow from B to A.
(b) E
1
is doing positive work
(c) As current flows from B to A through
resistor, B is at higher potential.
29. i R
2
2 5 = < W W
Clearly X is doing negative work.
(a) P Vi = Þ V
P
i
= = =
0.5
1.0
5.0 V
(b) E V iR = - = - = 5 2 3 V
(c) It is clear from figure that positive
terminal of X is towards left.
30. i =
-
+
=
150 50
3 2
20 A
V V i
P Q
- = + 50 3.0
V
Q
= - + 100 50 60 ( )
= - 10 V
31. (a) As volt me ter is ideal, it has in fi nite
re sis tance, there fore cur ren t is zero.
(b) V E i r = - Þ = E 5 . 0 V
(c) Reading of voltmeter Þ = V 5.0 V
32. V E i r
1 1
= - Þ E r - = 1.5 8.4          …(i)
V E i r
2 2
= + Þ E r + = 3.5 9.4 …(ii)
On solving, we get
r = 0.2W
E = 8.7 V
33. In case of charg ing
V E i r = + = + ´ = 2 5 0.1 2.5 V
34. Clearly cur rent through each branch is zero.
35. i
E
R G
1
=
+
On shunt i ng the gal va nom e t er with
re sis tan ce S ,
i
E
R R
GS
S G
2
=
+ ¢ +
+
As i i
1 2
=
E
R G
E
R R
GS
G S
+
=
+ ¢ +
+
Þ R R
GS
G S
R G + ¢ +
+
= +
R
G
G S
¢ =
+
2
36.
I
r
R r
I
V
R
2
=
+
=
11
8W
8W 4W
2W
2W
2V 2V
Þ
2V 2V 2V
4W 4W 4W
2V
G
i
2
E
R R'
S
R=2.0W
i A
E
B
+ +
– –
A B
E
1
E
2
+ +
– –
Q
3.0W
2.0W
150 V
50 V
P
i
A
A
V
I
2
B
I r
I
1
G
i
1
E
R
4V
0.5W
2V
6V 1W
1W
ß
0.5W 0.5W
2V
4V
Þ
r
R r
V
IR +
=
Þ
R
r
IR V
V
=
-
=
´ - 5 2500 100
100
r = ´ =
100
12400
2500 20.16W
37.
Let R be the resistance of voltmeter
As reading of voltmeter is 30 V,
1 1
400
1
300 R
+ = Þ R = 1200W
If voltmeter is connected across 300W
resistor,
Effective resistance of 300W resistor and
voltmeter
R¢=
´
+
=
300 1200
300 1200
240W
i=
+
60
400 240
=
60
640
A
=
3
32
A
\ Reading of voltmeter,
V iR = ¢ = ´
3
32
240
= 22.5 V
38. V
R
R R
V
2
1 2
=
¢
+ ¢
,
R
rR
r R
2
2
2
120
3
¢ =
+
=
=40W
V
2
40
60 40
120 =
+
=48V
39. S
i
i i
G R
g
g
=
-
+ ( )
R
i i
i
S G
g
g
=
-
-
=
-
´ -
-
-
20 10
10
20
3
3
0.005
= 79.995 W
40. r
L L
L
R =
-
=
-
´ =
1 2
2
5
0.52 0.4
0.4
1.5 W
41. Let R be the re sis tance of voltmeter
R
R
R
e
= + +
+
3 2
100
100
= +
+
5
100
100
R
R

i
R
R
=
+
+
=
3.4
.04
5
100
100
0
Þ      0.2 3.4 +
+
=
4
100
R
R

Þ         R=400W
Reading of voltmeter,
V i
R
R
= ´
+
100
100
= ´
´
+
0.04
100 400
100 400
=3.2V
If the voltmeter had been ideal,
Reading of voltmeter
= ´ =
100
105
3.4 3.24 V
42.
L
L
R
R
1
2
1
2
=
Þ
L
L
1
1
40
8
12 -
= (L L
1 2
40 + = cm)
Þ       L
1
16 = cm   from A.
12
G B
R
S
A
60 V
V
300W 400W
60 V
V
300W 400W
3.4 V
A
i
3W
V
100W
Page 5

22. i =
+ +
=
200
5 10 25
5 A
(i) V V
3 0
5 25 - = ´
Þ V
3
125 = V
(ii) V V
0 2
5 10 - = ´
V
2
50 = - V
(iii) V V
2 1
5 5 - = ´
V
1
75 = - V
(iv) V
3 2
5 35 175
-
= ´ = V
(v) V
1 2
5 5 25
-
= - ´ = - V
(vi) V
1 3
200
-
= - V
23. (a)
R R R R r
v a eff
= + + | |
=
´
+
+ +
50 200
50 200
2 1
= 43 W
i
E
R
= = =
eff
4.3
0.1 A
43
\ Reading of ammeter, i = 0.1 A
and reading of voltmeter = i R R
v
( | | )
= ´ = 0.1 V 40 4
(b)
R R R R r
a v eff
= + + ( )| |
=
´
+
+
52 200
52 200
1
= 42.26 W
i
E
R
= =
eff
0.102 W
Reading of voltmeter
V E ir = -
= - ´ 4.3 0.102 1
»4.2W
Reading of Ammeter,
i
V
R R
a
1
42
=
+
= =
4.2
0.08 A
24. Con sider the di rec tions of cur rent as shown
in fig ure.
Applying KVL in loop 1, 2 and 3, we
respectively get,
I I I I
1 1 2 1
6 5 42 + - + = ( )
Þ 12 6 42
1 2
I I - =
Þ 2 7
1 2
I I - = …(i)
4 6 8 10
2 2 1 2 3
I I I I I + - + + = ( ) ( )
Þ 9 3 4 5
2 1 3
I I I - + = …(ii)
8 16 4
2 3 3
( ) I I I + + =
2 6 1
2 3
I I + = …(iii)
On solving, we get,
I
1
= 4.7 A, I
2
= 2.4 A, I
3
= 0.5 A
Resistor 5W 1 W 4W 6W 8W 16W
Current 4.7 A 4.7 A 2.4 A 2.3A 2.9 A 0.5A
25.
9
r
A
V
E
i
i
1
R
i
2
42V 5W
4W
6W
1W
6W
8W
4V
D
B
10V
C
A
I
1
I
2
I
3
r
A
V
E
i
i
1
R
S
i
2
1
0
3
200 V
2
5W
10W
25W
i
1
V
R = 400W
400W
200W 100W
i
2
100W
100W
i
10 V
i
1
200W
100W
i
2
100W
100W
i
10 V
Þ
i
200W
D
B
A
C
As Wheatstone bridge is balanced, 100W
resistance between B and D can be removed,
ie,
\ i i
1 2
10
300
1
30
= = = A
Hence, reading of voltmeter
= Potential difference between B and C
= ´ = 200
20
3
2
i V
=6.67V
26. (a) (i) When S is open.
V
R
R R
E
V V
1
1
1 2
3000
5000
200 =
+
= ´
=120 V
V
R
R R
E
V
V V
2
2
1 2
2000
5000
200 =
+
= ´
= 80 V
(ii) When S is closed,
Now, R
1
and V
1
are in parallel and their
effective resistance
R
R R
R R
V
V
1
1
1
1
1
6000
5
1200 ¢ =
+
= = W
Similarly,
R
2
and V
2
are in parallel with their effective
resistance,
R
R R
R R
V
V
2
2
2
2
2
6000
5
1200 ¢ =
+
= = W
As  R R
1 2
¢ = ¢
Hence,
reading of V
1
= reading of V
2
=
+
´ =
1200
1200 1200
200 100 V
(b) Current distribution is shown in figure
i
E
R R
=
¢ + ¢
1 2
i = =
200
2400
1
12
A
i
R
R R
i
V
V
1
1
1
1
=
+
= ´
3000
5000
1
12
=
1
20
A
i
R
R R
i
V
2
1
1
1
2000
5000
1
12
=
+
= ´
=
1
30
A
\ Current flowing through
S i i = - = -
1 2
1
20
1
30
=
1
60
A
27. Ef fec tive emf of 2 V and 6 V bat ter ies
con nected in par al lel
E
E r E r
r r
¢ =
+
+
=
´ - ´
+
1 2 2 1
1 2
2 1 6 1
1 1
= - 2 V
and r
r r
r r
¢ =
+
=
1 2
1 2
1
2
W
= 0.5W
10
E
'
R
2
'
R
1
R
1
S
R
2
E
V
1
V
2
i
2
i
1
i
2 i
1
i
V
1
V
2
R
1 S
R
2
E
V
1
V
2
R
1 S
R
2
E
i
2
A
100W 200W
100W 200W
i
1
10 V
B
C
D
Net emf, E = - = 4 2 2 V
28. (a)
As E E
1 2
>
Current will flow from B to A.
(b) E
1
is doing positive work
(c) As current flows from B to A through
resistor, B is at higher potential.
29. i R
2
2 5 = < W W
Clearly X is doing negative work.
(a) P Vi = Þ V
P
i
= = =
0.5
1.0
5.0 V
(b) E V iR = - = - = 5 2 3 V
(c) It is clear from figure that positive
terminal of X is towards left.
30. i =
-
+
=
150 50
3 2
20 A
V V i
P Q
- = + 50 3.0
V
Q
= - + 100 50 60 ( )
= - 10 V
31. (a) As volt me ter is ideal, it has in fi nite
re sis tance, there fore cur ren t is zero.
(b) V E i r = - Þ = E 5 . 0 V
(c) Reading of voltmeter Þ = V 5.0 V
32. V E i r
1 1
= - Þ E r - = 1.5 8.4          …(i)
V E i r
2 2
= + Þ E r + = 3.5 9.4 …(ii)
On solving, we get
r = 0.2W
E = 8.7 V
33. In case of charg ing
V E i r = + = + ´ = 2 5 0.1 2.5 V
34. Clearly cur rent through each branch is zero.
35. i
E
R G
1
=
+
On shunt i ng the gal va nom e t er with
re sis tan ce S ,
i
E
R R
GS
S G
2
=
+ ¢ +
+
As i i
1 2
=
E
R G
E
R R
GS
G S
+
=
+ ¢ +
+
Þ R R
GS
G S
R G + ¢ +
+
= +
R
G
G S
¢ =
+
2
36.
I
r
R r
I
V
R
2
=
+
=
11
8W
8W 4W
2W
2W
2V 2V
Þ
2V 2V 2V
4W 4W 4W
2V
G
i
2
E
R R'
S
R=2.0W
i A
E
B
+ +
– –
A B
E
1
E
2
+ +
– –
Q
3.0W
2.0W
150 V
50 V
P
i
A
A
V
I
2
B
I r
I
1
G
i
1
E
R
4V
0.5W
2V
6V 1W
1W
ß
0.5W 0.5W
2V
4V
Þ
r
R r
V
IR +
=
Þ
R
r
IR V
V
=
-
=
´ - 5 2500 100
100
r = ´ =
100
12400
2500 20.16W
37.
Let R be the resistance of voltmeter
As reading of voltmeter is 30 V,
1 1
400
1
300 R
+ = Þ R = 1200W
If voltmeter is connected across 300W
resistor,
Effective resistance of 300W resistor and
voltmeter
R¢=
´
+
=
300 1200
300 1200
240W
i=
+
60
400 240
=
60
640
A
=
3
32
A
\ Reading of voltmeter,
V iR = ¢ = ´
3
32
240
= 22.5 V
38. V
R
R R
V
2
1 2
=
¢
+ ¢
,
R
rR
r R
2
2
2
120
3
¢ =
+
=
=40W
V
2
40
60 40
120 =
+
=48V
39. S
i
i i
G R
g
g
=
-
+ ( )
R
i i
i
S G
g
g
=
-
-
=
-
´ -
-
-
20 10
10
20
3
3
0.005
= 79.995 W
40. r
L L
L
R =
-
=
-
´ =
1 2
2
5
0.52 0.4
0.4
1.5 W
41. Let R be the re sis tance of voltmeter
R
R
R
e
= + +
+
3 2
100
100
= +
+
5
100
100
R
R

i
R
R
=
+
+
=
3.4
.04
5
100
100
0
Þ      0.2 3.4 +
+
=
4
100
R
R

Þ         R=400W
Reading of voltmeter,
V i
R
R
= ´
+
100
100
= ´
´
+
0.04
100 400
100 400
=3.2V
If the voltmeter had been ideal,
Reading of voltmeter
= ´ =
100
105
3.4 3.24 V
42.
L
L
R
R
1
2
1
2
=
Þ
L
L
1
1
40
8
12 -
= (L L
1 2
40 + = cm)
Þ       L
1
16 = cm   from A.
12
G B
R
S
A
60 V
V
300W 400W
60 V
V
300W 400W
3.4 V
A
i
3W
V
100W
43. S
i
i i
G R
g
g
=
-
+ ( )
Þ R
i i
i
S G
g
g
=
-
-
=
-
´ -
20 0.0224
0.0244
0.0250 9.36
=12.94W
44. (a) i
E
R r
V
=
+

V iR
E
R r
R
V
V
V
= =
+
(b)
r
R r
V
+
= = 1
1
100
%
R r
V
= = ´ 99 99 0.45
=44.55W
(c)
V
E
R
R r
V
V
=
+
As R
V
decreases, V decreases, decreasing
accuracy of voltmeter.
45. (a) When am me ter is con nected
I
E
R R r
A
A
=
+ +
When ammeter is removed
I
E
R r
R R r
R r
I
A
A
=
+
=
+ +
+
(b)
I
I
A
= 99%
R r
R R r
A
+
+ +
=
99
100

Þ R R r
A
= + = +
1
99
1
99
( ) (3.8 0.45)
R
A
= 0.043 W
(c) As
I
I
R r
R R r
A
A
=
+
+ +
, as R
A
increases, I
A
decreases, decreasing the accuracy of
ammeter.
46. I
R
max
max
= = =
r 36
15
2.4
A
For the given circuit
R R R R
e
= + =
1
2
3
2
Maximum power dissipated by the circuit
P I R
e
¢ =
max max
2
= ´ ´ = 15
3
2
54 2.4 W
47. To tal power of the cir cuit, P P P P = + +
1 2 3
= + + 40 60 75
= 175 W
As P
V
R
=
2
Þ R
V
P
=
2
= =
( ) 120
175
2
82.3 W
48. Ther mal power gen er ated in the battery
P i r i E V
1
2
= = - ( )
=0.6 W
Power development in the battery by electric
forces
P IE
2
= = 2.6 W
49. The given cir cuit can be con sid ered as the
sum of the cir cuit as shown.
\ P
1
7 2 14 = ´ = W,
P
2
1 1 1 = - ´ = W
50. (a) i
E E
R R
=
-
+
=
-
+
=
1 2
1 2
12 6
4 8
0.5 A
(b) Power dissipated in R I R
1
2
1
1 = = W
and power dissipated in R I R = =
2
2
2 W
(c) Power of battery E E I
1 1
=
= ´ = 12 6 0.5 W (supplied)
Power of battery E E I
2 2
=
= - ´ = - 6 3 0.6 W (absorbed)
13
E
i
r
R
3W
2W
2W
7V
21/6A
35
16
A
14/6A
+ 3W
2W
2/16A
5/16A
1V
ß
2W
2W
2A
7V
1A
3W 1V
3
16
A
E
V
R
v
i
r
```

## DC Pandey Solutions for NEET Physics

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## FAQs on DC Pandey Solutions: Current Electricity- 2 - DC Pandey Solutions for NEET Physics

 1. What is the formula for calculating electric current?
Ans. The formula for calculating electric current is I = Q/t, where I represents the current, Q represents the charge, and t represents the time.
 2. How is resistance related to current flow?
Ans. Resistance is directly proportional to the obstruction of current flow. Higher resistance leads to a decrease in current flow, while lower resistance allows for a higher current to pass through a circuit.
 3. What is Kirchhoff's current law?
Ans. Kirchhoff's current law states that the sum of currents entering a junction in an electrical circuit is equal to the sum of currents leaving the junction. It is based on the principle of conservation of electric charge.
 4. How does the length and cross-sectional area of a wire affect its resistance?
Ans. The length of a wire is directly proportional to its resistance, meaning that as the length increases, the resistance also increases. On the other hand, the cross-sectional area of a wire is inversely proportional to its resistance. A larger cross-sectional area results in lower resistance.
 5. What is the difference between series and parallel circuits in terms of current flow?
Ans. In a series circuit, the current remains the same at all points, while the voltage drops across each component. In a parallel circuit, the voltage remains the same across all components, but the current splits and flows through different paths.

## DC Pandey Solutions for NEET Physics

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