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# Chapter 20 - Current Electricity (Part - 2) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

## NEET : Chapter 20 - Current Electricity (Part - 2) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

``` 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
V E ir = -
= - ´ 4.3 0.102 1
»4.2W
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
V E ir = -
= - ´ 4.3 0.102 1
»4.2W
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
= 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,
1
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
V E ir = -
= - ´ 4.3 0.102 1
»4.2W
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
= 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,
1
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
V E ir = -
= - ´ 4.3 0.102 1
»4.2W
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
= 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,
1
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
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
V i
R
R
= ´
+
100
100
= ´
´
+
0.04
100 400
100 400
=3.2V
If the voltmeter had been ideal,
= ´ =
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
V E ir = -
= - ´ 4.3 0.102 1
»4.2W
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
= 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,
1
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
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
V i
R
R
= ´
+
100
100
= ´
´
+
0.04
100 400
100 400
=3.2V
If the voltmeter had been ideal,
= ´ =
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
```
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