DC Pandey Solutions: Current Electricity- 3 Notes | EduRev

DC Pandey Solutions for JEE Physics

NEET : DC Pandey Solutions: Current Electricity- 3 Notes | EduRev

 Page 1


6. E I R r = + ( )
Case I
E r = + 0.5 3.75 ( )
Case II
E r = + 0.4 (4.75 )
On solving
r = 0.25 W, E = 2 V
7.
I
I
g
=
50
20
 Þ I I
g
=
5
2
S
I
I I
G
g
g
=
-
 Þ G
I I
I
S
g
g
=
-
= ´
3
2
12
= 18 W
8. I I I
g
= = 2
1
50
% 
S
I
I I
G
G g
g
=
-
=
49
9. P
V
R
=
2
P P
V
R R
+ =
+
D
D
2
As R l µ
DR R = -10%
DP
V
R
V
R
= -
2 2
0.9
 = -
æ
è
ç
ö
ø
÷
1
1
0.9
P
= »
10
9
11 P P %
10. Potential difference between any two points
is zero.
11.        r
l l
l
R =
-
1 2
2
         =
-
´
75 60
60
10
         =2.5W
12. (b) By ap ply ing KCL at O
I I I
1 2 3
0 + + =
6
6
3
3
2
2
0
0 0 0
-
+
-
+
-
=
V V V
Þ 6 2 3 3 2 0
0 0 0
- + - + - = V V V ( ) ( )
V
0
3 = V
13. v
I
neA
I
ne r
d
= =
p
2
v
I
ne r
v v
d
d
¢ = = =
2
2 2 2
2
p( )
14. Voltmeter has higher resistance than
ammeter. 
Again higher the range of voltmeter, higher
will be its resistance.
15. I
R
R R R
I
2
2
1 2 3
1
1 1 1
=
+ +
Þ 
1 1 1
1 2 2 2 3
R
I
I R R R
= - +
æ
è
ç
ç
ö
ø
÷
÷
      =
´
- +
æ
è
ç
ö
ø
÷
0.8
0.3 20
1
20
1
15
            =
1
60
         R
1
60 = W
16. (d) I
E
R R
A V
1
=
+
, V I R
V 1 1
=
= - E I R
A 1
If resistance is connected in parallel with
voltmeter,
 18
A
I
1
I
I
3
I
2
R=15W
3
R=20W
2
R
1
A
B
I
2
I
1
3W
6W
V
0 O
2W
I
3
C
A V
I
1
E
Page 2


6. E I R r = + ( )
Case I
E r = + 0.5 3.75 ( )
Case II
E r = + 0.4 (4.75 )
On solving
r = 0.25 W, E = 2 V
7.
I
I
g
=
50
20
 Þ I I
g
=
5
2
S
I
I I
G
g
g
=
-
 Þ G
I I
I
S
g
g
=
-
= ´
3
2
12
= 18 W
8. I I I
g
= = 2
1
50
% 
S
I
I I
G
G g
g
=
-
=
49
9. P
V
R
=
2
P P
V
R R
+ =
+
D
D
2
As R l µ
DR R = -10%
DP
V
R
V
R
= -
2 2
0.9
 = -
æ
è
ç
ö
ø
÷
1
1
0.9
P
= »
10
9
11 P P %
10. Potential difference between any two points
is zero.
11.        r
l l
l
R =
-
1 2
2
         =
-
´
75 60
60
10
         =2.5W
12. (b) By ap ply ing KCL at O
I I I
1 2 3
0 + + =
6
6
3
3
2
2
0
0 0 0
-
+
-
+
-
=
V V V
Þ 6 2 3 3 2 0
0 0 0
- + - + - = V V V ( ) ( )
V
0
3 = V
13. v
I
neA
I
ne r
d
= =
p
2
v
I
ne r
v v
d
d
¢ = = =
2
2 2 2
2
p( )
14. Voltmeter has higher resistance than
ammeter. 
Again higher the range of voltmeter, higher
will be its resistance.
15. I
R
R R R
I
2
2
1 2 3
1
1 1 1
=
+ +
Þ 
1 1 1
1 2 2 2 3
R
I
I R R R
= - +
æ
è
ç
ç
ö
ø
÷
÷
      =
´
- +
æ
è
ç
ö
ø
÷
0.8
0.3 20
1
20
1
15
            =
1
60
         R
1
60 = W
16. (d) I
E
R R
A V
1
=
+
, V I R
V 1 1
=
= - E I R
A 1
If resistance is connected in parallel with
voltmeter,
 18
A
I
1
I
I
3
I
2
R=15W
3
R=20W
2
R
1
A
B
I
2
I
1
3W
6W
V
0 O
2W
I
3
C
A V
I
1
E
I
E
R
RR
R R
I
A
V
V
2 1
=
+
+
>
and      V E I R V
A 2 2 1
= - <
17. Be fore con nec tiv ity re sis tance is par al lel
with ammeter
I
E
R R
A V
1
=
+
, V I R
V 1 1
=
= - E I R
A 1
After connecting resistance in parallel to the
ammeter.
I
E
R
R
A
V
2
2
=
+
, 
Reading of ammeter =
1
2
2
I
=
+
>
E
R R
I
A V
2
1
2
1
  V I R
E
R R
V
V
A V
= =
+
<
2 1
2
2
2
18. R
R
n
e
=
2
P
V
R
=
2
P
V
R
n P
e
e
= =
2
2
19. As bulb A is in se ries with en tire cir cuit.
20. I
E E
R r r R
=
+
+ +
=
+
1 2
1 2
18
3
 
V E Ir
ab
= - =
2 2
0
3
18
3
1 0 -
+
´ =
R
Þ R = 3 W
21. I
R
R R
I
V
2
=
+
V
R
R
R R
I
V V
=
+
100
2500 2500
5 =
+
R
R
R R + = 2500 125
R =
2500
24
W »100W
22.
23.
R
R
1
2
20
80
1
4
= =                        …(i)
R
R
1
2
15 40
60
2
3
+
= =
19 
V
A
I
2
I
1
R
I
R/n
R/n
R/n
A V
I
2
E
R
A
V A
I
1
E
R/10 R/10
R/10 R/10
R/10 R/10
R/10 R/10
R/10 R/10
ß
R/5
R/5
R/5
R/5
R/5
Þ
R/25
V A
I
2
E
R
Page 3


6. E I R r = + ( )
Case I
E r = + 0.5 3.75 ( )
Case II
E r = + 0.4 (4.75 )
On solving
r = 0.25 W, E = 2 V
7.
I
I
g
=
50
20
 Þ I I
g
=
5
2
S
I
I I
G
g
g
=
-
 Þ G
I I
I
S
g
g
=
-
= ´
3
2
12
= 18 W
8. I I I
g
= = 2
1
50
% 
S
I
I I
G
G g
g
=
-
=
49
9. P
V
R
=
2
P P
V
R R
+ =
+
D
D
2
As R l µ
DR R = -10%
DP
V
R
V
R
= -
2 2
0.9
 = -
æ
è
ç
ö
ø
÷
1
1
0.9
P
= »
10
9
11 P P %
10. Potential difference between any two points
is zero.
11.        r
l l
l
R =
-
1 2
2
         =
-
´
75 60
60
10
         =2.5W
12. (b) By ap ply ing KCL at O
I I I
1 2 3
0 + + =
6
6
3
3
2
2
0
0 0 0
-
+
-
+
-
=
V V V
Þ 6 2 3 3 2 0
0 0 0
- + - + - = V V V ( ) ( )
V
0
3 = V
13. v
I
neA
I
ne r
d
= =
p
2
v
I
ne r
v v
d
d
¢ = = =
2
2 2 2
2
p( )
14. Voltmeter has higher resistance than
ammeter. 
Again higher the range of voltmeter, higher
will be its resistance.
15. I
R
R R R
I
2
2
1 2 3
1
1 1 1
=
+ +
Þ 
1 1 1
1 2 2 2 3
R
I
I R R R
= - +
æ
è
ç
ç
ö
ø
÷
÷
      =
´
- +
æ
è
ç
ö
ø
÷
0.8
0.3 20
1
20
1
15
            =
1
60
         R
1
60 = W
16. (d) I
E
R R
A V
1
=
+
, V I R
V 1 1
=
= - E I R
A 1
If resistance is connected in parallel with
voltmeter,
 18
A
I
1
I
I
3
I
2
R=15W
3
R=20W
2
R
1
A
B
I
2
I
1
3W
6W
V
0 O
2W
I
3
C
A V
I
1
E
I
E
R
RR
R R
I
A
V
V
2 1
=
+
+
>
and      V E I R V
A 2 2 1
= - <
17. Be fore con nec tiv ity re sis tance is par al lel
with ammeter
I
E
R R
A V
1
=
+
, V I R
V 1 1
=
= - E I R
A 1
After connecting resistance in parallel to the
ammeter.
I
E
R
R
A
V
2
2
=
+
, 
Reading of ammeter =
1
2
2
I
=
+
>
E
R R
I
A V
2
1
2
1
  V I R
E
R R
V
V
A V
= =
+
<
2 1
2
2
2
18. R
R
n
e
=
2
P
V
R
=
2
P
V
R
n P
e
e
= =
2
2
19. As bulb A is in se ries with en tire cir cuit.
20. I
E E
R r r R
=
+
+ +
=
+
1 2
1 2
18
3
 
V E Ir
ab
= - =
2 2
0
3
18
3
1 0 -
+
´ =
R
Þ R = 3 W
21. I
R
R R
I
V
2
=
+
V
R
R
R R
I
V V
=
+
100
2500 2500
5 =
+
R
R
R R + = 2500 125
R =
2500
24
W »100W
22.
23.
R
R
1
2
20
80
1
4
= =                        …(i)
R
R
1
2
15 40
60
2
3
+
= =
19 
V
A
I
2
I
1
R
I
R/n
R/n
R/n
A V
I
2
E
R
A
V A
I
1
E
R/10 R/10
R/10 R/10
R/10 R/10
R/10 R/10
R/10 R/10
ß
R/5
R/5
R/5
R/5
R/5
Þ
R/25
V A
I
2
E
R
R
R R
1
2 2
15 2
3
+ =
15 2
3
1
4
5
12
2
R
= - =
Þ R
2
36 = W,
R
R
1
2
4
9 = = W
24. (b) As V
V
1
2
= , R R
1 2
=
R
R
V
V
´
+
=
100
100
50
R
V
= 100 W
25. (d) 
E
R r
PB
+
=
+
=
2
4 1
0.4 W
V IR
AB AB
= = 1.6 W
K
V
L
PB
= = =
1.6
0.016
100
 V/cm
L
E
K
= = =
1
75
1.2
0.016
 cm
26. (d) V
AB
= ´ + + ´ - + ´ 3 2 3 1 4 2 6 1
       =17 V
27. (c) E
E r E r
r r
e
=
+
+
=
1 2 2 1
1 2
2 V
r
r r
r r
e
=
+
=
1 2
1 2
0.5 W
For maximum power R r
e
=
and P
E
r
e
e
max
( )
= =
´
=
2 2
4
2
4
2
0.5
 W
28. (a) V
R
R r
E =
+
 
r
E
V
R = -
æ
è
ç
ö
ø
÷
= -
æ
è
ç
ö
ø
÷
1 1 5
2.2
1.8
         =
10
9
W
29. (d) I
E E
R R r r
=
-
+ + +
1 2
1 2 1 2
    =
-
+ + +
10 5
25 15 2.5 2.5
=
1
9
A
V I
AB
= - + ( ) 25 15
= - ´ » -
1
9
40 4 V
30. (a) V kL
AB
= = ´ 0.2 100 = 20 mV
V
R
R R
E
AB
AB
AB
=
+
Þ  0.02 =
+
´
R
R
AB
AB
490
2
Þ R R
AB AB
+ = 490 100
R
AB
= »
490
99
4.9W
31. (c) When key is open,
I
E
R
1
2
3
=
When key is closed
I
E
R
2
3
4
=
\
I
I
1
2
8
9
=
 20
2W 2W
1W
1W 2W 2W
A
3V 2V
B
3A
1A 2A
A
E
I
1
R 2R
2R
R
Þ
A
E
3R
3R
I
1
ß
A
E
3R
3/2R
I
1
A
E
I
2
R 2R
2R
R
Þ
A
E
I
2
ß
A
4R/3
I
2
2R/3 2R/3
Page 4


6. E I R r = + ( )
Case I
E r = + 0.5 3.75 ( )
Case II
E r = + 0.4 (4.75 )
On solving
r = 0.25 W, E = 2 V
7.
I
I
g
=
50
20
 Þ I I
g
=
5
2
S
I
I I
G
g
g
=
-
 Þ G
I I
I
S
g
g
=
-
= ´
3
2
12
= 18 W
8. I I I
g
= = 2
1
50
% 
S
I
I I
G
G g
g
=
-
=
49
9. P
V
R
=
2
P P
V
R R
+ =
+
D
D
2
As R l µ
DR R = -10%
DP
V
R
V
R
= -
2 2
0.9
 = -
æ
è
ç
ö
ø
÷
1
1
0.9
P
= »
10
9
11 P P %
10. Potential difference between any two points
is zero.
11.        r
l l
l
R =
-
1 2
2
         =
-
´
75 60
60
10
         =2.5W
12. (b) By ap ply ing KCL at O
I I I
1 2 3
0 + + =
6
6
3
3
2
2
0
0 0 0
-
+
-
+
-
=
V V V
Þ 6 2 3 3 2 0
0 0 0
- + - + - = V V V ( ) ( )
V
0
3 = V
13. v
I
neA
I
ne r
d
= =
p
2
v
I
ne r
v v
d
d
¢ = = =
2
2 2 2
2
p( )
14. Voltmeter has higher resistance than
ammeter. 
Again higher the range of voltmeter, higher
will be its resistance.
15. I
R
R R R
I
2
2
1 2 3
1
1 1 1
=
+ +
Þ 
1 1 1
1 2 2 2 3
R
I
I R R R
= - +
æ
è
ç
ç
ö
ø
÷
÷
      =
´
- +
æ
è
ç
ö
ø
÷
0.8
0.3 20
1
20
1
15
            =
1
60
         R
1
60 = W
16. (d) I
E
R R
A V
1
=
+
, V I R
V 1 1
=
= - E I R
A 1
If resistance is connected in parallel with
voltmeter,
 18
A
I
1
I
I
3
I
2
R=15W
3
R=20W
2
R
1
A
B
I
2
I
1
3W
6W
V
0 O
2W
I
3
C
A V
I
1
E
I
E
R
RR
R R
I
A
V
V
2 1
=
+
+
>
and      V E I R V
A 2 2 1
= - <
17. Be fore con nec tiv ity re sis tance is par al lel
with ammeter
I
E
R R
A V
1
=
+
, V I R
V 1 1
=
= - E I R
A 1
After connecting resistance in parallel to the
ammeter.
I
E
R
R
A
V
2
2
=
+
, 
Reading of ammeter =
1
2
2
I
=
+
>
E
R R
I
A V
2
1
2
1
  V I R
E
R R
V
V
A V
= =
+
<
2 1
2
2
2
18. R
R
n
e
=
2
P
V
R
=
2
P
V
R
n P
e
e
= =
2
2
19. As bulb A is in se ries with en tire cir cuit.
20. I
E E
R r r R
=
+
+ +
=
+
1 2
1 2
18
3
 
V E Ir
ab
= - =
2 2
0
3
18
3
1 0 -
+
´ =
R
Þ R = 3 W
21. I
R
R R
I
V
2
=
+
V
R
R
R R
I
V V
=
+
100
2500 2500
5 =
+
R
R
R R + = 2500 125
R =
2500
24
W »100W
22.
23.
R
R
1
2
20
80
1
4
= =                        …(i)
R
R
1
2
15 40
60
2
3
+
= =
19 
V
A
I
2
I
1
R
I
R/n
R/n
R/n
A V
I
2
E
R
A
V A
I
1
E
R/10 R/10
R/10 R/10
R/10 R/10
R/10 R/10
R/10 R/10
ß
R/5
R/5
R/5
R/5
R/5
Þ
R/25
V A
I
2
E
R
R
R R
1
2 2
15 2
3
+ =
15 2
3
1
4
5
12
2
R
= - =
Þ R
2
36 = W,
R
R
1
2
4
9 = = W
24. (b) As V
V
1
2
= , R R
1 2
=
R
R
V
V
´
+
=
100
100
50
R
V
= 100 W
25. (d) 
E
R r
PB
+
=
+
=
2
4 1
0.4 W
V IR
AB AB
= = 1.6 W
K
V
L
PB
= = =
1.6
0.016
100
 V/cm
L
E
K
= = =
1
75
1.2
0.016
 cm
26. (d) V
AB
= ´ + + ´ - + ´ 3 2 3 1 4 2 6 1
       =17 V
27. (c) E
E r E r
r r
e
=
+
+
=
1 2 2 1
1 2
2 V
r
r r
r r
e
=
+
=
1 2
1 2
0.5 W
For maximum power R r
e
=
and P
E
r
e
e
max
( )
= =
´
=
2 2
4
2
4
2
0.5
 W
28. (a) V
R
R r
E =
+
 
r
E
V
R = -
æ
è
ç
ö
ø
÷
= -
æ
è
ç
ö
ø
÷
1 1 5
2.2
1.8
         =
10
9
W
29. (d) I
E E
R R r r
=
-
+ + +
1 2
1 2 1 2
    =
-
+ + +
10 5
25 15 2.5 2.5
=
1
9
A
V I
AB
= - + ( ) 25 15
= - ´ » -
1
9
40 4 V
30. (a) V kL
AB
= = ´ 0.2 100 = 20 mV
V
R
R R
E
AB
AB
AB
=
+
Þ  0.02 =
+
´
R
R
AB
AB
490
2
Þ R R
AB AB
+ = 490 100
R
AB
= »
490
99
4.9W
31. (c) When key is open,
I
E
R
1
2
3
=
When key is closed
I
E
R
2
3
4
=
\
I
I
1
2
8
9
=
 20
2W 2W
1W
1W 2W 2W
A
3V 2V
B
3A
1A 2A
A
E
I
1
R 2R
2R
R
Þ
A
E
3R
3R
I
1
ß
A
E
3R
3/2R
I
1
A
E
I
2
R 2R
2R
R
Þ
A
E
I
2
ß
A
4R/3
I
2
2R/3 2R/3
32. (b) S
I
I I
G
g
g
=
-
 = ´
1
34
33
34
3663
I
I
 
= 111 W
   R r
e
=
5
11
r = ´
11
5
1.5
= 3.3 W
As the circuit is symmetrical about
perpendicular bisector of AB, all points
lying on it are at same potential.
35. (c) R
L
L L
R
1
1
1 2
=
+
Þ R
R
1
6
3 = = W
R
l
l l
2
2
1 2
=
+
 Þ  R
2
15 = W
Hence R
1
 and R
2
 are in parallel
R
R R
R R
e
=
+
1 2
1 2
= 2.5W
36. (c) Let R R x
AD BC
= =
Clearly x < 1 as 1W resistor is in parallel
with some combination.
Now R x x
AB
= + + 1
= + 2 1 x
As x < 1
1 3 < < R
AB
37. (d) R R
R R R
R R
R
AB
= +
+
+
=
( )
0
0
0
2
Þ 2 2
2
0
2
0 0 0
2
R RR R RR RR R + + + = +
Þ 3
2
0
2
R R =
Þ R
R
=
0
3
38.
21 
33. B A
r
r
r
r
r
r
r
r
r
r
r
r
r
r
2r
B
A
Þ
ß
A
B
r
r
2r/3
2r/3
r
A
B
Þ
5/3r
5/3r
r
ß
5/11r
2r
(b)
34.
r
r
r
r
r
r
r
r
r
r
A
B
Þ
r
r
r
r
r
r
r
r
A
ß
B
2r
2r
2r
2r
B A
Þ B A
r/2
(b)
R
2
R
1
a
b
60°
1W
1W
1W
1W
1W
1W
1W 1W
1W 1W
1W 1W
1W
B A
P
Q
R
R
R
R
Þ
P
Q
R R/2
R/3
ß
P
Q
R
5
11
R Ü
5
6
R = 25W
R
R
Page 5


6. E I R r = + ( )
Case I
E r = + 0.5 3.75 ( )
Case II
E r = + 0.4 (4.75 )
On solving
r = 0.25 W, E = 2 V
7.
I
I
g
=
50
20
 Þ I I
g
=
5
2
S
I
I I
G
g
g
=
-
 Þ G
I I
I
S
g
g
=
-
= ´
3
2
12
= 18 W
8. I I I
g
= = 2
1
50
% 
S
I
I I
G
G g
g
=
-
=
49
9. P
V
R
=
2
P P
V
R R
+ =
+
D
D
2
As R l µ
DR R = -10%
DP
V
R
V
R
= -
2 2
0.9
 = -
æ
è
ç
ö
ø
÷
1
1
0.9
P
= »
10
9
11 P P %
10. Potential difference between any two points
is zero.
11.        r
l l
l
R =
-
1 2
2
         =
-
´
75 60
60
10
         =2.5W
12. (b) By ap ply ing KCL at O
I I I
1 2 3
0 + + =
6
6
3
3
2
2
0
0 0 0
-
+
-
+
-
=
V V V
Þ 6 2 3 3 2 0
0 0 0
- + - + - = V V V ( ) ( )
V
0
3 = V
13. v
I
neA
I
ne r
d
= =
p
2
v
I
ne r
v v
d
d
¢ = = =
2
2 2 2
2
p( )
14. Voltmeter has higher resistance than
ammeter. 
Again higher the range of voltmeter, higher
will be its resistance.
15. I
R
R R R
I
2
2
1 2 3
1
1 1 1
=
+ +
Þ 
1 1 1
1 2 2 2 3
R
I
I R R R
= - +
æ
è
ç
ç
ö
ø
÷
÷
      =
´
- +
æ
è
ç
ö
ø
÷
0.8
0.3 20
1
20
1
15
            =
1
60
         R
1
60 = W
16. (d) I
E
R R
A V
1
=
+
, V I R
V 1 1
=
= - E I R
A 1
If resistance is connected in parallel with
voltmeter,
 18
A
I
1
I
I
3
I
2
R=15W
3
R=20W
2
R
1
A
B
I
2
I
1
3W
6W
V
0 O
2W
I
3
C
A V
I
1
E
I
E
R
RR
R R
I
A
V
V
2 1
=
+
+
>
and      V E I R V
A 2 2 1
= - <
17. Be fore con nec tiv ity re sis tance is par al lel
with ammeter
I
E
R R
A V
1
=
+
, V I R
V 1 1
=
= - E I R
A 1
After connecting resistance in parallel to the
ammeter.
I
E
R
R
A
V
2
2
=
+
, 
Reading of ammeter =
1
2
2
I
=
+
>
E
R R
I
A V
2
1
2
1
  V I R
E
R R
V
V
A V
= =
+
<
2 1
2
2
2
18. R
R
n
e
=
2
P
V
R
=
2
P
V
R
n P
e
e
= =
2
2
19. As bulb A is in se ries with en tire cir cuit.
20. I
E E
R r r R
=
+
+ +
=
+
1 2
1 2
18
3
 
V E Ir
ab
= - =
2 2
0
3
18
3
1 0 -
+
´ =
R
Þ R = 3 W
21. I
R
R R
I
V
2
=
+
V
R
R
R R
I
V V
=
+
100
2500 2500
5 =
+
R
R
R R + = 2500 125
R =
2500
24
W »100W
22.
23.
R
R
1
2
20
80
1
4
= =                        …(i)
R
R
1
2
15 40
60
2
3
+
= =
19 
V
A
I
2
I
1
R
I
R/n
R/n
R/n
A V
I
2
E
R
A
V A
I
1
E
R/10 R/10
R/10 R/10
R/10 R/10
R/10 R/10
R/10 R/10
ß
R/5
R/5
R/5
R/5
R/5
Þ
R/25
V A
I
2
E
R
R
R R
1
2 2
15 2
3
+ =
15 2
3
1
4
5
12
2
R
= - =
Þ R
2
36 = W,
R
R
1
2
4
9 = = W
24. (b) As V
V
1
2
= , R R
1 2
=
R
R
V
V
´
+
=
100
100
50
R
V
= 100 W
25. (d) 
E
R r
PB
+
=
+
=
2
4 1
0.4 W
V IR
AB AB
= = 1.6 W
K
V
L
PB
= = =
1.6
0.016
100
 V/cm
L
E
K
= = =
1
75
1.2
0.016
 cm
26. (d) V
AB
= ´ + + ´ - + ´ 3 2 3 1 4 2 6 1
       =17 V
27. (c) E
E r E r
r r
e
=
+
+
=
1 2 2 1
1 2
2 V
r
r r
r r
e
=
+
=
1 2
1 2
0.5 W
For maximum power R r
e
=
and P
E
r
e
e
max
( )
= =
´
=
2 2
4
2
4
2
0.5
 W
28. (a) V
R
R r
E =
+
 
r
E
V
R = -
æ
è
ç
ö
ø
÷
= -
æ
è
ç
ö
ø
÷
1 1 5
2.2
1.8
         =
10
9
W
29. (d) I
E E
R R r r
=
-
+ + +
1 2
1 2 1 2
    =
-
+ + +
10 5
25 15 2.5 2.5
=
1
9
A
V I
AB
= - + ( ) 25 15
= - ´ » -
1
9
40 4 V
30. (a) V kL
AB
= = ´ 0.2 100 = 20 mV
V
R
R R
E
AB
AB
AB
=
+
Þ  0.02 =
+
´
R
R
AB
AB
490
2
Þ R R
AB AB
+ = 490 100
R
AB
= »
490
99
4.9W
31. (c) When key is open,
I
E
R
1
2
3
=
When key is closed
I
E
R
2
3
4
=
\
I
I
1
2
8
9
=
 20
2W 2W
1W
1W 2W 2W
A
3V 2V
B
3A
1A 2A
A
E
I
1
R 2R
2R
R
Þ
A
E
3R
3R
I
1
ß
A
E
3R
3/2R
I
1
A
E
I
2
R 2R
2R
R
Þ
A
E
I
2
ß
A
4R/3
I
2
2R/3 2R/3
32. (b) S
I
I I
G
g
g
=
-
 = ´
1
34
33
34
3663
I
I
 
= 111 W
   R r
e
=
5
11
r = ´
11
5
1.5
= 3.3 W
As the circuit is symmetrical about
perpendicular bisector of AB, all points
lying on it are at same potential.
35. (c) R
L
L L
R
1
1
1 2
=
+
Þ R
R
1
6
3 = = W
R
l
l l
2
2
1 2
=
+
 Þ  R
2
15 = W
Hence R
1
 and R
2
 are in parallel
R
R R
R R
e
=
+
1 2
1 2
= 2.5W
36. (c) Let R R x
AD BC
= =
Clearly x < 1 as 1W resistor is in parallel
with some combination.
Now R x x
AB
= + + 1
= + 2 1 x
As x < 1
1 3 < < R
AB
37. (d) R R
R R R
R R
R
AB
= +
+
+
=
( )
0
0
0
2
Þ 2 2
2
0
2
0 0 0
2
R RR R RR RR R + + + = +
Þ 3
2
0
2
R R =
Þ R
R
=
0
3
38.
21 
33. B A
r
r
r
r
r
r
r
r
r
r
r
r
r
r
2r
B
A
Þ
ß
A
B
r
r
2r/3
2r/3
r
A
B
Þ
5/3r
5/3r
r
ß
5/11r
2r
(b)
34.
r
r
r
r
r
r
r
r
r
r
A
B
Þ
r
r
r
r
r
r
r
r
A
ß
B
2r
2r
2r
2r
B A
Þ B A
r/2
(b)
R
2
R
1
a
b
60°
1W
1W
1W
1W
1W
1W
1W 1W
1W 1W
1W 1W
1W
B A
P
Q
R
R
R
R
Þ
P
Q
R R/2
R/3
ß
P
Q
R
5
11
R Ü
5
6
R = 25W
R
R
39. Wheatstone bridge is bal anced.
Þ   R R
e
=
7
6
40. (d) R
L
L L
R R
1
1
1 2
1
12
=
+
= 
= 3 W 
R
L
L L
R
2
2
1 2
11
12
=
+
= = 33 W
R
1
 and R
2
 are in parallel,
R
R R
R R
e
=
+
=
´
+
1 2
1 2
3 33
3 33
           =2.75W
41. (a) Re sis tance per unit length of wire
               =
4
2pr
R
r
r R
1 2
4
2
2 = ´ = =
p
p
        R
r
r
3
4
2
2
4
= ´ =
p p
\         
1 1 1 1
1 2 3
R R R R
e
= + +
= + + =
+ 1
2
1
2 4
4
4
p p
 
        R
e
=
+
4
4 p
W
42. (d) Points C and D are shorted hence the
por tion above line CD can be re moved.
43. (b) As AB is line of sym me try,
we can fold the network about AB.
 22
A
B
30°
O
A
R/2 R/2
R/2
R/2
R/2
R/2
B
Þ
R/2 R/2
B
A
R
R
ß
3R/2
A B
A
R
R
R
2R/3
2R/3 2R/3
2R/3
Ü
P
R
R R
R R
R
R
R
R
R
R
R
R
B
B
ß
Þ Þ
4R/3
4R/3
4R/3
4R/3
7R/3 7R/3
R R R
R
A
B
A
R R
R
R
R R
R
R
R
B
R
A
B
R
2 R
3
R
1
D
B A
C
ß
D
B A
C
R/2 R/2
R
B
A
Ü
R
R
A B
R/2
B A
Ü
ß
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