DC Pandey Solutions: Current Electricity- 3

# DC Pandey Solutions: Current Electricity- 3 | DC Pandey Solutions for NEET Physics PDF Download

``` 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
=
+
,
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
=
+
,
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
=
+
,
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
= =
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
=
+
,
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
= =
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
```

## DC Pandey Solutions for NEET Physics

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## FAQs on DC Pandey Solutions: Current Electricity- 3 - 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 current in amperes (A), Q is the charge in coulombs (C), and t is the time in seconds (s).
 2. How is resistance related to current in a circuit?
Ans. According to Ohm's Law, the resistance in a circuit is directly proportional to the current flowing through it. This means that as the resistance increases, the current decreases, and vice versa.
 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 changes its direction periodically, while DC flows in only one direction. AC is commonly used for household electricity, while DC is used in batteries and electronic devices.
 4. How is electric power calculated in a circuit?
Ans. Electric power in a circuit is calculated using the formula P = VI, where P is the power in watts (W), V is the voltage in volts (V), and I is the current in amperes (A).
 5. What is the unit of electric current?
Ans. The unit of electric current is the ampere (A). It is named after the French physicist André-Marie Ampère, who made significant contributions to the understanding of electromagnetism.

## DC Pandey Solutions for NEET Physics

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