NEET Exam > NEET Notes > DC Pandey Solutions for NEET Physics > DC Pandey Solutions: Current Electricity- 3
DC Pandey Solutions: Current Electricity- 3 | DC Pandey Solutions for NEET Physics PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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
Read More
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.
About this Document
|
4.75/5 Rating |
|
Nov 23, 2024 Last updated |
Document Description: DC Pandey Solutions: Current Electricity- 3 for NEET 2024 is part of DC Pandey Solutions for NEET Physics preparation.
The notes and questions for DC Pandey Solutions: Current Electricity- 3 have been prepared according to the NEET exam syllabus. Information about DC Pandey Solutions: Current Electricity- 3 covers topics
like and DC Pandey Solutions: Current Electricity- 3 Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for DC Pandey Solutions: Current Electricity- 3.
Introduction of DC Pandey Solutions: Current Electricity- 3 in English is available as part of our DC Pandey Solutions for NEET Physics
for NEET & DC Pandey Solutions: Current Electricity- 3 in Hindi for DC Pandey Solutions for NEET Physics course.
Download more important topics related with notes, lectures and mock test series for NEET
Exam by signing up for free. NEET: DC Pandey Solutions: Current Electricity- 3 | DC Pandey Solutions for NEET Physics
Description
Full syllabus notes, lecture & questions for DC Pandey Solutions: Current Electricity- 3 | DC Pandey Solutions for NEET Physics - NEET | Plus excerises question with solution to help you revise complete syllabus for DC Pandey Solutions for NEET Physics | Best notes, free PDF download
Information about DC Pandey Solutions: Current Electricity- 3
In this doc you can find the meaning of DC Pandey Solutions: Current Electricity- 3 defined & explained in the simplest way possible. Besides explaining types of
DC Pandey Solutions: Current Electricity- 3 theory, EduRev gives you an ample number of questions to practice DC Pandey Solutions: Current Electricity- 3 tests, examples and also practice NEET
tests
|
Explore Courses for NEET exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Extra Questions
,Exam
,Previous Year Questions with Solutions
,ppt
,DC Pandey Solutions: Current Electricity- 3 | DC Pandey Solutions for NEET Physics
,Important questions
,DC Pandey Solutions: Current Electricity- 3 | DC Pandey Solutions for NEET Physics
,Viva Questions
,shortcuts and tricks
,Sample Paper
,Objective type Questions
,DC Pandey Solutions: Current Electricity- 3 | DC Pandey Solutions for NEET Physics
,MCQs
,Semester Notes
,practice quizzes
,video lectures
,study material
,mock tests for examination
,past year papers
,Summary
,Free
;
Additional Information about DC Pandey Solutions: Current Electricity- 3 for NEET Preparation
DC Pandey Solutions: Current Electricity- 3 Free PDF Download
The DC Pandey Solutions: Current Electricity- 3 is an invaluable resource that delves deep into the core of the NEET exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the DC Pandey Solutions: Current Electricity- 3 now and kickstart your journey towards success in the NEET exam.
Importance of DC Pandey Solutions: Current Electricity- 3
The importance of DC Pandey Solutions: Current Electricity- 3 cannot be overstated, especially for NEET aspirants.
This document holds the key to success in the NEET exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
DC Pandey Solutions: Current Electricity- 3 Notes
DC Pandey Solutions: Current Electricity- 3 Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to DC Pandey Solutions: Current Electricity- 3.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, DC Pandey Solutions: Current Electricity- 3 Notes on EduRev are your ultimate resource for success.
DC Pandey Solutions: Current Electricity- 3 NEET Questions
The "DC Pandey Solutions: Current Electricity- 3 NEET Questions" guide is a valuable resource for all aspiring students preparing for the
NEET exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study DC Pandey Solutions: Current Electricity- 3 on the App
Students of NEET can study DC Pandey Solutions: Current Electricity- 3 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the DC Pandey Solutions: Current Electricity- 3,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of DC Pandey Solutions: Current Electricity- 3 is prepared as per the latest NEET syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup