Current Electricity Practice Questions - DPP for JEE

 Page 1


1. (a)
I = 
Potential difference across second cell
= 
 – 
= 0
 R = 
2. (a)
hence R
eq
 = 3/2;
3. (a) Let, we connect 24 cells in n rows of m cells, then if I is the
current in external circuit then
Page 2


1. (a)
I = 
Potential difference across second cell
= 
 – 
= 0
 R = 
2. (a)
hence R
eq
 = 3/2;
3. (a) Let, we connect 24 cells in n rows of m cells, then if I is the
current in external circuit then
...(1)
For I to be maximum, (mr + nR) should be minimum.
It is minimum for ...(2)
So maximum current in external circuit is
...(3)
here R =3, r = 0.5 so equation (2) become 
so n = 2, m = 12
4. (d)
Applying Kirchhoff ’s rule in loop abcfa
e
1
 – (i
1
 + i
2
) R – i
1
  r
1
 = 0.
5. (b) Current, × 1.6 × 10
–19
 = 0.66A towards
right.
6. (d) From the curve it is clear that slopes at points A, B, C, D have
following order A > B > C > D.
And also resistance at any point equals to slope of the V-i curve.
So order of resistance at three points will be
R
A
 > R
B
 > R
C
 > R
D
 
Page 3


1. (a)
I = 
Potential difference across second cell
= 
 – 
= 0
 R = 
2. (a)
hence R
eq
 = 3/2;
3. (a) Let, we connect 24 cells in n rows of m cells, then if I is the
current in external circuit then
...(1)
For I to be maximum, (mr + nR) should be minimum.
It is minimum for ...(2)
So maximum current in external circuit is
...(3)
here R =3, r = 0.5 so equation (2) become 
so n = 2, m = 12
4. (d)
Applying Kirchhoff ’s rule in loop abcfa
e
1
 – (i
1
 + i
2
) R – i
1
  r
1
 = 0.
5. (b) Current, × 1.6 × 10
–19
 = 0.66A towards
right.
6. (d) From the curve it is clear that slopes at points A, B, C, D have
following order A > B > C > D.
And also resistance at any point equals to slope of the V-i curve.
So order of resistance at three points will be
R
A
 > R
B
 > R
C
 > R
D
 
7. (d)
R
net
 between AB = 
8. (d) . But m = pr
2
 ld 
, , 
Page 4


1. (a)
I = 
Potential difference across second cell
= 
 – 
= 0
 R = 
2. (a)
hence R
eq
 = 3/2;
3. (a) Let, we connect 24 cells in n rows of m cells, then if I is the
current in external circuit then
...(1)
For I to be maximum, (mr + nR) should be minimum.
It is minimum for ...(2)
So maximum current in external circuit is
...(3)
here R =3, r = 0.5 so equation (2) become 
so n = 2, m = 12
4. (d)
Applying Kirchhoff ’s rule in loop abcfa
e
1
 – (i
1
 + i
2
) R – i
1
  r
1
 = 0.
5. (b) Current, × 1.6 × 10
–19
 = 0.66A towards
right.
6. (d) From the curve it is clear that slopes at points A, B, C, D have
following order A > B > C > D.
And also resistance at any point equals to slope of the V-i curve.
So order of resistance at three points will be
R
A
 > R
B
 > R
C
 > R
D
 
7. (d)
R
net
 between AB = 
8. (d) . But m = pr
2
 ld 
, , 
9. (a) Consider an element part of solid at a distance x from left end of
width dx.
Resistance of this elemental part is,
Current through cylinder is, I = 
Potential drop across element is, dV = I dR = 
10. (a) The simplified circuit is
Page 5


1. (a)
I = 
Potential difference across second cell
= 
 – 
= 0
 R = 
2. (a)
hence R
eq
 = 3/2;
3. (a) Let, we connect 24 cells in n rows of m cells, then if I is the
current in external circuit then
...(1)
For I to be maximum, (mr + nR) should be minimum.
It is minimum for ...(2)
So maximum current in external circuit is
...(3)
here R =3, r = 0.5 so equation (2) become 
so n = 2, m = 12
4. (d)
Applying Kirchhoff ’s rule in loop abcfa
e
1
 – (i
1
 + i
2
) R – i
1
  r
1
 = 0.
5. (b) Current, × 1.6 × 10
–19
 = 0.66A towards
right.
6. (d) From the curve it is clear that slopes at points A, B, C, D have
following order A > B > C > D.
And also resistance at any point equals to slope of the V-i curve.
So order of resistance at three points will be
R
A
 > R
B
 > R
C
 > R
D
 
7. (d)
R
net
 between AB = 
8. (d) . But m = pr
2
 ld 
, , 
9. (a) Consider an element part of solid at a distance x from left end of
width dx.
Resistance of this elemental part is,
Current through cylinder is, I = 
Potential drop across element is, dV = I dR = 
10. (a) The simplified circuit is
We have to find I.
Let potential of point P be 0.  Potential at other points are shown in the
figure apply Kirchoff’s current law at B where potential is assume
to be x volt.
?  x – 10 + 2x – 20 + x – 20 + 2x – 20 = 0 
? 6x = 70  ?  volt
?   
11. (c) In series, R
s
 = nR
In parallel, 
? R
s
/R
p
 = n
2
 /1 = n
2
12. (c) By principle of symmetry and superposition,
(Current  in AB is due to division in current entering at A and current 
is due to current returning from infinity of grid).