Current Electricity– 2 Practice Questions - DPP for NEET

 Page 1


(1) (d). Let R is resistance of the voltmeter. The effective
resistance across points A, B is
r = 
60R
60R
´
+
..............(1)
The current in the circuit is I = 12/(50 + r)
The p.d. across AB points is V = Ir
or   6 = 
12
50r +
 × r or 50 + r = 2r
or   r = 50 W ..............(2)
using it in (1),
we get 50 = 
60
60R +
300 + 5R = 6R
or  R = 300 W
(2) (c).  S = 
G GG
n 1 100 1 99
= =W
--
(3) (b). S = 
g
g
i
ii -
 G = 
10
100 10 -
 × 99 = 11 1W
(4) (c). R = 20 + 
80 80
80 80
´
+
 = 60W
i = 
V 21
R 60 30
== amp.
\  V = iR' = 
1
30
 × 40 = 1.33 volt.
(5) (a).  According to Kirchhoff's first law
At  junection A, i
AB
 = 2 + 2 = 4A
At junection B, i
AB
 = i
BC
 – 1 = 3A
2 A
2 A
1 A
1.3 A
i
C
B A
At junection C, i = i
BC
 – 1.3 = 3 – 1.3 = 17 amp
(6) (b). The current required for a full-scale deflection of the
galvanometer is
i = 4.0 x 10
-4
 x 25 = 10
-2
 A
Let a resistance R W is to be connected in series
Then by the ohm's law, we have i = 
V
GR +
Here G = 50 W, V = 2.5 V and i = 10
-2
 A
\  G + R = 
2
V 2.5
i
10
-
=
 = 250
Þ  R = 250 – G = 250 – 50 = 200W.
(7) (a). \ i = 
V 25
R 1000
= AA
Let R' be the required resistance to be connected in
series with voltmeter.
So  i = 
V'
R R' +
Here V' = 250, R = 1000 W and  i = 
25
1000
 AA
\ 
25
1000
 = 
250
1000 R ' +
 Þ R' = 9000 W.
(8) (d). The potential difference between A and B in the absence
of voltmeter = 2 volt.
Current flowing in the circuit
1V
2
2
1V
EE
I
RR
R R'
R
RR
==
+
+
+
42
I
44
3
4
44
==
´
+
+
 ampere
Potential difference measured by voltmeter
V'
AB
 = IR' = 
2
3
 × 2 = 
4
3
Error in the reading of voltmeter
= V
AB
 – V'
AB
 = 2 – 
4
3
 = 
2
3
 volt
The error in voltmeter reading for 2 volt p.d. = 
2
3
 volt
The error in voltmeter reading for 1 volt p.d.
= 
2 11
3 23
´= volt
the error in voltmeter reading for 100 volt p.d.
= 
100
3
 = 33.3% volt
4kW 4kW
K
( )
+  –
4V
A
B
R
2
R
1
4kW
V
R
V
V
AB
Page 2


(1) (d). Let R is resistance of the voltmeter. The effective
resistance across points A, B is
r = 
60R
60R
´
+
..............(1)
The current in the circuit is I = 12/(50 + r)
The p.d. across AB points is V = Ir
or   6 = 
12
50r +
 × r or 50 + r = 2r
or   r = 50 W ..............(2)
using it in (1),
we get 50 = 
60
60R +
300 + 5R = 6R
or  R = 300 W
(2) (c).  S = 
G GG
n 1 100 1 99
= =W
--
(3) (b). S = 
g
g
i
ii -
 G = 
10
100 10 -
 × 99 = 11 1W
(4) (c). R = 20 + 
80 80
80 80
´
+
 = 60W
i = 
V 21
R 60 30
== amp.
\  V = iR' = 
1
30
 × 40 = 1.33 volt.
(5) (a).  According to Kirchhoff's first law
At  junection A, i
AB
 = 2 + 2 = 4A
At junection B, i
AB
 = i
BC
 – 1 = 3A
2 A
2 A
1 A
1.3 A
i
C
B A
At junection C, i = i
BC
 – 1.3 = 3 – 1.3 = 17 amp
(6) (b). The current required for a full-scale deflection of the
galvanometer is
i = 4.0 x 10
-4
 x 25 = 10
-2
 A
Let a resistance R W is to be connected in series
Then by the ohm's law, we have i = 
V
GR +
Here G = 50 W, V = 2.5 V and i = 10
-2
 A
\  G + R = 
2
V 2.5
i
10
-
=
 = 250
Þ  R = 250 – G = 250 – 50 = 200W.
(7) (a). \ i = 
V 25
R 1000
= AA
Let R' be the required resistance to be connected in
series with voltmeter.
So  i = 
V'
R R' +
Here V' = 250, R = 1000 W and  i = 
25
1000
 AA
\ 
25
1000
 = 
250
1000 R ' +
 Þ R' = 9000 W.
(8) (d). The potential difference between A and B in the absence
of voltmeter = 2 volt.
Current flowing in the circuit
1V
2
2
1V
EE
I
RR
R R'
R
RR
==
+
+
+
42
I
44
3
4
44
==
´
+
+
 ampere
Potential difference measured by voltmeter
V'
AB
 = IR' = 
2
3
 × 2 = 
4
3
Error in the reading of voltmeter
= V
AB
 – V'
AB
 = 2 – 
4
3
 = 
2
3
 volt
The error in voltmeter reading for 2 volt p.d. = 
2
3
 volt
The error in voltmeter reading for 1 volt p.d.
= 
2 11
3 23
´= volt
the error in voltmeter reading for 100 volt p.d.
= 
100
3
 = 33.3% volt
4kW 4kW
K
( )
+  –
4V
A
B
R
2
R
1
4kW
V
R
V
V
AB
DPP/ P 37
106
(9) (d). E = V + I r = I R + I r
Þ E = 0.25 x 10 + 0.25 × r
In second stage
Þ E = 0.5 × 4 + 0.5 r
Subtracting eq. (b) from eq. (a)
2.5 + 0.25 r – 2.0 – 0.5 r = 0
0.5 = 0.25 r
r = 
0.5
0.25
 = 2W.
(10) (d) Suppose current through different paths of the circuit
is as follows.
28 W 54 W
6 V
I
3
8 V
12 V
1 2
After applying KVL for loop (1) and loop (2)
We get 28i
1
 = – 6 – 8 Þ i
1
 = A
2
1
-
and 54i
2
 = – 6 – 12 Þ i
2
 = 
A
3
1
-
Hence i
3
 = i
1
 + i
2
 = 
5
A
6
-
(11) (d) V
AB
 = 4 = 
5X + 2 × 10
X + 10
 Þ =  20 W
12. (b) The circuit can be simplified as follows
i
1
30 W
40 W
40 W
40 V
80 V
i
3
i
2
B C
A D
F E
i
3
Applying KCL at junction A
i
3
 = i
1
 + i
2
...(i)
Applying Kirchoff's voltage law for the loop ABCDA
– 30i
1
 – 40i
3
 + 40 = 0
Þ – 30i
1
 – 40(i
1
 + i
2
) + 40 = 0
Þ 7i + 4i
2
 = 0 ...(ii)
Applying Kirchoff's voltage law for the loop ADEFA.
– 40i
2
 – 40i
3
 + 80 + 4 = 0
Þ – 40i
2
 – 40(i
1
 + i
2
) = – 120
Þ i
2
 + 2i
2
 = 3 ...(iii)
On solving equation (ii) and (iii) i
1
 = – 0.4 A.
13. (b) Cells area joined in parallel when internal resistance is
higher then a external resistance. (R << r)
i = 
E
R+
r
n
14. (b). Current in the ammeter I = 
E
R'
R' r1
R
éù
++
êú
ëû
On increasing the value of R, the denominator will
decrease and consequently the value of I will increase.
15. (a)
Let ABCDEFGH be skeleton cube formed of twelve
equal wires each of resistance R. Let a battery of e.m.f.
E be connected across A and G. Let the total current
entering at the corner A and leaving the diagonally
opposite corner G be I. By symmetry the distribution
of currents in wires of cube, according to Kirchoff's I
st
law is shown in fig. ApplyingKirchoff's II
nd
 law to mesh
ADCGEA, we get
– 
1
3
 R – 
1
6
 R – 
1
3
 R + E = 0
or   E = 
5
6
 IR  ......(a)
If R
AB
 is equivalent resistance between comers A and B,
then from Ohm’s law comparing (a) and (b), we get
IR
AB
 = 
5
6
IR
16. (d).
E
x + 2y
I = x + 2y
2 (y–z)
z
(y – z)
(y – z)
y
y
x
(y – z)
(y – z)
F
z
y
A
D
H
G
E
y B C
I/3
I/6
I/6
I
I/3
I/3
I/3
E
I/6
I/6
I/3
I/3
I/3
I/6
I
Page 3


(1) (d). Let R is resistance of the voltmeter. The effective
resistance across points A, B is
r = 
60R
60R
´
+
..............(1)
The current in the circuit is I = 12/(50 + r)
The p.d. across AB points is V = Ir
or   6 = 
12
50r +
 × r or 50 + r = 2r
or   r = 50 W ..............(2)
using it in (1),
we get 50 = 
60
60R +
300 + 5R = 6R
or  R = 300 W
(2) (c).  S = 
G GG
n 1 100 1 99
= =W
--
(3) (b). S = 
g
g
i
ii -
 G = 
10
100 10 -
 × 99 = 11 1W
(4) (c). R = 20 + 
80 80
80 80
´
+
 = 60W
i = 
V 21
R 60 30
== amp.
\  V = iR' = 
1
30
 × 40 = 1.33 volt.
(5) (a).  According to Kirchhoff's first law
At  junection A, i
AB
 = 2 + 2 = 4A
At junection B, i
AB
 = i
BC
 – 1 = 3A
2 A
2 A
1 A
1.3 A
i
C
B A
At junection C, i = i
BC
 – 1.3 = 3 – 1.3 = 17 amp
(6) (b). The current required for a full-scale deflection of the
galvanometer is
i = 4.0 x 10
-4
 x 25 = 10
-2
 A
Let a resistance R W is to be connected in series
Then by the ohm's law, we have i = 
V
GR +
Here G = 50 W, V = 2.5 V and i = 10
-2
 A
\  G + R = 
2
V 2.5
i
10
-
=
 = 250
Þ  R = 250 – G = 250 – 50 = 200W.
(7) (a). \ i = 
V 25
R 1000
= AA
Let R' be the required resistance to be connected in
series with voltmeter.
So  i = 
V'
R R' +
Here V' = 250, R = 1000 W and  i = 
25
1000
 AA
\ 
25
1000
 = 
250
1000 R ' +
 Þ R' = 9000 W.
(8) (d). The potential difference between A and B in the absence
of voltmeter = 2 volt.
Current flowing in the circuit
1V
2
2
1V
EE
I
RR
R R'
R
RR
==
+
+
+
42
I
44
3
4
44
==
´
+
+
 ampere
Potential difference measured by voltmeter
V'
AB
 = IR' = 
2
3
 × 2 = 
4
3
Error in the reading of voltmeter
= V
AB
 – V'
AB
 = 2 – 
4
3
 = 
2
3
 volt
The error in voltmeter reading for 2 volt p.d. = 
2
3
 volt
The error in voltmeter reading for 1 volt p.d.
= 
2 11
3 23
´= volt
the error in voltmeter reading for 100 volt p.d.
= 
100
3
 = 33.3% volt
4kW 4kW
K
( )
+  –
4V
A
B
R
2
R
1
4kW
V
R
V
V
AB
DPP/ P 37
106
(9) (d). E = V + I r = I R + I r
Þ E = 0.25 x 10 + 0.25 × r
In second stage
Þ E = 0.5 × 4 + 0.5 r
Subtracting eq. (b) from eq. (a)
2.5 + 0.25 r – 2.0 – 0.5 r = 0
0.5 = 0.25 r
r = 
0.5
0.25
 = 2W.
(10) (d) Suppose current through different paths of the circuit
is as follows.
28 W 54 W
6 V
I
3
8 V
12 V
1 2
After applying KVL for loop (1) and loop (2)
We get 28i
1
 = – 6 – 8 Þ i
1
 = A
2
1
-
and 54i
2
 = – 6 – 12 Þ i
2
 = 
A
3
1
-
Hence i
3
 = i
1
 + i
2
 = 
5
A
6
-
(11) (d) V
AB
 = 4 = 
5X + 2 × 10
X + 10
 Þ =  20 W
12. (b) The circuit can be simplified as follows
i
1
30 W
40 W
40 W
40 V
80 V
i
3
i
2
B C
A D
F E
i
3
Applying KCL at junction A
i
3
 = i
1
 + i
2
...(i)
Applying Kirchoff's voltage law for the loop ABCDA
– 30i
1
 – 40i
3
 + 40 = 0
Þ – 30i
1
 – 40(i
1
 + i
2
) + 40 = 0
Þ 7i + 4i
2
 = 0 ...(ii)
Applying Kirchoff's voltage law for the loop ADEFA.
– 40i
2
 – 40i
3
 + 80 + 4 = 0
Þ – 40i
2
 – 40(i
1
 + i
2
) = – 120
Þ i
2
 + 2i
2
 = 3 ...(iii)
On solving equation (ii) and (iii) i
1
 = – 0.4 A.
13. (b) Cells area joined in parallel when internal resistance is
higher then a external resistance. (R << r)
i = 
E
R+
r
n
14. (b). Current in the ammeter I = 
E
R'
R' r1
R
éù
++
êú
ëû
On increasing the value of R, the denominator will
decrease and consequently the value of I will increase.
15. (a)
Let ABCDEFGH be skeleton cube formed of twelve
equal wires each of resistance R. Let a battery of e.m.f.
E be connected across A and G. Let the total current
entering at the corner A and leaving the diagonally
opposite corner G be I. By symmetry the distribution
of currents in wires of cube, according to Kirchoff's I
st
law is shown in fig. ApplyingKirchoff's II
nd
 law to mesh
ADCGEA, we get
– 
1
3
 R – 
1
6
 R – 
1
3
 R + E = 0
or   E = 
5
6
 IR  ......(a)
If R
AB
 is equivalent resistance between comers A and B,
then from Ohm’s law comparing (a) and (b), we get
IR
AB
 = 
5
6
IR
16. (d).
E
x + 2y
I = x + 2y
2 (y–z)
z
(y – z)
(y – z)
y
y
x
(y – z)
(y – z)
F
z
y
A
D
H
G
E
y B C
I/3
I/6
I/6
I
I/3
I/3
I/3
E
I/6
I/6
I/3
I/3
I/3
I/6
I
DPP/ P 37
107
Let I = x + 2y current enter at point A, when a battery of
e.m.f. E and no internal resistance is connected across edge
AB. The edges AD and AH are symmetrically connected
to A, therefore they will carry equal currents. The
distribution of currents according to Kirchoff's I
st
 law is
shown in fig.
If R
AB
 is equivalent resistance, then from Ohm's law ,
E = R
AB
 I = R
AB
 (x + 2y) ....(a)
and from Kirchoff's law applied to mesh containing AB
and cell E is
R x = E .....(b)
(since R is resistance of each wire)
Applying Kirchoff's II law to mesh AHEB
yR + zR + yR – xR = 0 or x – 2y – z = 0 .....(c)
Applying Kirchoff's II law to mesh DGFC
(y – z) R + 2 (y – z) R – zR = 0
or 4(y – z) –z = 0 or 4y = 5 z .....(d)
i.e. z = (4/5) y
 .....(E)
Substituting this value in (c), we get
x – 2y – 
4
5
 y = 0
or  
14
5
 y = x i.e. y = 
5
14
x
Substituting value of y in (a), we get
E = R
AB
 
10
xx
14
æö
+
ç÷
èø
E = R
AB
 
24
14
 x = R . x
R
AB
 = 
24
14
 R  \ R
AB
 = 
7
12
 R.
17. (a).
Let a battery of e.m.f.  E is applied between points A and B.
Let a current I, enter through point A.
If R
AB
 is equivalent resistance between points A and B,
then from Ohm's law
R
AB
 I = E
E
I
I/2-I
1
y
F
A
D
H
G
B
I/2-I
1
E
I/2
I-2I
1
(I/2-I
1
)
I/2-I
1 I
1
I/2
·
I
I
1
I/2 C
The distribution of currents, keeping in mind symmetry
condition, is shown in fig.
Let R (= 2W) be the resistance of each wire.
Applying Kirchoff's II law to mesh DGFC, we get
1
1
I
2
æö
-
ç÷
èø
 R + (I – 2I
1
)
R + 1
1
I
2
æö
-
ç÷
èø
  R – I
1
 R = 0
or  2 1
1
I
2
æö
-
ç÷
èø
 + (I – 2I
1
) – I
1
 = 0
or 2I – 5I
1
 = 0 or I
1
 = 
2
5
I ...... (b)
Applying Kirchoff's II
nd
 law to external circuit AHEBE',
we get
1
2
R + I
1 
R + 
1
2
 R = E
 IR + 
2
5
 IR = E'  [Using (b)]
or
7
5
 IR = E ......(c)
Comparing (a) and (c), wet get
R
AB
 I = 
7
5
 IR i.e. R
AB
 = 
7
5
R = 
7
5
 × 2 = 2.8 W
18. (c). In the first case I = E/(r + R) and in the second case
I' = E/(r + R/2) = 2E/(2r + R)
Using E = I(r + R), we get
I' = I 
2r 2R
2rR
+ æö
ç÷
èø +
 = I 
R
1
2rR
æö
+
ç÷
èø +
Thus the term in bracket is greater than 1 but less than
2. Thus 2I > I' > I
19. (b). Let R be the combined resistance of galvanometer and
an unknown resistance and r the internal resistance of
each battery. When the batteries, each of e.m.f. E are
connected in series, the net e.m.f. = 2E and net internal
resistance = 2r
\ Current i
1
 = 
2E
R 2r +
   or    1.0 = 
2 15
R 2r
´
+
\ R + 2r = 3.0. ...(i)
When the batteries are connected in parallel, the e.m.f.
remains E and net internal resistance becomes r/2. therefore
Current i
2
 =  
E 2E
r
2Rr
R
2
=
+
+
\ 2R + r = 
2
2E 2 15
i 0.6
´
= = 5.0 ...(i)
Solving (i) and (ii), we get r = 1/3 W.
Page 4


(1) (d). Let R is resistance of the voltmeter. The effective
resistance across points A, B is
r = 
60R
60R
´
+
..............(1)
The current in the circuit is I = 12/(50 + r)
The p.d. across AB points is V = Ir
or   6 = 
12
50r +
 × r or 50 + r = 2r
or   r = 50 W ..............(2)
using it in (1),
we get 50 = 
60
60R +
300 + 5R = 6R
or  R = 300 W
(2) (c).  S = 
G GG
n 1 100 1 99
= =W
--
(3) (b). S = 
g
g
i
ii -
 G = 
10
100 10 -
 × 99 = 11 1W
(4) (c). R = 20 + 
80 80
80 80
´
+
 = 60W
i = 
V 21
R 60 30
== amp.
\  V = iR' = 
1
30
 × 40 = 1.33 volt.
(5) (a).  According to Kirchhoff's first law
At  junection A, i
AB
 = 2 + 2 = 4A
At junection B, i
AB
 = i
BC
 – 1 = 3A
2 A
2 A
1 A
1.3 A
i
C
B A
At junection C, i = i
BC
 – 1.3 = 3 – 1.3 = 17 amp
(6) (b). The current required for a full-scale deflection of the
galvanometer is
i = 4.0 x 10
-4
 x 25 = 10
-2
 A
Let a resistance R W is to be connected in series
Then by the ohm's law, we have i = 
V
GR +
Here G = 50 W, V = 2.5 V and i = 10
-2
 A
\  G + R = 
2
V 2.5
i
10
-
=
 = 250
Þ  R = 250 – G = 250 – 50 = 200W.
(7) (a). \ i = 
V 25
R 1000
= AA
Let R' be the required resistance to be connected in
series with voltmeter.
So  i = 
V'
R R' +
Here V' = 250, R = 1000 W and  i = 
25
1000
 AA
\ 
25
1000
 = 
250
1000 R ' +
 Þ R' = 9000 W.
(8) (d). The potential difference between A and B in the absence
of voltmeter = 2 volt.
Current flowing in the circuit
1V
2
2
1V
EE
I
RR
R R'
R
RR
==
+
+
+
42
I
44
3
4
44
==
´
+
+
 ampere
Potential difference measured by voltmeter
V'
AB
 = IR' = 
2
3
 × 2 = 
4
3
Error in the reading of voltmeter
= V
AB
 – V'
AB
 = 2 – 
4
3
 = 
2
3
 volt
The error in voltmeter reading for 2 volt p.d. = 
2
3
 volt
The error in voltmeter reading for 1 volt p.d.
= 
2 11
3 23
´= volt
the error in voltmeter reading for 100 volt p.d.
= 
100
3
 = 33.3% volt
4kW 4kW
K
( )
+  –
4V
A
B
R
2
R
1
4kW
V
R
V
V
AB
DPP/ P 37
106
(9) (d). E = V + I r = I R + I r
Þ E = 0.25 x 10 + 0.25 × r
In second stage
Þ E = 0.5 × 4 + 0.5 r
Subtracting eq. (b) from eq. (a)
2.5 + 0.25 r – 2.0 – 0.5 r = 0
0.5 = 0.25 r
r = 
0.5
0.25
 = 2W.
(10) (d) Suppose current through different paths of the circuit
is as follows.
28 W 54 W
6 V
I
3
8 V
12 V
1 2
After applying KVL for loop (1) and loop (2)
We get 28i
1
 = – 6 – 8 Þ i
1
 = A
2
1
-
and 54i
2
 = – 6 – 12 Þ i
2
 = 
A
3
1
-
Hence i
3
 = i
1
 + i
2
 = 
5
A
6
-
(11) (d) V
AB
 = 4 = 
5X + 2 × 10
X + 10
 Þ =  20 W
12. (b) The circuit can be simplified as follows
i
1
30 W
40 W
40 W
40 V
80 V
i
3
i
2
B C
A D
F E
i
3
Applying KCL at junction A
i
3
 = i
1
 + i
2
...(i)
Applying Kirchoff's voltage law for the loop ABCDA
– 30i
1
 – 40i
3
 + 40 = 0
Þ – 30i
1
 – 40(i
1
 + i
2
) + 40 = 0
Þ 7i + 4i
2
 = 0 ...(ii)
Applying Kirchoff's voltage law for the loop ADEFA.
– 40i
2
 – 40i
3
 + 80 + 4 = 0
Þ – 40i
2
 – 40(i
1
 + i
2
) = – 120
Þ i
2
 + 2i
2
 = 3 ...(iii)
On solving equation (ii) and (iii) i
1
 = – 0.4 A.
13. (b) Cells area joined in parallel when internal resistance is
higher then a external resistance. (R << r)
i = 
E
R+
r
n
14. (b). Current in the ammeter I = 
E
R'
R' r1
R
éù
++
êú
ëû
On increasing the value of R, the denominator will
decrease and consequently the value of I will increase.
15. (a)
Let ABCDEFGH be skeleton cube formed of twelve
equal wires each of resistance R. Let a battery of e.m.f.
E be connected across A and G. Let the total current
entering at the corner A and leaving the diagonally
opposite corner G be I. By symmetry the distribution
of currents in wires of cube, according to Kirchoff's I
st
law is shown in fig. ApplyingKirchoff's II
nd
 law to mesh
ADCGEA, we get
– 
1
3
 R – 
1
6
 R – 
1
3
 R + E = 0
or   E = 
5
6
 IR  ......(a)
If R
AB
 is equivalent resistance between comers A and B,
then from Ohm’s law comparing (a) and (b), we get
IR
AB
 = 
5
6
IR
16. (d).
E
x + 2y
I = x + 2y
2 (y–z)
z
(y – z)
(y – z)
y
y
x
(y – z)
(y – z)
F
z
y
A
D
H
G
E
y B C
I/3
I/6
I/6
I
I/3
I/3
I/3
E
I/6
I/6
I/3
I/3
I/3
I/6
I
DPP/ P 37
107
Let I = x + 2y current enter at point A, when a battery of
e.m.f. E and no internal resistance is connected across edge
AB. The edges AD and AH are symmetrically connected
to A, therefore they will carry equal currents. The
distribution of currents according to Kirchoff's I
st
 law is
shown in fig.
If R
AB
 is equivalent resistance, then from Ohm's law ,
E = R
AB
 I = R
AB
 (x + 2y) ....(a)
and from Kirchoff's law applied to mesh containing AB
and cell E is
R x = E .....(b)
(since R is resistance of each wire)
Applying Kirchoff's II law to mesh AHEB
yR + zR + yR – xR = 0 or x – 2y – z = 0 .....(c)
Applying Kirchoff's II law to mesh DGFC
(y – z) R + 2 (y – z) R – zR = 0
or 4(y – z) –z = 0 or 4y = 5 z .....(d)
i.e. z = (4/5) y
 .....(E)
Substituting this value in (c), we get
x – 2y – 
4
5
 y = 0
or  
14
5
 y = x i.e. y = 
5
14
x
Substituting value of y in (a), we get
E = R
AB
 
10
xx
14
æö
+
ç÷
èø
E = R
AB
 
24
14
 x = R . x
R
AB
 = 
24
14
 R  \ R
AB
 = 
7
12
 R.
17. (a).
Let a battery of e.m.f.  E is applied between points A and B.
Let a current I, enter through point A.
If R
AB
 is equivalent resistance between points A and B,
then from Ohm's law
R
AB
 I = E
E
I
I/2-I
1
y
F
A
D
H
G
B
I/2-I
1
E
I/2
I-2I
1
(I/2-I
1
)
I/2-I
1 I
1
I/2
·
I
I
1
I/2 C
The distribution of currents, keeping in mind symmetry
condition, is shown in fig.
Let R (= 2W) be the resistance of each wire.
Applying Kirchoff's II law to mesh DGFC, we get
1
1
I
2
æö
-
ç÷
èø
 R + (I – 2I
1
)
R + 1
1
I
2
æö
-
ç÷
èø
  R – I
1
 R = 0
or  2 1
1
I
2
æö
-
ç÷
èø
 + (I – 2I
1
) – I
1
 = 0
or 2I – 5I
1
 = 0 or I
1
 = 
2
5
I ...... (b)
Applying Kirchoff's II
nd
 law to external circuit AHEBE',
we get
1
2
R + I
1 
R + 
1
2
 R = E
 IR + 
2
5
 IR = E'  [Using (b)]
or
7
5
 IR = E ......(c)
Comparing (a) and (c), wet get
R
AB
 I = 
7
5
 IR i.e. R
AB
 = 
7
5
R = 
7
5
 × 2 = 2.8 W
18. (c). In the first case I = E/(r + R) and in the second case
I' = E/(r + R/2) = 2E/(2r + R)
Using E = I(r + R), we get
I' = I 
2r 2R
2rR
+ æö
ç÷
èø +
 = I 
R
1
2rR
æö
+
ç÷
èø +
Thus the term in bracket is greater than 1 but less than
2. Thus 2I > I' > I
19. (b). Let R be the combined resistance of galvanometer and
an unknown resistance and r the internal resistance of
each battery. When the batteries, each of e.m.f. E are
connected in series, the net e.m.f. = 2E and net internal
resistance = 2r
\ Current i
1
 = 
2E
R 2r +
   or    1.0 = 
2 15
R 2r
´
+
\ R + 2r = 3.0. ...(i)
When the batteries are connected in parallel, the e.m.f.
remains E and net internal resistance becomes r/2. therefore
Current i
2
 =  
E 2E
r
2Rr
R
2
=
+
+
\ 2R + r = 
2
2E 2 15
i 0.6
´
= = 5.0 ...(i)
Solving (i) and (ii), we get r = 1/3 W.
DPP/ P 37
108
20. (a). The circuit with current distribution is shown in fig.
Applying Kirchoff's second law to the loop DEFGHID,
we have i
1
 × 100 – (i – i
1
) × 200 = 0
300 i
1
 – 200 i = 0                    ......(1)
Now applying Kirchoff's second law to loop ADIHGCBA,
we have. (i – i
1
) 200 + i × 300 = 110
500i – 200 i
1
 = 110                    .....(2)
Solving eqs. (1) and (2), we get
i = 
3
10
 amp and i
1
 = 
1
5
amp.
Current in 100 ohm resistance i
1
 = 
1
5
 amp.
Current in 200 ohm resistance i – i
1
 = 
1
10
Current in 300 ohm resistance i = 
3
10
 amp.
Potential difference between A and C
= Potential difference across 100 ohm
resistance
or potential difference across 200 ohm resistance
\ V
A 
 – V
C
 = current × resistance
= i
1
 × 100 = 
1
5
 × 100 = 20 volt.
Potential difference between C and B is given by
V
C
 – V
B
 = i × 300 = 
3
10
 × 300 = 90 volt.
21. (a). After full charging, the steady current in the condenser
is zero, hence no current will flow in 4W resistance.
E 66
I
23 R R ' 28 12
28
23
= ==
´ ++ æö
+
ç÷
èø +
 =1.5 A A
Let current flowing in 2W resistance is I
1
\ 2W and 3W resistance are connected in parallel
\ 2I
1
 = (1.5 – I
1
) x 3
 5I
1
 = 4.5
I
1
 = 0.9 amp.
24. (d). 
G
R
I
g
I–I
g
For Ammeter I
g
G = (I – I
g
) R
50 × 10
–6
 × 100 = 5 × 10
–3
 × (R)  Þ R » 1W
For voltmeter I
g 
(R + G) = V
50 mA (R + G) = 10V Þ R + G = 200 kW Þ R » 200kW
25. (a) Potential at A = 6V
V
A
 – V
C
 = 4
Þ  V
C
 = 2V
26. (d)
AC AD
AB AB
V V AD 42
;
V V AB 63
= = ==
200
AD cm.
3
=
27. (a) D is balance point, hence no current
28. (a) V oltameter measures current indirectly in terms of mass
of ions deposited and electrochemical equivalent of
the substance 
m
I.
Zt
æö
=
ç÷
èø
 Since value of m and Z are
measured to 3rd decimal place and 5th decimal place
respectively . The relative error in the emasurement of
current by voltmeter will be very small as compared to
that when measured by ammeter directly .
29. (c) The e.m.f. of a dry cell is dependent upon the electrode
potential of cathode and anode which in  turn is
dependent upon the reaction involved as well as
concentration of the electrolyte. It has nothing to do
with size of the cell.
So, statement-1 is false and statement-2 is true.
30. (d) V = E – ir = 4 – 2 × 2 = 0,  During charging V > E.
C
A
i
D
200W
100W
G
F
E
i
i
300W
H
110V
B
D
i
1
i - i
1