Current Electricity– 1 Practice Questions - DPP for NEET

 Page 1


1. (b). The current in 1W resistance is 3A. The current in 3W
resistance is
 I
1
 =
2
12
R
I
RR +
 =
6
36 +
 × 3 = 2A.
Therefore the ratio is 
2
3
.
2. (d).  R = 
12
(R R) 1
22
+
+ [(R
1
 + R
2
)
2
 + 4 R
3
(R
1
 + R
2
)]
1/2
.....(a)
R
1
 = 1W, R
2
 = 0. R
3
 = 2W. ......(b)
From eqs. (a) and (b)
R = 
1
2
 + 
1
2
 [1 + 4 × 2 × 1]
1/2 
= 
2
1
 [1 + 3] = 2W.
3. (b). Since Q is connected in parallel the net resistance
becomes R/2, so the current I = 2V/R, double the value.
4. (b). Since there is no current in edcb part, the p.d. across
be should be 2V . Let current in 500 W is I, then same current
flows through X (think). Therefore, for loop abefa,
12 = I (500) + IX
or  12 = I (500) + 2 (\ IX = 2 volt)
Thus I = (1/50) A or from IX = 2,
X = 2 × 50 = 100 W.
5. (b). Let l
1
 be the initial length of the wire. Then the new
length will be
l
2
 = 
110
100
 l
1
 = 
11
10
 l
1
Since, the volume remains constant
A
1
l
1
 = A
2
l
2
 or A
1
 /A
2
 = l
2
/l
1
 = 
11
10
(where A
1
 and A
2
 are initial and final area of cross–section
of the wire).
If R
1
 and R
2
 are the initial and final resistances, then
2
1 12
2 21
RA 10 10 10
R A 11 11 11
æö
= = ´=
ç÷
èø
l
l
or  
2
2
1
R 11
R 10
æö
=
ç÷
èø
Now , percentage change in resistance is
1
R
R
D
 × 100 =
21
1
RR
R
æö -
ç÷
èø
× 100
  =
2
11
1
10
éù
æö
- êú
ç÷
èø
êú
ëû
× 100 = 21%
6. (a). The circuit is equivalent to Fig. It is a balanced
wheatstone bridge between abcd, and then in parallel (2R)
resistances. Thus ignoring resistance between bd arm. The
circuit is equivalent to three (2R) resistances in parallel
(abc, adc, aRRc).
i.e.  
eq
1 111 3
R 2R 2R 2R 2R
=+ +=
Þ  R
eq
 = 
2
3
 R
7. (c).
Imagine, A being pulled on the left side, then abcd becomes
a balanced wheatstone bridge Fig. The arm bd can be
ignored. Then resistance between A, B becomes = r.
i.e. 
eq
1 1 11
R 2r 2rr
=+=  Þ  R
eq
 = r
8. ( d) R =  91´ 10
2
 » 9.1 kW.
9. ( d) Spacific resistance doesn't depend upon length and area.
10. (b). The diagram can be redrawn as shown in fig.
· ·
b
d
R R
a
c
R
R
· ·
c
b
d
r r
r
r
a
r
B A
Page 2


1. (b). The current in 1W resistance is 3A. The current in 3W
resistance is
 I
1
 =
2
12
R
I
RR +
 =
6
36 +
 × 3 = 2A.
Therefore the ratio is 
2
3
.
2. (d).  R = 
12
(R R) 1
22
+
+ [(R
1
 + R
2
)
2
 + 4 R
3
(R
1
 + R
2
)]
1/2
.....(a)
R
1
 = 1W, R
2
 = 0. R
3
 = 2W. ......(b)
From eqs. (a) and (b)
R = 
1
2
 + 
1
2
 [1 + 4 × 2 × 1]
1/2 
= 
2
1
 [1 + 3] = 2W.
3. (b). Since Q is connected in parallel the net resistance
becomes R/2, so the current I = 2V/R, double the value.
4. (b). Since there is no current in edcb part, the p.d. across
be should be 2V . Let current in 500 W is I, then same current
flows through X (think). Therefore, for loop abefa,
12 = I (500) + IX
or  12 = I (500) + 2 (\ IX = 2 volt)
Thus I = (1/50) A or from IX = 2,
X = 2 × 50 = 100 W.
5. (b). Let l
1
 be the initial length of the wire. Then the new
length will be
l
2
 = 
110
100
 l
1
 = 
11
10
 l
1
Since, the volume remains constant
A
1
l
1
 = A
2
l
2
 or A
1
 /A
2
 = l
2
/l
1
 = 
11
10
(where A
1
 and A
2
 are initial and final area of cross–section
of the wire).
If R
1
 and R
2
 are the initial and final resistances, then
2
1 12
2 21
RA 10 10 10
R A 11 11 11
æö
= = ´=
ç÷
èø
l
l
or  
2
2
1
R 11
R 10
æö
=
ç÷
èø
Now , percentage change in resistance is
1
R
R
D
 × 100 =
21
1
RR
R
æö -
ç÷
èø
× 100
  =
2
11
1
10
éù
æö
- êú
ç÷
èø
êú
ëû
× 100 = 21%
6. (a). The circuit is equivalent to Fig. It is a balanced
wheatstone bridge between abcd, and then in parallel (2R)
resistances. Thus ignoring resistance between bd arm. The
circuit is equivalent to three (2R) resistances in parallel
(abc, adc, aRRc).
i.e.  
eq
1 111 3
R 2R 2R 2R 2R
=+ +=
Þ  R
eq
 = 
2
3
 R
7. (c).
Imagine, A being pulled on the left side, then abcd becomes
a balanced wheatstone bridge Fig. The arm bd can be
ignored. Then resistance between A, B becomes = r.
i.e. 
eq
1 1 11
R 2r 2rr
=+=  Þ  R
eq
 = r
8. ( d) R =  91´ 10
2
 » 9.1 kW.
9. ( d) Spacific resistance doesn't depend upon length and area.
10. (b). The diagram can be redrawn as shown in fig.
· ·
b
d
R R
a
c
R
R
· ·
c
b
d
r r
r
r
a
r
B A
DPP/ P 36
101
X Y
3 W
4 W
8 W
4 W
2 W
3 W
B
C
6 W
+ –
A
2V
The effective resistance R
AC
 between A and C
AC
1 1 13
R 2 44
= +=
 \R
AC
 = 
3
4
 ohm
The effective resistance R
CB
 between C and B
R
CB 
=
1 13
4 88
+= \R
CB 
= 
8
3
 ohm.
Now, R
ACB
 = R
AC
 + R
CB
 = 
48
33
+ = 4ohm.
Corresponding to points X and Y , the resistances 3 ohm, 4
ohm and 6 ohm are in parallel, hence effective resistance
R
XY
 is
XY
1 1 1 1 4 3 29
R 3 4 6 12 12
++
=++==
\ R
XY
 = 
124
93
= ohm.
Total resistance R of the circuit  =
42
33
+ = 2W.
Current in the circuit = 
2
2
 = 1 AA
Power dissipated in the circuit = i
2
 R = 1 × 2 = 2 watts
Potential difference between X and
Y = i × R
XY
 = 1 ×  
4
3
= 
4
3
V ..
\ Potential difference across 3 ohm resistor  = 
4
3
 V ..
Current in 3 ohm resistor = 
4/34
39
= = 0.44 amp.
11. (c). Requivalent = 
( )
( )
30 30 30
30 30 30
+
++
 = 
60 30
90
´
= 20 W
\ i = 
V
R
 = 
2
20
=
1
10
ampere
12. (c). R = 2 + 2 + 
2R
2 + R
´
 Þ  2R + R
2
 = 8 + 4R + 2R
Þ R
2
 – 4R – 8 = 0 Þ R = 
4 16 32
2
±+
 = 
2 23 ±
R cannot be negative, hence R = 
2 23 ±
 = 5.46 W
13. (c). P = 
2
V
R
. If resistance of heater coil is R, then resistance
of parallel combination of two halves will be 
4
R
So
1
2
P
P
 = 
2
1
P
P
 = 
/4 R
R
= 
1
4
14. (c). Total kWh consumed = 
60 8 301
1000
´´
 = 14.4
Hence cost = 14.4 ´ 1.25 = ` 18
15. (d). Since all bulbs are identical they have the same
resistances. The current I flowing through 1 branches at
A. So current in 2 and 3, as well as in 4 will be less than I.
The current through 5 is also I. Thus 1 and 5 glow equally
brightly and more than 2, 3 or 4.
16. (b). Let R
1
 and R
2
 be the resistances of the coils, V the
supply voltage, Q the heat required to boil the water.
Heat produced by first coil of resistance R
1
 in time t
1
(= 6 min) = Q = 
2
1
1
Vt
JR
 = 
2
1
V 6 60
4.2R
´´
 cal      ......(a)
Heat produced in second coil of resistance R
2
 in time
t
2
  (= 8 min)
= Q = 
2 2
1
22
Vt V 6 60
JR 4.2R
´´
=
.....(b)
Equating (a) and (b), we get
22
68
RR
= i.e. 
2
1
R 84
R 63
==
or R
2
 = 
4
3
R
1
 .....(c)
(i) When the two heating coils are in series, the effective
resistance is
R' = R
1
 + R
2
 = R
1
 + 
4
3
. R
1
 = 
7
3
 R
1
.
with two coils in series, let the kettle take t' time to boil.
22
1
V t' V t'
Q
7 JR'
4.2R
3
==
æö
´
ç÷
èø
      .....(d)
Page 3


1. (b). The current in 1W resistance is 3A. The current in 3W
resistance is
 I
1
 =
2
12
R
I
RR +
 =
6
36 +
 × 3 = 2A.
Therefore the ratio is 
2
3
.
2. (d).  R = 
12
(R R) 1
22
+
+ [(R
1
 + R
2
)
2
 + 4 R
3
(R
1
 + R
2
)]
1/2
.....(a)
R
1
 = 1W, R
2
 = 0. R
3
 = 2W. ......(b)
From eqs. (a) and (b)
R = 
1
2
 + 
1
2
 [1 + 4 × 2 × 1]
1/2 
= 
2
1
 [1 + 3] = 2W.
3. (b). Since Q is connected in parallel the net resistance
becomes R/2, so the current I = 2V/R, double the value.
4. (b). Since there is no current in edcb part, the p.d. across
be should be 2V . Let current in 500 W is I, then same current
flows through X (think). Therefore, for loop abefa,
12 = I (500) + IX
or  12 = I (500) + 2 (\ IX = 2 volt)
Thus I = (1/50) A or from IX = 2,
X = 2 × 50 = 100 W.
5. (b). Let l
1
 be the initial length of the wire. Then the new
length will be
l
2
 = 
110
100
 l
1
 = 
11
10
 l
1
Since, the volume remains constant
A
1
l
1
 = A
2
l
2
 or A
1
 /A
2
 = l
2
/l
1
 = 
11
10
(where A
1
 and A
2
 are initial and final area of cross–section
of the wire).
If R
1
 and R
2
 are the initial and final resistances, then
2
1 12
2 21
RA 10 10 10
R A 11 11 11
æö
= = ´=
ç÷
èø
l
l
or  
2
2
1
R 11
R 10
æö
=
ç÷
èø
Now , percentage change in resistance is
1
R
R
D
 × 100 =
21
1
RR
R
æö -
ç÷
èø
× 100
  =
2
11
1
10
éù
æö
- êú
ç÷
èø
êú
ëû
× 100 = 21%
6. (a). The circuit is equivalent to Fig. It is a balanced
wheatstone bridge between abcd, and then in parallel (2R)
resistances. Thus ignoring resistance between bd arm. The
circuit is equivalent to three (2R) resistances in parallel
(abc, adc, aRRc).
i.e.  
eq
1 111 3
R 2R 2R 2R 2R
=+ +=
Þ  R
eq
 = 
2
3
 R
7. (c).
Imagine, A being pulled on the left side, then abcd becomes
a balanced wheatstone bridge Fig. The arm bd can be
ignored. Then resistance between A, B becomes = r.
i.e. 
eq
1 1 11
R 2r 2rr
=+=  Þ  R
eq
 = r
8. ( d) R =  91´ 10
2
 » 9.1 kW.
9. ( d) Spacific resistance doesn't depend upon length and area.
10. (b). The diagram can be redrawn as shown in fig.
· ·
b
d
R R
a
c
R
R
· ·
c
b
d
r r
r
r
a
r
B A
DPP/ P 36
101
X Y
3 W
4 W
8 W
4 W
2 W
3 W
B
C
6 W
+ –
A
2V
The effective resistance R
AC
 between A and C
AC
1 1 13
R 2 44
= +=
 \R
AC
 = 
3
4
 ohm
The effective resistance R
CB
 between C and B
R
CB 
=
1 13
4 88
+= \R
CB 
= 
8
3
 ohm.
Now, R
ACB
 = R
AC
 + R
CB
 = 
48
33
+ = 4ohm.
Corresponding to points X and Y , the resistances 3 ohm, 4
ohm and 6 ohm are in parallel, hence effective resistance
R
XY
 is
XY
1 1 1 1 4 3 29
R 3 4 6 12 12
++
=++==
\ R
XY
 = 
124
93
= ohm.
Total resistance R of the circuit  =
42
33
+ = 2W.
Current in the circuit = 
2
2
 = 1 AA
Power dissipated in the circuit = i
2
 R = 1 × 2 = 2 watts
Potential difference between X and
Y = i × R
XY
 = 1 ×  
4
3
= 
4
3
V ..
\ Potential difference across 3 ohm resistor  = 
4
3
 V ..
Current in 3 ohm resistor = 
4/34
39
= = 0.44 amp.
11. (c). Requivalent = 
( )
( )
30 30 30
30 30 30
+
++
 = 
60 30
90
´
= 20 W
\ i = 
V
R
 = 
2
20
=
1
10
ampere
12. (c). R = 2 + 2 + 
2R
2 + R
´
 Þ  2R + R
2
 = 8 + 4R + 2R
Þ R
2
 – 4R – 8 = 0 Þ R = 
4 16 32
2
±+
 = 
2 23 ±
R cannot be negative, hence R = 
2 23 ±
 = 5.46 W
13. (c). P = 
2
V
R
. If resistance of heater coil is R, then resistance
of parallel combination of two halves will be 
4
R
So
1
2
P
P
 = 
2
1
P
P
 = 
/4 R
R
= 
1
4
14. (c). Total kWh consumed = 
60 8 301
1000
´´
 = 14.4
Hence cost = 14.4 ´ 1.25 = ` 18
15. (d). Since all bulbs are identical they have the same
resistances. The current I flowing through 1 branches at
A. So current in 2 and 3, as well as in 4 will be less than I.
The current through 5 is also I. Thus 1 and 5 glow equally
brightly and more than 2, 3 or 4.
16. (b). Let R
1
 and R
2
 be the resistances of the coils, V the
supply voltage, Q the heat required to boil the water.
Heat produced by first coil of resistance R
1
 in time t
1
(= 6 min) = Q = 
2
1
1
Vt
JR
 = 
2
1
V 6 60
4.2R
´´
 cal      ......(a)
Heat produced in second coil of resistance R
2
 in time
t
2
  (= 8 min)
= Q = 
2 2
1
22
Vt V 6 60
JR 4.2R
´´
=
.....(b)
Equating (a) and (b), we get
22
68
RR
= i.e. 
2
1
R 84
R 63
==
or R
2
 = 
4
3
R
1
 .....(c)
(i) When the two heating coils are in series, the effective
resistance is
R' = R
1
 + R
2
 = R
1
 + 
4
3
. R
1
 = 
7
3
 R
1
.
with two coils in series, let the kettle take t' time to boil.
22
1
V t' V t'
Q
7 JR'
4.2R
3
==
æö
´
ç÷
èø
      .....(d)
DPP/ P 36
102
Comparing (a) and (d), we get  
t'
(7 / 3)
 = 6 × 60
or   t' = 
7
3
  × 6 × 60 sec = 14 min.
(ii) When the two heating coils are in parallel, the effective
resistance is,
11
12
1
12
11
4
RR
RR 4 3
RR
4 RR7
RR
3
æö
ç÷
èø
= == ¢¢
+ æö
+
ç÷
èø
In parallel arrangement of heating coils, let t" be the time
taken by kettle to boil, so
22
1
Vt" Vt''
Q
4 JR"
4.2R
7
==
æö
´
ç÷
èø
.....(5)
Comparing (a) and (5), we get
t"
(4 / 7)
 = 6 × 60 or t" = 
4
7
 × 6 × 60 sec = 3.43 min.
17. (c).
24V
2kW
1.2kW
7.5amp
15V
9V
I=
v
R
eq
 p
1.56
R
7.5
´ éù
=
êú
ëû
240 60
I 7.5mA
328
= Þ=
(1) Currrent I is 7.5mA
(2) V oltage drop across R
L
 is 9 volt
(3) 
2
1 12
2
21
2
P vR 2256
16.66
P R 2 81
v
´
= Þ=
´
(4) After intercharging the two resistor R
1
 and R
2
eq
v 24
I 7 3.5mA
R (48)
= = ´=
24V
21V
3V
2
2 2
1 1 L1
2
2L2
2
P vRv 9
9
PR v3
v
æö
æö
= Þ ==
ç÷ ç÷
èø
èø
Sol. (18-20).
12V
I 2A
(1 5)
==
+W
5
1 12 V
W
W
Þ  Rate of chemical energy conversion = EI = 12 × 2 = 24 W
and P (in battery) = I
2
r = 4 W
Also, P (in resistor) = I
2
r = 20 W
18. (a)      19.   (c)      20.   (a)
21. (d) Resistivity of a semiconductor decreases with the
temperature. The atoms of a semiconductor vibrate with
larger amplitudes at higher temperatures thereby
increasing it's conductivity not resistivity.
22. (d) It is quite clear that in a battery circuit, the point of
lowest potential is the negative terminal of the battery
and the current flows from higher potential to lower
potential.
23. (b) The temperature co-efficient of resistance for metal is
positive and that for semiconductor is negative.
In metals free electrons (negative charge) are charge
carriers while in p-type semiconductors, holes (positive
charge) are majority charge carriers.