Electricity

1. Current flows through a conductor connected 4 10

across a voltage source. Now the resistance of the (a) r (b) r
3 3
conductor is reduced to one fourth to its initial 5
value and connected across the same voltage (c) r (d) 10r
3
source. The heating effect in the conductor will
become
6. Following graph was plotted between V and I
(a) Half (b) Double
values, across a wire. Which of the following
(c) Four times (d) One fourth
statements) is/are correct regarding this?
2. A strip of copper and another of silicon are cooled
from room temperature to 32 K. Which of the
following is true regarding this?
(a) Resistance of copper strip decreases because
it has positive temperature coefficient of
resistance.
(b) Resistance of copper strip increases because it
has negative temperature coefficient of resistance.
(c) Resistance of silicon strip decreases because it
has negative temperature coefficient of resistance.
(d) Resistance of silicon strip increases because it V
(a) Value of ratio when the potential difference
has positive temperature coefficient of resistance. l
V
is 0.8 V is not equal to the value of ratio when
3. An air conditioner is rated 260 V, 2.0 kW. The air l
conditioner is switched on for 10 hours each day. the potential difference is 1.2 V.
What is electrical energy consumed in 30 days? (b) This graph illustrates the non-ohmic law.
(a) 20 kW h (b) 2000 kW h (c) While plotting this graph, the temperature
(c) 600 kW h (d) 420 kW h remains constant.
(d) All of these
4. Match the column I with column II and select the
correct option from the codes given. 7. In which of the following network of resistors the
Column I Column II equivalent resistance between points A and B is
(A) Electric current (i) volt highest?
(B) E.m.f. (ii) ohm
(C) Resistance (iii) ohm-metre
(D) Resistivity (iv) ampere

(a) (A)-(iv), (B)-(ii), (C)-(i), (D)-(iii)

(b) (A)-(iii), (B)-(iv), (C)-(i), (D)-(ii)
(c) (A)-(iv), (B)-(i), (C)-(ii), (D)-(iii)
(d) (A)-(iii), (B)-(i), (C)-(ii), (D)-(iv)

5. What is the equivalent resistance between points

A and B in the given circuit diagram?

(a) Network (i)

(b) Network (ii)
(c) Network (iii)
(d) All have equal equivalent resistance.

8. Which of the following statements about the

current in given circuit is correct?

2
 12. Arrange the order of power dissipated in the given
circuits, if the same current enters at point A in all
the circuits and resistance of each resistor is R.

(a) Current at Q is greater than the current at R.

(b) Current at R is greater than the current at P.
(c) P has maximum current.
(d) R has maximum current.

9. An electric bulb rated 220 V, 60 W is working at

full efficiency. Another identical bulb is connected (a) (ii) > (iii) > (iv) > (i)
in the same circuit having power supply of 220 V. (b) (iii) > (ii) > (iv) > (i)
(i) If both the bulbs are connected in series then (c) (iv) > (iii) > (ii) > (i)
the total power consumption will be 60 W. (d) (i) > (ii) > (iii) > (iv)
(ii) If only one bulb is connected then the total
power consumption will be 30 W. 13. While carrying out an experiment, to find the
(iii) If the both bulbs are connected in parallel then equivalent resistance of two resistances
the total power consumption will be 120 W. connected in parallel, a student sets up the circuit
Which of the above statement(s) is/are correct as shown. The teacher checks it and tells him that
regarding the circuit? his circuit has one or more of the following 'faults'.

(a) Only (i) and (ii) (b) Only (ii) and (iii)
(c) Only (iii) (d) Only (i)

10. How will be the reading in the ammeter A affected

if another identical bulb Q is connected in parallel
to P? (The voltage in the mains is maintained at a
constant value.)
(i) The resistors R1 and R2 have not been correctly
connected in parallel
(ii) The voltmeter has not been correctly
connected in the circuit
(iii) The ammeter and the key have not been
(a) The reading will be reduced to one-half. correctly connected in the circuit.
(b) The reading will not be affected. Out of these three, the actual fault in hi circuit
(c) The reading will be doubled. is/are
(d) The reading will be increased four-fold.
(a) Both (i) and (ii) (b) Both (ii) and (iii)
11. Six equal resistances are connected between (c) (i) only (d) (ii) only
points P, Q and R as shown in the figure. The net
resistance will be maximum between Direction (Q. No. 14 and 15): Observe the given
circuit diagram and answer the following questions.

(a) P and Q (b) Q and R

(c) P and R (d) Any two points. 14. What is the value of current shown by the
ammeter?

3
 (a) 2 A (b) 0.4 A
(c) 1.6 A (d) 5 A

15. What is the potential difference across 2 ?

(a) 2.5 V (b) 1 V
(c) 1.6V (d) 5V

Achievers Section (HOTS) l1 1 l1 1

(a)  (b) 
l2 4 l2 2
16. The resistors R1, R2 , R3 and R4 in the given circuit
l1 2 l1 3
are all equal in value and connected with a (c)  (d) 
negligible resistance wire. Which of the following l2 3 l2 5
is correct relationship between the voltmeters
readings V1, V2 and V3 19. Consider a thin square sheet of side x and
thickness y made of a material of resistivity p. The
resistance between two opposite faces, shown by
the shaded areas in the figure is

(a) V1  V2  V3 (b) V1  V2  V3
(c) V1  V2  V3 (d) V1  V2  V3

17. In the circuit shown here, the ammeter A reads 5

A and the voltmeter V reads 20 V. The correct (a) Directly proportional to x
value of resistance R is (b) Directly proportional to y
(c) Independent of x
(d) Independent of y.

20. The plot represents the flow of current through a

wire at three different time intervals. The ratio of
(a) Exactly 4
charges flowing through the wire at different
(b) Slightly greater than 4
intervals is
(c) Slightly less than 4
(d) Zero

18. A ring is made of a wire having a resistance 12.

The points P and Q as shown in figure, at which
a current carrying conductor should be
connected, so that the resistance of the sub circuit
8
between these points is equal to , divide the
3
ring in lengths l1 and l2. Then
(a)1 : 2 : 3 (b) 1 : 1 : 1
(c) 3 : 2 : 2 (d) 2 : 3 : 3

4
 Answer key
1. C 2. A 3. C 4. C 5. B

6. C 7. C 8. C 9. C 10. C

11. A 12. A 13. A 14. C 15. C

16. D 17. B 18. B 19. C 20. B

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 HINTS & EXPLANATIONS (ii)
1

1

1

1

7
Rii 1000 200 1000 1000
1000
1. (c) Heating effect is the wattage of the conductor.  Rii   142.85
7
V2
So, H 1 1 1 2
R (iii)   
R1 1000 1000 1000
If resistance is reduced to four times then
V2  1000
V2 R1   500
H'  or H '  4   2
(R/ 4) R
Riii  500  200  700
 H '  4H
 Riii  Ri  Rii

2. (a) Not Available

8. (c) Current at P is maximum because it is
3. (c) V = 260 V, P = 2 kW = 2000 W connected in series with battery and hence shows
Time for which air conditioner is switched on t = total current drawn from battery.
10 hours  30 = 300 hours
Electrical energy consumed in 30 days 9. (c) V  220V , P  60 W
 p  t  2000  300 V2
P …(i)
 600000 Wh  600 kW h R
As bulbs are connected in series, then
4. (c) Not Available V2 V2 V2
P'   
Req R  R 2R
5. (b) Resistance of wires CD and ED are in series, 1
R1  2r  2r  4r or P'  P (from (i))
2
 Resistance of wire CE is in parallel with R1 60
or P '   30 W
1 1 1 1 2 2
  
R2 4r 2r 4r As bulbs are connected in parallel, then
1 1 1 R
1 3 4   or Req 
  R2  r Req R R 2
R2 4r 3
 Resistance of wire AC, R2 and resistance of V2 V2
So power Peq   2
Req R
wire EB are in series
4r 3r  4r  3r 10 or Peq  2P  2  60  120 W
r r   r
3 3 3
10. (c) Let, resistance of bulb P and Q be R. So
6. (c) The ratio of potential difference applied to the reading of ammeter before connecting Q to the
wire and current passing through it is a constant. circuit is
Ohm's law is illustrated by this graph. Hence, V
I ….(i)
temperature must be constant while calculating R
values. After connecting Q to the circuit, resultant
resistance of the circuit is
7. (c) (i) R1  1k  200 1 1 1 2 R
   or Req  
 1000   200  1200 Req R R R 2
1 1 1 11 Thus, new reading of ammeter will be
  
Ri 1200 1000 6000 V 2V
l'  
6000 R/2 R
 Ri   545.45
11 or l '  2l (From (i))

6
11. (a) Let resistance of each resistor be r.
(ii)
Net resistance, RAB  3R
Power dissipated.
PAB  l 2 RAB  l 2 (3R)  3l 2 R

(iii)

Resistance in arm PQ, R1  r

r r r
Resistance in arm QR, R2  
r r 2
R 3
Resistance in arm PR, Net resistance, RAB  R R
2 2
1 r
R3   Power dissipated.
1 1 1 3
  3  3
r r r PAB  l 2 RAB  l 2  R   l 2 R
Let net resistance between P and Q is RPQ . 2  2

1 1 1 1 6 11
     (iv)
RPQ r r  r r 5r 5r
3 2
5r
 RPQ 
11
Net resistance between Q and R RQR , 1 1 1 3
Net resistance   
RAB 2R R 2R
1 1 1 2 3 11
     2
RQR r r  r r 4r 4r or RAB  R
2 3 3
Power dissipated.
4r
 RQR  2  2
11 PAB  l 2 RAB  l 2  R   l 2 R
Net resistance between P and R, RPR 3  3
Thus, (ii) > (iii) > (iv) > (i)
1 1 1 3 2 11
    
RPQ r / 3 r  r / 2 r 3r 3r
13. (a) Resistors R1 and R2 should be connected in
3r parallel with each other.
 RPR 
11 Voltmeter should be connected in parallel to both
Hence RPQ  RQR  RPR R1 and R2 .
The ammeter and the key have been correctly
12. (a) (i) connected in the circuit.

R
Net resistance, RAB 
3
Power dissipated. 14. (c)
R
PAB  l 2 RABl 2
3

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15. (c) Voltage drop in arm containing 2 and 3 is l x
R 
4 V. Let Voltage drop across each resistor be 2x A xy
and 3 x. 
or, R 
 2x  3  4  x  4 / 5 y
 Voltage across 2 is 2x Thus, resistance between two opposite faces,
i.e., 2  4 / 5  1.6 V shown by the shaded areas in the figure is
independent of x.
16. (d) Total potential difference across resistances is
V  V1  V2  V3 20. (b) We know, current flowing through the wire,
Since the resistances are connected in series, the dq Q
l 
same current flow through each resistance and dt t
R1  R2  R3  R or Q  lt
Thus, area under the graph between current and
 V1  lR1  lR
time will give charge flowing through the wire.
V2  lR2  lR
For first time interval, (t1  2  1  1s)
V3  l(R3  R4 )  l(R R)  2l R
Q1  l1  t1  2  1  2C
 V1  V2  V3
For second time interval, (t 2  5  3  2 s)
Q2  l2  t2  1  2  2C
17. (b) Voltmeter is not ideal so a small amount of
current will pass through it. Hence voltage drop For third time interval, (t 3  8  6  2s)
across R will 20 V if R is lightly greater than 4. 1 1
Q3   l3  t3   2  2  2C
2 2
18. (b) Here, resistance of wire  R  12   Ratio of charges flowing through the wire at
different intervals is
8
Equivalent resistance of lengths l1 and l2 
 Q1 : Q2 : Q3  2 : 2 : 2 :  1 : 1 : 1
3
Let, resistance of length l1 and l 2 are R1 R2
respectively.
R1 R2
 Req 
R1  R2
We know, R1  R2  R  12  …(i)
8 R1 R2
So, 
3 12
or, R1 R2  32 …(ii)
From (i) and (ii)
R1  4  and R2  8 
l
We know, R  
A
or Rl
R l l 1
So, 1  1 or 1 
R2 l2 l2 2

l
19. (c) We know, R  
A
Here, resistivity  , l  x, A  xy