Electricity
TOP 25 QUESTIONS
-by Prashant Bhaiya
MULTIPLE CHOICE QUESTIONS
1. The expressions that relate (i) Q. I and t and (ii) Q. V and W respectively
are (Here the symbols have their usual meanings)
a) (i) I = Q/t (ii) W = V/Q
b) (i) Q = I*t (ii) W = V*Q
c) (i) Q = I/t (ii) W = V/Q
d) (i) I = Q/t (ii) Q = V/W
2. A student plots V-I graphs for three samples of nichrome wire with
resistances R1, R2 and R3. Choose from the following the statements that
holds true for this graph.
a) R1 = R2 = R3
b) R1 > R2 > R3
c) R3 > R2 > R1
d) R2 > R1 > R3
3. A cylindrical conductor of length 'T' and uniform area of cross section
'A' has resistance 'R’. The area of cross section of another conductor of
same material and same resistance but of length '2l' is
a) A/2
b) 3A/2
c) 2A
d) 3A
4. Assertion (A): Alloys are commonly used in electrical heating devices
like electric iron and heater.
Reason (R): Resistivity of an alloy is generally higher than that of its
constituent metals but the alloys have low melting points then their
constituent metals.
a) Both (A) and (R) are true and (R) is the correct explanation of (A).
b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
c) (A) is true but (R) is false.
d) (A) is false but (R) is true.
5. If a person has five resistors each of value 1/5 ohm, then the maximum
resistance he can obtain by connecting them is
a) 1 ohm
b) 5 ohm
c) 10 ohm
d) 25 ohm
6. Two LED bulbs of 12 W and 6 W are connected in series. If the current
through 12 W bulb is 0.06 A the current through 6 W bulb will be
a) 0.04 A
b) 0.06 A
c) 0.08 A
d) 0.12 A
7. The resistance of a resistor is reduced to half of its initial value. If other
parameters of the electrical circuit remain unaltered, the amount of heat
produced in the resistor will become
a) four times
b) two times
c) half
d) one fourth
8. The resistance of a resistor is reduced to half of its initial value. In doing
so, if other parameters of the circuit remain unchanged, the heating
effects in the resistor will become
a) two times
b) half
c) one-fourth
d) four times
9. An electric kettle consumes 1 kW of electric power when operated at
220 V. The minimum rating of the fuse wire to be used for it is
a) 1A
b) 2A
c) 4A
d) 5A
10. Two bulbs of 100 W and 40 W are connected in series. The current
through the 100 W bulb is 1 A. The current through the 40 W bulb will be
a) 0.4A
b) 0.6 A
c) 0.8 A
d) 1 A
SHORT AND LONG QUESTIONS
11. (a) Define the term ‘coulomb’.
(b) State the relationship between the electric current, the charge
moving through a conductor and the time of flow.
Calculate the charge passing through an electric bulb in 20 minutes if the
value of current is 200 mA.
12. Study the following electric circuit and find (i) the current flowing in
the circuit and (ii) the potential difference across 10 Ω resistor.
13. (a) An electric iron of 1 kW is operated at 220 V. Find which of the
following fuses that respectively rated at 1 A,3 A and 5 A
can be used in it.
(b) An electrical fuse is rated at 2 A. What is meant by this statement?
14. Find the current flowing through the following electric circuit.
15. Two devices of rating 44 W, 220 V and 11 W, 220 V are connected in
series. The combination is connected across a 440 V mains. The fuse of
which of the two devices is likely to burn when the switch is ON? Justify
your answer.
16. Consider the circuit shown in the diagram. Find the current in 3Ω
resistor.
17. For the circuit shown in this diagram, calculate
(i) the resultant resistance.
(ii) the total current.
(iii) the voltage across 7 Ω resistor.
18. A 9 Ω resistance is cut into three equal parts and connected in parallel.
Find the equivalent resistance of the combination.
19. (a) How is the direction of electric current related to the direction of
flow of electrons in a wire?
(b) Calculate the current in a circuit if 500 C of charge passes through it in
10 minutes.
20. An electric iron has a rating of 750 W, 220 V. Calculate the (i) current
flowing through it, and (ii) its resistance when in use.
21. (a) List the factors on which the resistance of a conductor in the shape
of a wire depends.
(b) Why are metals good conductors of electricity whereas glass is a
bad conductor of electricity? Give reason.
(c) Why are alloys commonly used in electrical heating devices? Give
reason.
22. Study the following electric circuit in which the resistors are arranged
in three arms A, B and C:
(a) Find the equivalent resistance of arm A.
(b) Calculate the equivalent resistance of the parallel combination of the
arms B and C.
(c) (i) Determine the current that flows through the ammeter.
OR
(ii) Determine the current that flows in the ammeter when the arm B is
withdrawn from the circuit.
23. (a) List two disadvantages of using a series circuit in homes.
(b) Calculate the effective resistance between A and B in the circuit given
below:
24. (i) State Joule's law of heating. Express it mathematically when an
appliance of resistance R is connected to a source of voltage V and the
current / flows through the appliance for a time t.
(ii) A 52 resistor is connected across a battery of 6 volts. Calculate the
energy that dissipates as heat in 10 s.
25. Compare the power used in 2 ohm resistor in each of the following
circuits.
SOLUTIONS:
1. (b): As, I=Q/t , V=W/Q or Q=It, W=VQ
2. (d): The inverse of the slope of I-V graph gives the resistance of the
material. Here the slope of R3 is highest. Thus, R2>R1>R3.
3. (c): The resistance of a conductor of length I, and area of cross section, A
is
4. (c): Alloy : These are mixture of metal with other elements in nature.
5. (a): The maximum resistance can be obtained from a group of resistors
by connecting them in series. Thus, Rs = 1/5 + 1/5 + 1/5 + 1/5 + 1/5 = 1 ohm
6. (b): As they are connected in series, so current is same in both the bulbs.
7. (b): As voltage is same, resistance is halved,
8. (a): We know,
9. (d): Given: P = 1kW and V = 220 V
P = VI
I = P/V = 4.54 A
So, minimum rating of fuse wire is 5 A.
10. (d) : Given power of first bulb = 100 W and second buld P2 = 40 W.
Current through 100 W bulb, I1 = 1 A
Current through 40 W bulb, l2 =?
Since both the bulbs are connected in series, the electric current passing
through both the bulbs are same i.e., I2 = 1 A.
11. (a) When 1 A current flows across the wire in 1 second, the charge
transfer across its ends is said to be 1 coulomb.
(b) The relationship between the electric current I, the charge q and
time t is: I = q/t
t = 20 * 60 = 1200 s
I = 200 mA = 0.2 A
Charge passing = q = It = 240 C
12. 10 ohm and 20 ohm are connected in series, their eq. resistance is
Rs = R1 + R2 = 30 ohm
(i) Current flowing in the circuit
I = V/Rs = 3/30 = 0.1 A
(ii) Potential difference across 10 ohm resistor
V = IR = 0.1 * 10 = 1 V
13. (a) Given: P = 1 kW = 1000 W and V = 220 V
Current drawn, I = P/V = 4.54 A
To run iron of 1 kW, rated fuse of 5 A should be used.
(b) It means, the maximum current will flow through it is only 2 A. Fuse
wire will melt if the current exceeds 2 A value through it.
14. Series combination of 1 ohm, 2 ohm and 3 ohm resistance is in parallel
combination with 6 ohm. Their equivalent resistance is
Now, 3.6 ohm, 2.4 ohm and 3 ohm are in series, their equivalent resistance
be
Hence, the current flowing through the circuit is:
I = V/R = 4.5/9 = 0.5 A
15.
16. 3 Ω and 6 Ω are in parallel.
1/ Rp = 1/3 + 1/6 = 1/2 ohm
Rp = 2 ohm
Rp and 10 Ω are connected in series. So, Rs = Rp + 10 Ω = 12 Ω
Total current in circuit, I = V/Rs = 12/12 = 1 A
Potential difference across Rp = I*Rp = 2 V
So, Potential difference across 3 ohm = 2 V
Current throught 3 ohm, I1 = V/R1 = 2/3 = 0.67 A
17. (b)
18. Resistance of each part = R/3 = 3 ohm
R1 = R2 = R3 = 3 ohm
In parallel combination,
1/ Rp = 1/R1 + 1/R2 + 1/R3 = 1 ohm
19. (a) Conventional direction of electric current is opposite to the
direction of flow of electrons in a wire.
(b) q = 500 C, t = 10 X 60 = 600 s.
I = 500/600 = 5/6A
20. Given: P = 750 W, V = 220 V
(i) P = VI
750 = 220 * I
I = 750/220 = 3.40 A
(ii) P = V2/R
R = 220*220/750 = 64.53 ohm
21. (a) Resistance of a conductor depends upon the following factors:
(1) Length of the conductor: Greater the length (l) of the conductor more
will be the resistance (R).
(2) Area of cross-section of the conductor: Greater the cross-sectional area
of the conductor, less will be the resistance.
(3) Nature of conductor
(b) Metal have very low resistivity and hence they are good conductors
of electricity. Whereas glass has very high resistivity so glass is a bad
conductor of electricity.
(c) Alloys are commonly used in electrical heating devices due to the
following reasons
(i) Alloys have higher resistivity than metals
(ii) Alloys do not get oxidised or burn readily.
22.
23. (a) Two disadvantages of using a series circuit in homes:
(i) If all devices are connected in series, the greater's the resistance.
(ii) If a device have faulted then all devices of home stop working.
(b) Equivalent resistance between A and B
We know that 6 ohm and 4 ohm are in series, then
R1 =6+4 = 10 ohm
and 3 ohm and 3 ohm are in series, R3 = 3 +3 = 6 ohm
Now equivalent resistance between A and B,
1/ Rab = 1/10 + 1/15 + 1/6 = 1 / 3
Rab = 3 ohm
24. (i) According to the Joule's law of heating, when a current (I) is passed
through a conductor of resistance (R) for a certain time (t), the conductor
gets heated up and the amount of energy released is given by
H = I2Rt or V2t/R
(ii) Here, V = 6 volt, R=5 ohm, t = 10 sec
H = 6*6*10/5
H = 72 J
25. In circuit A,
Total resistance, R = 1+2 =3 ohm
Voltage across 2 ohm = Vtotal/Rtotal * 2 = 6/3 * 2 = 4 V
Power used in 2 ohm resistor, P = V2/R = 8 W
In circuit B, Voltage across both the resistance is same i.e. 4V and both are
connected in parallel combination.
Power used in 2 ohm resistor = V2/R = 4*4/2 = 8 W
:. Power used in 2 ohm resistor in each case is same i.e. 8 W.
