ELECTRICITY

QUESTION BANK

1 MARK QUESTIONS

1. If the charge on an electron be 1.6 X 10-19 C, Find the approximate

number of electrons in 1C.
A-q=ne,n=q/e=1/1.6 X 10-19 =6X1018
2. Why tungsten is used for making filament of electric lamp?
A- Because of low resistivity and high melting point.
3. Name the instrument used to measure electric current in a circuit.
A-Ammeter
4. What is potential difference? Give its S I unit.
A-Amount of work done in moving an electric charge between two given
points.
5. What is meant by resistance?
A-Opposition offered to the flow of electrons by a conductor.
6. What are the factors on which resistivity of a material depends?
A- Nature of material.
7. What will happen to the resistance of a circuit if the current through it is
doubled?
A-No change
8. Define electric power. Write its S. I unit.
A-Rate of doing work / Rate of dissipation of energy, Watt
9. State which has a higher resistance a 50 W or a 25 W lamp bulb and how
many times.
A-R=V2/t or Rα1/P
Thus 25 W lamp has double the resistance of 50 W lamp.
10.What is the lowest resistance that can be obtained by combining 4 coils of
resistances 4 Ω, 8 Ω, 12 Ω and 24 Ω.
A- By combining them in parallel connection. Rp=2 Ω
11.In the following circuit calculate the current flowing through the resistor.

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Ans-V=12 V, R=2 Ω

I=V/R=12/2=6 A

12.In the following circuit find out the potential difference across each
resistor.

12V

Ans- Rs=2+2=4 Ω

Total I=V/R=12/4=3 A

V1=IXR=3 X2=6 V

13. What is the maximum resistance which can be made using five resistors
each of 1/5 Ω?
Ans- Rs=R1+......R5=1/5+......5 times=1 Ω
14.A cylindrical conductor of length l and uniform area of cross section A
has resistance R. What will be the area of cross section of another
conductor of length 2l and resistance R of the same material?
Ans-A/2
15.A student carries out an experiment and plots the V-I graph of three
samples of nichrome wire with resistances R1, R2 and R3 respectively.

Arrange the resistances in their decreasing order.

Ans- R3>R2>R1
2 MARK QUESTIONS

1. What is voltmeter? How is it connected in a circuit?

A- Device used to measure potential difference, parallel

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 2. Mention the condition under which charges can move in a conductor.
Name the device which is used to maintain this condition in an electric
circuit.
A- If there is a potential difference across the two ends of the conductor.
Electric cell / battery.
3. Name the effect of electric current which is utilised in the working of an
electric fuse. How is a fuse connected in a domestic circuit?
A- Heating effect, series.
4. A bulb is rated at 5 V, 100 mA. Calculate its power and resistance.
A-V=5 V, I=100 mA=0.1 A, P=VI=5 X 0.1=0.5 W
R=V/I=5/0.q=50 Ω
5. List two differences between a voltmeter and an ammeter.
A- Voltmeter- a) Measures potential difference
b) Connected in parallel

B- Ammeter- a) Measures current

b) Connected in series

3 MARK QUESTIONS

1. Find the number of electrons transferred between two points kept at a

potential difference of 20 V if 40 J of work is done.
A- V=20 V, W=40 J
V=W/q
q=W/V=40/20=2C
n=q/e=2/1.6X 10-19=1.25 X 1019
2. A piece of wire of resistance 6 Ω is connected to a battery of 12 V. Find
the amount of current flowing through it. Now the same wire is redrawn
by stretching it to double its length. Find the resistance of the new wire.
A- R=6 Ω, V=12 V
I=V/R=12/6=2 A
Original Resistance R=ρL/A
L’=2L,, A’=A/2
R’=ρL’/A’=4ρL/A=4R
3. Define electric current and its S. I unit. With the help of ohm’s law
explain the meaning of 1 Ω resistance.
A- Rate of flow of electric charge

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 1 A is the amount of current flowing through the conductor for which
1 C of charge transferred per second.
From Ohm’s law-
V=IR
R=V/I
1Ω=1V /1A
1 Ω is the resistance of a conductor when potential difference across it
is 1 V and current flowing through it is 1 A.
4. Resistivity of two elements A & B are 1.62X10-8 Ωm and 520 X10-8 Ωm
respectively. Out of these two, name the element that can be used –
i) Filament of electric bulb
ii) Wires of electrical transmission lines. Justify your answer.
A- i) Element B – it has more resistivity.

ii)Element A- it has less resistivity , less dissipation of energy.

5. An electric lamp, whose resistance is 20 W, and a conductor of 4 W

resistance are connected to a 6 V battery .Calculate –
(a) the total resistance of the circuit,
(b) the current through the circuit, and
(c) the potential difference across the electric lamp and conductor.

Ans- The resistance of electric lamp, R1 = 20 W,

The resistance of the conductor connected in series, R2 = 4 W.
Then the total resistance in the circuit
R = R 1 + R2
Rs = 20 W + 4 W = 24 W.
The total potential difference across the two terminals of the battery
V = 6 V.
Now by Ohm’s law, the current through the circuit is given by
I = V/Rs

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 = 6 V/24 W
= 0.25 A.

Applying Ohm’s law to the electric lamp and conductor separately,

we get potential difference across the electric lamp,
V1 = 20 W × 0.25 A
= 5 V;
and,
that across the conductor, V2 = 4 W × 0.25 A
= 1 V.
Suppose that we like to replace the series combination of electric
lamp and conductor by a single and equivalent resistor. Its resistance
must be such that a potential difference of 6 V across the battery
terminals will cause a current of 0.25 A in the circuit. The resistance
R of this equivalent resistor would be
R = V/I
= 6 V/ 0.25 A
= 24 W.
This is the total resistance of the series circuit; it is equal to the sum
of the two resistances.

6. If in the following Figure R1 = 10 Ω, R2 = 40 Ω, R3 = 30 Ω, R4 = 20 Ω,

R5 = 60 Ω, and a 12 V battery is connected to the arrangement. Calculate
(a) the total resistance in the circuit, and
(b) the total current flowing in the circuit.

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Ans- Suppose we replace the parallel resistors R1 and R2 by an
equivalent resistor of resistance, R’. Similarly we replace
the parallel resistors R3, R4 and R5 by an equivalent single
resistor of resistance R². Then,

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 1/ R’ = 1/10 + 1/40 = 5/40; that is R’ = 8 Ω.
Similarly
1/ R’’ = 1/30 + 1/20 + 1/60 = 6/60;
that is, R’’ = 10 Ω.
Thus, the total resistance, R = R’ + R’’ = 18 Ω.
To calculate the current, we use Ohm’s law, and get
I = V/R = 12 V/18 Ω = 0.67 A.

5 MARK QUESTIONS

1. (a) State Ohm’s law. Does Ohm’s law hold good under all conditions.
(b)Two resistors, with resistances 5 Ω and 10 Ω respectively are
connected to a battery of e.m.f 6 V so as to obtain

(i) Minimum current flowing (ii) maximum current flowing

How will you connect resistances in each case show with circuit
diagram?

Ans- (a) At constant temperature potential difference across two ends of a

conductor varies directly with the current.
No
(b) i) Minimum current when connected in series ii) Maximum current
when connected in parallel.
2. (a) Name the factors on which resistance of a conductor depends.
(b)Find out an expression for resistivity from resistance of a conductor.
(c) A piece of wire of resistance 20 Ω is drawn out so that its length is
increased to twice its original length. Calculate the resistance of the wire
in the new situation.

Ans- a) length , Area of cross section and nature of material.

b) ρ=RA/l

c) R=ρl/A

R’=2ρl/A=2R=2 X20=40 Ω

3. (a) Draw a circuit diagram of a circuit consisting of a battery of two

cells, a lamp, an ammeter and a plug key all connected in series.

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 (b) A current of 0.5 A is drawn by a filament of an electric bulb for
10minutes. Find the amount of electric charge that flows through any point of
the electric circuit.

Ans- a)

b) I=0.5 A, t=10 minute=10 X 60=600 s

Q=IX t=0.5 X 600=300 C

4. (a) Write down any two advantages of parallel combination of resistors over
series combination.

(b) Two resistors when connected in parallel give resultant value of 2 Ω, when
connected in series the value becomes 9 Ω. Calculate the value of each
resistance.

Ans- a) 1. Total current flowing through the circuit is more than series
connection

2. Appliances connected in parallel can be operated independently.

b) R1 +R2=9Ω, R1 X R2/R1+R2=2Ω
R1 XR2=2 X9=18
R2=9-R1
R1(9-R1)=18
9R1-R12=18
R12-9R1+18=0
(R1-6)(R1-3)=0
R1=3Ω R2=6 Ω
Or R1=6 Ω R2=3 Ω

5. (a) What is joule’s law of heating? List any two of its practical application.

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(b) An electric iron has a rating of 750 W, 220 V. Calculate-

i) The current passing through it.

ii) Its resistance.

Ans- a) H=I2Rt

1. Electric fuse

2. Electric Geyser

b) i) I=P/V=750/220=3.4 A

ii)R=V/I=220/3.4=64.71 Ω

6. (a) What is the commercial unit of electrical energy? Represent it in terms of

joules.

(b) A geyser is rated 1500 W, 250 V. It is connected to 250 V mains. Calculate

– i) The current drawn

ii) The energy consumed in 50 hours

iii) The cost of energy consumed at Rs 2.20 per kWh.

Ans- a) kWh, 1kWh=1kW X1 h

=1000W X 3600 s

=3.6 X106J

b) i) I=P/V=1500/250=6 A

ii) E=P X t=1500 X 50 =75kWh

iii) Cost=75 X 2.20=Rs.165/-

7. (a) Why alloys are preferred over metals to make heating elements?

(b) Two resistors of 4 Ω and 6 Ω are connected in parallel. The combination is

connected across a 6 V battery of negligible battery. Calculate –

i) The power supplied by the battery.

ii) The power dissipated in each resistor.

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Ans-a) Alloys have more resistivity than their constituent metals and do not get
oxidised easily.

b) Rp=R1XR2/R1+R2=2.4 Ω

i) P=V2/R=36/2.4=15 W

ii) P1=V2/R1=36/4=9 W

P2= V2/R2=36/6=6W

8. (a) Why does the cord of an electric heater not glow while the heating
element does?

(b) Two lamps one rated 100 W at 220 V and the other 60 W at 220 V are
connected in parallel to electric mains supply. What current is drawn from the
line if the supply voltage is 220 V?

Ans- a) Cord is made of good conductor while heating element is made of alloys
having high resistivity which do not get oxidised.

b) R1=220X220/100=484 Ω

R2=220X220/60=2420/3Ω

Rp=2420/8

I=V/R=220X8/2420=0.73 A

9. (a) Why is the series arrangement not used for domestic circuits?

(b) Compare the power used in the 2 Ω resistor in each of the following circuits-

i) A 6 V battery in series with 1 Ω and 2 Ω resistors

ii) A 4 V battery in parallel with 12 Ω and 2 Ω resistor.

Ans- a) Because if one of the component in the circuits stops working the other
appliances will not work .

b) i)Rs=1+2=3, I=V/R=6/3=2 A,P=I2R=8 W

ii) P=V2/R=4X4/2=8 W

10. (a) Why are copper and Aluminium wires usually employed for electricity
transmission?

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(b) A hot plate of an electric oven connected a 220 V line has two resistance
coils A and B, each of 24 Ω resistance which may be used separately in series or
in parallel. What are the currents in the three cases?

Ans- a) Because they are pure metals having low resistivity and good
conductors of electricity.

b) Case-1

I=V/R=220/24=9.2 A

Case-2

I=V/Rs=220/24+24=4.6 A

Case-3

I=V/Rp=220/12=18.3 A

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