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ELECTRICITY 12

TEXTBOOK QUESTIONS AND THEIR ANSWERS

Q.1. Name and define SI unit of electric current?

Ans. The name of SI unit of electric current is ampere. An

ampere is the amount of current flowing in a

conductor when a charge of 1 coulomb flows through

it in one second.

Q.2. What does an electric circuit mean?

Ans. Electric circuit is a continuous conducting path

between the terminals of a source of electric energy,

conducting wires and other electrical components

through which electric current flows.

Q.3. Name a device that helps to maintain a potential

difference across a conductor.

Ans. Electric cell or electric battery is a device in which

chemical energy changes into electrical energy, and



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hence, helps in maintaining potential difference across

a conductor.

Q.4. What is meant by saying that potential difference

between two points is 1 V?

Ans. It means one joule of work is done when a charge of

1 coulomb flows between the two given points.

Q.5. How much energy is given to each coulomb of

charge passing through a 6 V battery?

Ans. Energy = Work = P.D. × Charge

W = V × Q = 6V × 1C = 6 J

Q.6. Draw a schematic diagram of a circuit consisting of

a battery of three cells of 2V each, a 5Ω resistor an

8Ω resistor and a 12Ω resistor, a plug key and an

ammeter to measure the current through the

resistors, all connected in series. A voltmeter to

measure the potential difference across the 12Ω

resistor is also connected. What would be the

readings in the ammeter and the voltmeter?



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Ans.

Total voltage of 3 cells = 3 × 2V = 6V

Total resistance in series = 5Ω + 8 Ω + 12Ω

= 25Ω

∴ Current flowing through the circuit and hence

ammeter (I)

= 6V25 Ω

= 0.24A

Voltage drop across 12 Ω resistor = IR = 0.24 × 12

= 2.88 V

∴ Reading of voltmeter across 12 Ω resistor = 2.88V

Q.7. On what factors does the resistance of a conductor

depend?

Ans. (i) Resistance of a conductor is directly proportional

to the length of the conductor.



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(ii) Resistance of a conductor is inversely

proportional to the area of cross-section of the

conductor.

(iii) Resistance of a conductor depends upon its nature

of molecules.

Q.8. Will the current flow more easily through a thin

wire or a thick wire of the same material, when

connected to the same source? Why?

Ans. The current will flow more easily through thick wire.

It is because the resistance of a conductor is inversely

proportional to its area of cross-section. Thus, thicker

the wire, less is the resistance and hence more easily

the current flows.

Q.9. (a) Which among the iron and mercury is a better

conductor?

(b) Which material is the best conductor?

Ans. (a) Mercury is a better conductor than iron as it has

lower resistivity than iron.



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(b) Silver is the best conductor of electricity.

Q.10. A copper wire has diameter 0.5 mm and resistivity

of 1.6 × 10–8 Ωm. What will be the length of this

wire to make its resistance of 10Ω? How much will

the resistance change, if the diameter is doubled?

Ans. (i) R = 10 Ω ; ρ = 1.6 × 10–8 Ω m;

diameter (d) = 0.5 mm = 0.0005 m

∴ Length of wire l =

2

2dR

RA⎛ ⎞× π ⎜ ⎟⎝ ⎠=

ρ ρ

= –8

10 × 22 × 0.0005m × 0.0005 m7 × 4 × 1.6 × 10 m

WW

= –4 –4

–8

220 × 5 × 10 × 5 × 10 m7 × 4 × 1.6 × 10

= 220 × 2544.8

= 122.76 m



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(ii) 1

2

RR

=

22

22

1 1

2

2

dAA d

æ öp ç ÷è ø=

æ öp ç ÷è ø

∴ 2

10R

W = 2

2

(2×0.0005 m)(0.0005 m)

(∵ d2 = 2d1)

R2 = 104Ω = 2.5 Ω

Q.11. Let the resistance of an electrical component

remains constant, while the potential difference

across the two ends of the component decreases to

half of its former value. What change will occur in

the current through it?

Ans. The current will decrease to half, because according to

Ohm’s law, the current in a circuit is directly

proportional to potential difference.

Q.12. Why are coils of electric toasters and electric irons

made of an alloy, rather than a pure metal.



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Ans. 1. Alloys offer more resistance to the passage of

electric current than pure metals.

2. Alloys do not get oxidised easily as compared to

metals.

Q.13. Judge the equivalent resistance when the following

are connected in parallel : (a) 1 Ω and 106 Ω;

(b) 1 Ω, 103 Ω and 106Ω.

Ans. In either of the cases (a) and (b), the resistance will be

less than 1Ω, because the equivalent resistance in

parallel is less than the individual resistances.

Q.14. Which of the following terms does not represent

electrical power in a circuit :

(a) I2R (b) IR2 (c) VI (d) 2V

R

Ans. (b) is the correct answer.

Q.15. An electric bulb is rated 220 V and 100 W. When

operated on 110 V, the power consumed will be :



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(a) 100 W (b) 75 W (c) 50 W (d) 25 W

Ans. (d) is the correct answer.

Reason : Resistance of the bulb (R)

= 2V 220 × 220=

P 100 = 484 Ω.

∴ Power consumed at 110 V,

P = 2V 110 × 0=

R 48411 = 25W.

Q.16. Two conducting wires of the same material and of

equal lengths and equal diameters are first

connected in series and then parallel in an electric

circuit. The ratio of heat produced in series and

parallel combinations would be :

(a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1

Ans. (c) is the correct answer.

Reason : Let V be the potential difference at the ends

of conducting wires in series and parallel.



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(i) Resistance of 2 conducting wires in series

= R + R = 2R.

∴ Heat produced in series circuit H1 =2V

2R

(ii) Resistance of 2 conducting wires in parallel

Rp = R2

∴ Heat produced in parallel circuit H2

= 2 2V 2V=

R/2 R

∴ H1 : H2 = 2 2V 2V

2R R: = 1 : 4

Q.17. Why are copper and aluminium wires usually

employed for electricity transmission?

Ans. Copper and aluminium have very low resistivity.

Thus, large amount of current is transmitted through

them without any wastage.



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Q.18. What is (a) the highest, (b) lowest resistance that

can be secured by combinations of four coils of

resistance 4Ω, 8 Ω, 12 Ω and 24 Ω?

Ans. (i) For highest resistance, the coils should be

connected in series.

∴ Rs = r1 + ..... r4

= (4 + 8 + 12 + 24) Ω = 48 Ω

(ii) For lowest resistance, the coils should be

connected in parallel.

p

1R

= 1 4

1 1+ .....+r r

p

1R

= 1 1 1 1+ + +4 8 12 24

= 6 + 3 + 2+1 12 1= =24 24 2

∴ Rp = 2 Ω



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Q.19. An electric motor takes 5 A from a 220 V line.

Determine the power of the motor and the energy

consumed in 2 h.

Ans. Power of an electric motor (P) = I × V

= 5 × 220 = 1100 W.

Energy consumed by motor in 2 h

= P × t

= 1100W × 2h = 2200Wh

= 2.2 kWh.

Q.20. When a 12 V battery is connected across an

unknown resistor, there is a current of 2.5 mA in

the circuit. Find the value of the resistance of the

resistor.

Ans. Potential difference (V) = 12 V

Current (I) = 2.5 mA = 2.51000

A



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∴ Resistance (R) = V 12 × 1000=I 2.5

= 4800 Ω.

Q.21. A battery of 9 V is connected in series with resistors

of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω respectively.

How much current would flow through the 12 Ω

resistor?

Ans. Total resistance in series (R)

= (0.2 + 0.3 + 0.4 + 0.5 + 12) Ω

= 13.4 Ω

Potential difference (V) = 9 V

∴ Current in circuit (I) = V 9=R 13.4

= 0.67 A.

As all resistors in series circuit have same magnitude

of current, therefore, current in 12 Ω resistor = 0.67A.

Q.22. Two lamps, one rated 100 W at 220 V, and other

60 W at 220 V are connected in parallel to electric



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mains supply. What current is drawn from the

line, if the supply voltage is 220 V?

Ans. Total power of 2 bulbs in parallel (W)

= 100 + 60 = 160 W.

Supply voltage = 220 V

∴ Current drawn from line (I) = PV

= 160 W220 V

= 0.727 A

Q.23. Why is the tungsten used almost exclusively for the

filament of electric bulbs?

Ans. Tungsten has melting point of 3380°C and can be

drawn in fine wires. Furthermore, it offers very large

amount of resistance. Thus, when electric current is

passed through the tungsten filament, its temperature

rises above 2000°C, and hence, large amount of heat

energy produced, which is converted into light energy.



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Q.24. Why are the conductors of electric heating devices,

such as bread-toasters and electric irons, made of

an alloy, rather than a pure metal?

Ans. Manganin and nichrome are the alloys commonly

used in electric heating devices, because they offer

fairly large amount of electric resistance and do not

get oxidised till 1000°C.

Q.25. Why is the series arrangement not used for

domestic circuits?

Ans. In series arrangement, the magnitude of resistance is

maximum, and hence, very small amount of current

flows. Furthermore, the potential difference is not

constant in series circuit. Now, as the most of the

electric devices work at constant potential, but

consume different amounts of current, therefore, series

circuit is unsuitable. Furthermore, various devices in

series circuit cannot be operated independently with

independent switches.



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Q.26. An electric lamp of 100 Ω, a toaster of resistance

50 Ω and a water filter of resistance 500 Ω are

connected in parallel to a 220 V source. What is the

resistance of an electric iron connected to the same

source that takes as much current as all three

appliances, and what is the current through it?

Ans. Resistance of electric iron = Resistance of

appliances connected

in parallel

p

1R

= 1 1 1+ +100 50 500

= 5 + 10 + 1 16=500 500

∴ Rp = 50016

= 31.25 Ω.

∴ Current drawn by electric iron

I = V 220=R 31.25

= 7.04 A



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Q.27. How can three resistors of resistances 2 Ω, 3 Ω and

6 Ω be connected to give a total resistance of :

(a) 4 Ω (b) 1 Ω?

Ans. (a)

In the above circuit diagram resistance of 3 Ω and

6 Ω in parallel

p

1R

= 1 1 3+ = =3 6 6

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∴ Rp = 2 Ω

Therefore, resistance of 2 Ω and the parallel

segment

R = 2 + 2 = 4 Ω

(b)

Resistance of 2 Ω, 3 Ω, 6 Ω in parallel.



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∴ p

1R

= 1 1 1 6+ + = =2 3 6 6

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∴ Rp = 1 Ω

Q.28. An electric iron of resistance 20 Ω takes a current

of 5 A. Calculate the heat developed in 30 s.

Ans. Power of electric iron

P = I2 × R = (5)2 × 20 = 500 W

∴ Heat developed in 30 s

H = P × t = 500 W × 30 s = 15,000 J

Q.29. How many 176 Ω resistors (in parallel) are

required to carry 5 A on 220 V line?

Ans. Current in circuit (I) = 5 A


∴ Resistance of parallel circuit (Rp) = VI



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= 220 V5A

= 44 Ω

Now, p

1R

= 1 2

1 +r r

1 + …n times

⇒ 144

= 1 +176 176

1 + …n times

⇒ 144

= 176n

∴ n = 17644

= 4

Thus, 4 resistors of 176 Ω should be connected in

parallel.

Q.30. Show how you would connect three resistors, each

of resistance 6 Ω, so that the combination has a

resistance of : (i) 9 Ω ; (ii) 4 Ω.

Ans. (i)



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Resistance of B and C in parallel

p

1R

= 1 1 2+ = =6 6 6

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∴ Rp = 3 Ω.

∴ Combined resistance of A and Rp in series.

Rs = 6 Ω + 3 Ω = 9 Ω.

(ii)

Resistance of A and B in series

Rs = 6 + 6 = 12 Ω

∴ Resistance of Rs in parallel with C

∴ Rp = 1 1 3 1+ = =12 6 12 4

∴ Rp = 4 Ω



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Q.31. Several electric bulbs designed to be used on a

200 V electric supply line, are rated 10 W. How

many lamps can be connected in parallel with each

other across the two wires of 220 V line, if the

maximum allowable current is 5 A?

Ans. Maximum current (I) = 5 A


∴ Maximum available power (P) = I × V

= 5 × 220 = 1100 W

∴ Required number of lamps

= Maximum powerPower of one lamp

= 1100 W10 W

= 110 lamps

Q.32. A hot plate of an electric oven connected to a 220 V

line has two resistance coils A and B, each of 24 Ω

resistance, which may be used separately, in series



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or in parallel. What are the currents in the three

cases?

Ans. Constant voltage (V) = 220 V

(i) When single coil is used, then

R = 24 Ω

∴ Current in coil (I) = V 220V=R 24 W

= 9.17 A

(ii) When two coils are used in series, then

Rs = 24 Ω + 24 Ω = 48 Ω

∴ Current in coils in series

I = s

V 220=R 48

VW

= 4.58 A

(iii) When two coils are used in parallel, then

p

1R

= 1 1 2+ = =24 24 24 12

1

∴ Rp = 12 Ω



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∴ Current in coils in parallel

I = p

V 220=R 12

VW

= 18.33 A

Q.33. 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

resistors.

Ans. (i) Resistance of 1 Ω and 2 Ω in series

R = (1 + 2) Ω = 3 Ω

∴ Potential difference (V) = 6 V

∴ Current in series circuit

I = V = 2 A 6 V=R 3W

As current in series circuit is a constant quantity



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∴ Current in 2 Ω resistor = 2 A

∴ Power in 2 Ω resistor in series

P1 = I2 R = (2)2 × 2 = 8 W

(ii) P.D across 2 Ω resistor = 4 V.

∴ Power in 2 Ω resistor in parallel

(P2) = 2 2V ( = 8 W 4)=

R 2

∴ P1 : P2 = 8 W : 8 W

= 1 : 1

Q.34. Which uses more energy, a 250 W TV set in 1 hour

or a 1200 W toaster in 10 minutes?

Ans. Energy consumed by T.V. set = P × t

= 250 W × 1 h



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= 250 Wh

Energy consumed by toaster = P × t

= 1200 W × 1060

h

= 200 Wh.

TV set used more energy than toaster.

Q.35. Calculate the number of electrons constituting one

coulomb of charge.

Ans. When the charge is 1.602 × 10–19 C, the number of

electron = 1

When the charge is 1 C, the number of electrons

= –19

11.602 × 10

= 6.24 × 1018 electrons.

Q.36. Draw a schematic diagram of a circuit consisting of

a battery of four 2 V cells, a 5 ohm resistor, an

8 ohm resistor, and a 12 ohm resistor and a plug

key, all connected in series.



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Ans.

Q.37. Why does the connecting cord of an electric heater

not glow while the heating element does?

Ans. Heat produced in a conductor is given by the

expression I2.R.t. Now for a given circuit ‘I’ and ‘t’

are constant quantities. Thus, heat produced is directly

proportional to the resistance of the conductor.

Now the resistance of conducting wires is very small,

and hence, practically no heat is produced. However,

the resistance of heating element is very large. Thus, it

gets red hot.

Q.38. A piece of wire of resistance R is cut into five equal

parts. These parts are then connected in parallel. If

the equivalent resistance of this combination is R′,

then the ratio R/R′ is :

(a) 1/25 (b) 1/5 (c) 5 (d) 25

Ans. (d) is the correct answer.

Reason : Resistance of each small piece



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= R5

∴ Equivalent resistance of 5 pieces in parallel

1R¢

= 5 5 5 5+ + + +R R R R R

5

1R¢

= 25R

∴ RR¢

= 25.

Q.39. An electric heater of resistance 8 Ω draws 15 A

from the service mains in 2 hours. Calculate the

rate at which heat is developed in the heater.

Ans. Rate of heat development = Power of heater

= I2.R = (15 A)2 × 8 Ω

= 1800 W

Q.40. What are the advantages of connecting electrical

devices in parallel with the battery instead of

connecting them in series?



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Ans. 1. In parallel circuit, each appliance maintains a

steady potential difference and the current divides

in the inverse ratio of the resistances. Thus, each

appliance receives current at steady voltage, and

hence, they work efficiently. This is not possible

in series circuit.

2. In parallel circuit, if one appliance goes out of

order or is switched off, the other appliances

continue working. This is not possible in series

circuit, where the complete circuit stops working

if one appliance goes out of order or is switched

off.

3. In parallel circuit overall resistance decreases, and

hence, adequate current flows through every

appliance. In series circuit, the overall resistance

increases, and hence, magnitude of current

decreases.

Q.41. The values of current I flowing in a given resistor

for the corresponding values of potential difference

V across the resistor are given below :



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I (amperes) 0.5 1.0 2.0 3.0 4.0

V (volts) 1.6 3.4 6.7 10.2 13.2

Plot a graph between V and I and calculate the

resistance of that resistor.

Ans.

Resistance = Coordinates on Y - axisCoordinates on X - axis

Resistance = 2 1

2 1

Y – YX – X



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= (13.2 – 1.6) V(4.0 – 0.5) A

= 11.6V3.5 A

= 3.31 Ω

Q.42. Compute the heat generated while transferring

96,000 coulomb of charge in one hour through a

potential difference of 50 V.

Ans. Here, Q = 96000 C, t = 1 hour = 3600 sec, V = 50 V

Heat generated H = ?

H = VQ = 50V × 96000 C

= 4800000 J

Q.43. What determines the rate at which energy is

delivered by a current?

Ans. Resistance of the circuit determines the rate at which

the energy is delivered by a current.

ELECTRICITY - Testlabz · Name and define SI unit of electric current? ... Ans. Electric circuit is...

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