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8/19/2019 Q Electricity

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hence, helps in maintaining potential difference across

a conductor.

Q.4. What is meant by saying that potential differencebetween 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 5 resistor an

8 resistor and a 12 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 12

resistor is also connected. What would be thereadings 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)

=6V

25 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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(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

m. What will be the length of thiswire to make its resistance of 10 ? 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

2

d R

RA

⎛ ⎞× π ⎜ ⎟⎝ ⎠=

ρ ρ

= –8

10 × 22 × 0.0005 m × 0.0005 m

7 × 4 × 1.6 × 10 m

W

W

= –4 –4

–8

220 × 5 × 10 × 5 × 10 m

7 × 4 × 1.6 × 10

=220 × 25

44.8 = 122.76 m


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(ii) 1

2

R

R =

2

2

2

2

1 1

2

2

d

A

A d

æ ö p ç ÷è ø

=æ ö

p ç ÷

è ø

∴ 2

10

R

W =

2

2

(2×0.0005 m)

(0.0005 m) (

d2 = 2d1)

R2 =10

4

Ω = 2.5

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 tometals.

Q.13. Judge the equivalent resistance when the following

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

(b) 1 , 103 and 10

6 .

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) IR

2 (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 484

11 = 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

R p =R

2

∴ 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.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 =2.5

1000

A


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∴ Resistance (R) = V 12 × 1000

=I 2.5

= 4800 .

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

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

How much current would flow through the 12

resistor?

Ans. Total resistance in series (R)

= (0.2 + 0.3 + 0.4 + 0.5 + 12) Ω

= 13.4

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) = P

V

= 160 W

220 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.26. An electric lamp of 100 , a toaster of resistance

50 and a water filter of resistance 500 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

1

R =

1 1 1+ +

100 50 500

= 5 + 10 + 1 16

=500 500

∴ R p = 500

16 = 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 , 3 and

6 be connected to give a total resistance of :

(a) 4 (b) 1 ?

Ans. (a)

In the above circuit diagram resistance of 3 Ω and

6 Ω in parallel

p

1R

= 1 1 3+ = =3 6 6

12

∴ R p = 2

Therefore, resistance of 2 Ω and the parallel

segment

R = 2 + 2 = 4

(b)

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


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∴ p

1

R =

1 1 1 6+ + = =

2 3 6 6

1

1

∴ R p = 1

Q.28. An electric iron of resistance 20 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 resistors (in parallel) are

required to carry 5 A on 220 V line?

Ans. Current in circuit (I) = 5 A


∴ Resistance of parallel circuit (R p) = V

I


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= 220 V

5A = 44

Now, p

1

R = 1 2

1

+r r

1

+ …n times

⇒ 1

44 =

1+

176 176

1 + …n times

⇒ 1

44 =

176

n

∴ n =176

44 = 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 , so that the combination has a

resistance of : (i) 9 ; (ii) 4 .

Ans. (i)


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Resistance of B and C in parallel

p

1

R =

1 1 2+ = =

6 6 6

1

3

∴ R p = 3 .

∴ Combined resistance of A and R p in series.

R s = 6 Ω + 3 Ω = 9 .

(ii)

Resistance of A and B in series

R s = 6 + 6 = 12

∴ Resistance of R s in parallel with C

∴ R p =1 1

3 1

+ = =12 6 12 4

∴ R p = 4


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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 power

Power of one lamp

=1100 W

10 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

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

∴ Current in coil (I) =V 220V

=R 24 W

= 9.17 A

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

R s = 24 Ω + 24 Ω = 48

∴ Current in coils in series

I =s

V 220=

R 48

V

W = 4.58 A

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

p

1

R =

1 1 2+ = =

24 24 24 12

1

∴ R p = 12


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∴ Current in coils in parallel

I = p

V 220=

R 12

V

W = 18.33 A

Q.33. Compare the power used in the 2 resistor in each

of the following circuits :

(i) a 6 V battery in series with 1 and 2

resistors,

(ii) a 4 V battery in parallel with 12 and 2

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 A6V

=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 ×10

60 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

1

1.602 × 10 = 6.24 × 10

18 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 plugkey, 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. Ifthe 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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=R

5

∴ Equivalent resistance of 5 pieces in parallel

1

R¢ =

5 5 5 5+ + + +

R R R R R

5

1

R¢ =

25

R

∴ R R¢ = 25.

Q.39. An electric heater of resistance 8 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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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 theresistance of that resistor.

Ans.

Resistance =Coordinates on Y - axis

Coordinates on X - axis

Resistance = 2 1

2 1

Y – Y

X – X


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=(13.2 – 1.6) V

(4.0 – 0.5) A

=11.6V

3.5 A = 3.31

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.

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