11. Electricity Class 10 Science [LATEST] Solutions TextBook Exercise in english - CBSE Study

NCERT Solutions Exercise

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Q1. 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) 25

Q2. Which of the following terms does not represent electrical power in a circuit?

(a) I 2R

(b) IR2

(c) VI

(d) V2/R

Ans: (b) IR2

P = VI = I2R = V2/R

Q3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –

(a) 100 W

(b) 75 W

(c) 50 W

(d) 25 W

Ans: (d) 25 W

Q4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. 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) 1:4

Q5. How is a voltmeter connected in the circuit to measure the potential difference between two points?

Ans: Voltmeter is always connected in parallel.

Chapter 11: Electricity

NCERT Textbook Questions & Solutions

Q6. A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10^–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Solution:

Given:

• Resistance, R = 10 Ω
• Resistivity, ρ = 1.6 × 10^–8 Ω m
• Diameter, d = 0.5 mm = 0.5 × 10^–3 m
• Radius, r = 0.25 × 10^–3 m

Cross-sectional area,

A = πr²

= 3.14 × (0.25 × 10^–3)²

= 1.963 × 10^–7 m²

Using,

R = ρL/A

L = RA/ρ

= (10 × 1.963 × 10^–7)/(1.6 × 10^–8)

L ≈ 122.7 m

If the diameter is doubled, the area becomes four times.

Since resistance is inversely proportional to area,

New resistance = 10/4 = 2.5 Ω

Answer:

• Length of the wire = 122.7 m
• New resistance = 2.5 Ω
• Decrease in resistance = 7.5 Ω

Q7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below. Plot a graph between V and I and calculate the resistance of that resistor.

Solution:

Plot the following points on a graph with current (I) on the X-axis and potential difference (V) on the Y-axis.

Current (A)	0.5	1.0	2.0	3.0	4.0
Voltage (V)	1.6	3.4	6.7	10.2	13.2

The graph obtained is approximately a straight line passing through the origin, showing that the resistor obeys Ohm's law.

Using any point,

R = V/I

= 13.2/4

R = 3.3 Ω

Answer: The resistance of the resistor is approximately 3.3 Ω.

Q8. 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.

Solution:

Given:

• Potential difference, V = 12 V
• Current, I = 2.5 mA = 0.0025 A

Using Ohm's law,

R = V/I

= 12/0.0025

R = 4800 Ω = 4.8 kΩ

Answer: The resistance of the resistor is 4800 Ω (4.8 kΩ).

Q9. 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?

Solution:

Total resistance,

R = 0.2 + 0.3 + 0.4 + 0.5 + 12

= 13.4 Ω

Using Ohm's law,

I = V/R

= 9/13.4

I ≈ 0.67 A

Since the resistors are connected in series, the same current flows through every resistor.

Answer: Current through the 12 Ω resistor is 0.67 A.

Q10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Solution:

Given:

• Voltage, V = 220 V
• Total current, I = 5 A
• Each resistor = 176 Ω

Equivalent resistance required,

R = V/I

= 220/5

R = 44 Ω

For n equal resistors in parallel,

R = 176/n

176/n = 44

n = 4

Answer: 4 resistors of 176 Ω connected in parallel are required.

Chapter 11: Electricity

NCERT Textbook Questions & Solutions

Q11. 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.

Solution:

(i) To obtain 9 Ω:

Connect two 6 Ω resistors in series.

Equivalent resistance = 6 + 6 = 12 Ω

Now connect this 12 Ω combination in parallel with the third 6 Ω resistor.

Equivalent resistance,

1/R = 1/12 + 1/6

= 1/12 + 2/12 = 3/12 = 1/4

R = 4 Ω

Since this gives 4 Ω, for 9 Ω:

Connect two 6 Ω resistors in parallel.

Equivalent resistance = 3 Ω

Now connect this combination in series with the third 6 Ω resistor.

Total resistance = 3 + 6 = 9 Ω

(ii) To obtain 4 Ω:

Connect two 6 Ω resistors in series (12 Ω), then connect this combination in parallel with the third 6 Ω resistor.

Equivalent resistance = 4 Ω.

Answer:

• (i) 6 Ω || 6 Ω, then in series with 6 Ω → 9 Ω
• (ii) 6 Ω + 6 Ω, then in parallel with 6 Ω → 4 Ω

Q12. Several electric bulbs designed to be used on a 220 V electric supply line are rated 10 W. How many lamps can be connected in parallel across a 220 V line if the maximum allowable current is 5 A?

Solution:

Given:

• Power of each lamp = 10 W
• Voltage = 220 V
• Maximum current = 5 A

Current drawn by one lamp,

I = P/V = 10/220 = 0.045 A

Number of lamps,

n = 5/0.045

n ≈ 110

Answer: A maximum of 110 lamps can be connected in parallel.

Q13. 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, or in parallel. What are the currents in the three cases?

Solution:

(i) When one coil is used:

R = 24 Ω

I = V/R = 220/24

I = 9.17 A

(ii) When both coils are connected in series:

R = 24 + 24 = 48 Ω

I = 220/48

I = 4.58 A

(iii) When both coils are connected in parallel:

Equivalent resistance,

R = (24 × 24)/(24 + 24) = 12 Ω

I = 220/12

I = 18.33 A

Answer:

• One coil: 9.17 A
• Series: 4.58 A
• Parallel: 18.33 A

Q14. 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.

Solution:

(i) Series circuit:

Total resistance = 1 + 2 = 3 Ω

Current,

I = 6/3 = 2 A

Power in 2 Ω resistor,

P = I²R

= 2² × 2

P = 8 W

(ii) Parallel circuit:

The 2 Ω resistor gets the full 4 V.

Power,

P = V²/R

= 4²/2

= 16/2

P = 8 W

Answer:

• Power in case (i) = 8 W
• Power in case (ii) = 8 W
• Hence, the power consumed by the 2 Ω resistor is the same in both circuits.

Chapter 11: Electricity

NCERT Textbook Questions & Solutions

Q15. 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?

Solution:

Given:

• Lamp 1: 100 W, 220 V
• Lamp 2: 60 W, 220 V
• Supply voltage = 220 V

Current drawn by the 100 W lamp,

I₁ = P/V = 100/220 = 0.455 A

Current drawn by the 60 W lamp,

I₂ = P/V = 60/220 = 0.273 A

Total current drawn from the mains,

I = I₁ + I₂

= 0.455 + 0.273 = 0.728 A

Answer: The current drawn from the line is approximately 0.73 A.

Q16. Which uses more energy, a 250 W TV set in 1 hour, or a 1200 W toaster in 10 minutes?

Solution:

Energy used by the TV:

Power = 250 W = 0.25 kW

Time = 1 hour

Energy = Power × Time

= 0.25 × 1

= 0.25 kWh

Energy used by the toaster:

Power = 1200 W = 1.2 kW

Time = 10 minutes = 1/6 hour

Energy = 1.2 × 1/6

= 0.20 kWh

Answer: The 250 W TV uses more energy (0.25 kWh) than the 1200 W toaster (0.20 kWh).

Q17. An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.

Solution:

Given:

• Resistance, R = 8 Ω
• Current, I = 15 A

The rate of heat developed is equal to the electrical power.

P = I²R

= 15² × 8

= 225 × 8

P = 1800 W

Answer: The rate at which heat is developed in the heater is 1800 W (1.8 kW).

Q18. Explain the following.

(a) Why is tungsten used almost exclusively for the filament of electric lamps?

Answer: Tungsten has a very high melting point (about 3400°C) and high resistivity. It can become white hot without melting, making it suitable for lamp filaments.

(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?

Answer: Alloys have high electrical resistance, high melting point and do not oxidise easily even at high temperatures. Therefore, they are ideal for heating elements.

(c) Why is the series arrangement not used for domestic circuits?

Answer: In a series circuit, the same current flows through all appliances. If one appliance fails or is switched off, the entire circuit breaks and all other appliances stop working. Hence, domestic wiring uses parallel connections.

(d) How does the resistance of a wire vary with its area of cross-section?

Answer: The resistance of a wire is inversely proportional to its cross-sectional area. As the area increases, the resistance decreases.

(e) Why are copper and aluminium wires usually employed for electricity transmission?

Answer: Copper and aluminium have very low resistivity and are good conductors of electricity. They allow current to flow with minimum energy loss and are therefore widely used for electrical transmission.