12. Electricity Class 10 NCERT Solutions (CBSE 2026 Guide)

Chapter 12: Electricity – NCERT Solutions for Class 10 Science CBSE

Chapter 12: Electricity

Internal Questions and Answers

Example 12.1: A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.
Solution: Here, I = 0.5 A , t = 10 min = 600 second and Q = ?
We know that , Q = It
Q = 0.5 A × 600 s
Q = 300 C

Internal Questions and Answers

1. What does an electric circuit mean ?

Answers: A continuous and closed path of an electric current is called an electric circuit.

2. Define the unit of current.

Answers: One ampere (the unit of current) is constituted by the flow of one coulomb of charge per second .
3. Calculate the number of electrons constituting one coulomb of charge .

Solution: Here, ,

The number of electron

Example 12.2: How much work is done in moving a charge of 2 C across two points having a potential difference 12 V?

Solution : Here, Q = 2 Coulomb , V = 12 volt and W = ?

We have, W = VQ
= 12 V × 2 C
= 24 J

Internal Questions and Answers

1. Name a device that helps to maintain a potential difference across a conductor.

Answer: A battery (or cell) is a common device that helps maintain a potential difference (voltage) across a conductor.

2. What is meant by saying that the potential difference between two points is 1 V ?
Answer: The potential difference between two points is 1 volt that mean a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.

3. How much energy is given to each coulomb of charge passing through a 6 V battery?

Solution: Here, V = 6 volt , Q = 1 Coulomb and W = ?

We know that , W = VQ = 6 × 1 J = 6 J

Example 12.3: (a) How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 Ω?

Solution: Here, V = 220 V; R = 1200 Ω.
Using Ohm’s, law,

(b) How much current will an electric heater coil draw from a 220 V source, if the resistance of the heater coil is 100 Ω?
Solution: Here , V = 220 V, R = 100 Ω and I = ?

Using Ohm’s law,

Example 12.4 : The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?
Solution: Here, , current
According to Ohm’s law,

When the potential difference is increased to 120 volt .

Again, , ,

The current through the heater becomes 8 A.

Example 12.5 : Resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? Using Table 12.2, predict the material of the wire.
Solution: Here, ,, , ,

We have,

The resistivity of the metal at 20°C is .

Example 12.6 : A 4 Ω resistance wire is doubled on it. Calculate the new resistance of the wire.

Solution: Here,

We have,

When a wire is doubled on it, its length would become half and area of cross-section would double.

Here, ,

We have ,

The new resistance of the wire is 1 Ω.

Internal Questions and Answers :

1. On what factors does the resistance of a conductor depend ?

Answer: The resistance of a conductor depends on the factors are length, area of the cross-section and the nature of the materials.

2. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer: Current flows more easily through a thick wire. This is because a thick wire has lower resistance compared to a thin wire of the same material and length. Lower resistance allows more current to pass for the same potential difference. So, thicker wires conduct electricity better.

3. 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?
Answer: Using Ohm’s law,

If the potential difference becomes half while resistance stays constant, then

So, the current also becomes half of its original value.

4. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer: Coils of electric toasters and electric irons are made of an alloy rather than a pure metal because alloys have higher electrical resistance, which allows them to generate more heat when an electric current passes through them, making them more suitable for heating applications.

5. Use the data in Table 12.2 to answer the following –
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?

Answer: (a) iron is a better conductor than mercury .
(b) Silver is the best conductor.

Example 12.7: An electric lamp, whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery (Fig. 12.9). 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.

Fig. 12.9

Solution: (a) Here, , and

The total resistance

(b) We have,

For the conductor : Here,

1. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Answer: The diagram of a circuit : Here,

2. Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

Solution: The diagram of a circuit:

Here, ,

We have,

Using Ohm’s law,

Potential difference across the 12 Ω resistor:

Example 12.8: In the circuit diagram given in Fig. 12.10, suppose the resistors , and have the values 5 Ω, 10 Ω, 30 Ω, respectively, which have been connected to a battery of 12 V. Calculate (a) the current through each resistor, (b) the total current in the circuit, and (c) the total circuit resistance.

Fig. 12.10

Solution: Here, , , and

(a) Using Ohm’s law ,

(b) The total current in the circuit

Example 12.9: If in Fig. 12.12, ,,, , 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.

Solution: Here, ,,, , , and

(a) The total resistance in the circuit : For resistance and :

For resistance , and :

(b) The total current flowing in the circuit : Here,

We have,

Internal Question and Answer:

1. Judge the equivalent resistance when the following are connected in parallel – (a) 1 Ω and 10 6 Ω, (b) 1 Ω and 10 3 Ω, and 10 6 Ω.

Solution: (a) Here, and

We have,

(b) Here, , and

We have,

So, the equivalent resistances are: (a) 1 Ω (b) 0.998 Ω (approximately)

2. 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?
Solution: Here, , , and volt

We have,

The total current (I) drawn by these three appliances when connected in parallel to the 220 V source.

Using Ohm's law ,

So, the resistance of the electric iron is 31.25 Ω, and the current through it is 7.04 A.

3. What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

Answer: When electrical devices are connected in parallel across a battery, each device receives the full battery voltage, so they operate properly even if other devices are switched off or fail. In a parallel circuit, the effective resistance decreases, allowing each device to draw the required current without affecting others. In contrast, in a series connection, the voltage gets divided and if one device fails, the entire circuit gets broken.

4. 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?

Solution: (a) Total Resistance of 4 Ω:

Here, , ,

We have,

and

(b) Total Resistance of 1 Ω :

We have,

5. What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Solution: (a) Highest Total Resistance: Here, , , and

In a series connection, the total resistance is the sum of the individual resistances.

So, the highest total resistance that can be obtained is 48 Ω when all the coils are connected in series.

(b) Lowest Total Resistance:

So, the lowest total resistance that can be obtained is 2 Ω when all the coils are connected in parallel

Example 12.10: An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V. What are the current and the resistance in each case?

Solution: (a) When heating is at the maximum rate : Here, P = 840 watt , V = 220 volt

We know that ,

The current

and the resistance of the electric iron is

(b) When heating is at the minimum rate : Here, P = 360 watt , V = 220 volt

and the resistance of the electric iron is

Example 12.11: 100 J of heat are produced each second in a 4 Ω resistance. Find the potential difference across the resistor.

Solution: Here, H = 100 J, R = 4 Ω, t = 1 s, V = ?

We have,

Thus the potential difference across the resistor, V is

Internal Question and Answer:

1. Why does the cord of an electric heater not glow while the heating element does?

Answer: The heating element of an electric heater has high resistance, so it gets very hot and glows due to heating effect of electric current. The cord is made of copper/aluminium with very low resistance, so it allows current to pass without producing much heat, hence it does not glow.

2. Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.

Solution: Given, Electric charge (Q) = 96000 C , Potential difference (V) = 50 volts , Time (t) = 1 hour = 3600 seconds .

We have, H = Q × V

H = 96000 C × 50 V = 4,800,000 J = 4.8 × 10⁶ J

Thus, the heat generated is 4.8 × 10⁶ J .

3. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.

Solution: Given: Resistance (R) = 20 Ω , Current (I) = 5 A , Time (t) = 30 s

Using the formula,

Therefore, the heat developed in 30 seconds by an electric iron with a resistance of 20 Ω, carrying a current of 5 A, is 250 joules.

Example 12.12 : An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?
Solution: We have, P = VI
= 220 V × 0.50 A = 110 J/s = 110 W

Example 12.13: An electric refrigerator rated 400 W operates 8 hour/day. What is the cost of the energy to operate it for 30 days at Rs 3.00 per kW h?

Solution: Given, P=400 W = 0.4 kW , Time per day = 8 hours/day =8hours/day , Number of days = 30 , Cost = Rs 3.00 per kWh

Total hours (t) = 8 h/day × 30 days = 240 h

Total energy consumed (H) = P × t = 0.4 kW × 240 h = 96 kW h

Total cost = 96 kW h × 3.00 Rs/kW h = Rs 288

Therefore, the cost of energy is Rs 288.

Internal Question and Answer :

1. What determines the rate at which energy is delivered by a current?

Answer: The rate at which energy is delivered by a current is determined by electric power.

2. An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.

Solution: Given, Current I=5 A , Voltage (V) = 220 V , Time (t) =2 hours = 2 × 3600 = 7200 s

Power of the motor (P) = V× I = 220 V × 5 A =1100 W

So, power = 1100 W = 1100W = 1.1kW.

Exercises of Chapter 12: Electricity

1. 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) (b) (c) 5 (d) 25

Answer: (d) 25

[ When a wire of resistance R is cut into five equal parts and these parts are connected in parallel,

Let R be the resistance of each of the five equal parts .

Since the five equal parts are cut from the original wire of resistance R, each part's resistance will be R/5.

A/Q,

Therefore, the correct answer is (d) 25. ]
2. Which of the following terms does not represent electrical power in a circuit?
(a) (b) (c) (d)

Answer: (b)

3. 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

Answer: (d) 25 W

[ The electric bulb is rated at 220 V and 100 W.

Let, R be the first resistance.

We have,

The power consumed (P') when the bulb is operated at 110 V:

So, the power consumed by the electric bulb when operated on 110 V is 25 W. ]

4. 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

Answer: (c) 1:4

[ Let R be the resistance of each conducting wire .

When the wires are connected in series, the total resistance,

When the wires are connected in parallel, the total resistance ,

The formula for heat produced in a resistor is given by

The ratio of heat produced in series to parallel combinations would be:

and

]

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

Answer: A voltmeter is always connected in parallel across the two points between which the potential difference is to be measured. This ensures that the voltmeter has the same potential difference as the component, and due to its high resistance, it draws very little current from the circuit.

6. A copper wire has diameter 0.5 mm and resistivity of 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, Diameter

Resistivity

Resistance

We know that ,

Therefore, the length of the copper wire required to have a resistance of 10 Ω is 122.5 meters.

The resistance changes if the diameter of the wire is doubled.

Given: New diameter and

We have ,

Now,

Therefore, the resistance is time when the diameter of the wire is doubled.

7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –

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.

Solution: Plot the data points (V, I) on a graph.

We draw the graph with I (Ampere) on the y-axis and V (volt) on the x-axis.

The relationship between voltage (V) and current (I) for a resistor is given by Ohm's law: V = IR, where R is the resistance.

We take two data points, ( 6.7 V ,2.0 A) and (13.2 V , 4.0 A)

The resistance

So, the resistance of the given resistor is 3.25 ohms.

8. 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: V = 12 volt , Current (I) = 2.5 mA = 0.0025 A

Using Ohm's Law:

Therefore, the value of the resistance of the unknown resistor is 4800 Ω .

9. 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: Given, V = 9 volt , , , , and

In a series circuit, the total resistance is the sum of the individual resistances:

Using Ohm's Law,

Therefore, 0.672 A (or 672 mA) of current would flow through the 12 Ω resistor in the given circuit.

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

Solution: Given, Resistance of each resistor (R) = 176 Ω

Current (I) = 5 A Voltage (V) = 220 V

The formula for total resistance in a parallel circuit is given by:

Let N be the number of resistors required .

For N resistors connected in parallel, the total resistance is

Again,

A/Q,

Therefore, you would need 4 resistors of 176 Ω each, connected in parallel, to carry 5 A on a 220 V line.

11. 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) For a combination resistance of 9 Ω:

Here, , and

We have,

So,

(ii) For a combination resistance of 4 Ω :

Again,

12. 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 with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Solution: Given: V = 220 volt ,

Power rating of each lamp (P) = 10 W ,

Maximum allowable current ( ) = 5 A

We know that ,

For each lamp: P = 10 W , V = 220 volt

The number of lamp

Therefore, 110 lamps can be connected in parallel with each other across the two wires of the 220 V line, given a maximum allowable current of 5 A

13. 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: Given: V = 220 volt ,

Resistance (R) of each coil (A and B) = 24 Ω

Using coil A : Resistance () = 24 Ω , Voltage (V) = 220 V

The current flowing through coil A when used separately is 9.17 A.

Using coils A and B : Total resistance () , Voltage (V) = 220 volt

The current flowing through the series combination of coil A and coil B is 4.58 A.

Using coils A and B in parallel:

Total resistance ()=12 Ω , Voltage (V) = 220 V

The current flowing through the parallel combination of coil A and coil B is 18.33 A.

14. 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, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Solution: (i) Circuit with a 6 V battery in series with 1 Ω and 2 Ω resistors: Here, V = 6 volt , ,

We have,

Again,

(ii) Circuit with a 4 V battery in parallel with 12 Ω and 2 Ω resistors: Here, V = 4 volt ,

Again,

15. 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: For the first lamp: Here, Power () = 100 W , Voltage (V) = 220 V

For the second lamp: Here, Power () = 60 W , Voltage (V) = 220 V

Therefore,

Therefore, the total current drawn from the line when the supply voltage is 220 V is 0.73 A.

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

Solution: We know that, Energy (H) = Power (P) × Time (t)

For the TV set: Here, Power () = 250 W , Time () = 1 hour = 60 × 60 seconds = 3600 seconds

Energy for the TV set J

For the toaster: Here, Power () = 1200 W , Time () = 10 minutes = 10 × 60 seconds = 600 seconds

Energy for the toaster J

Therefore ,the toaster uses less energy compared to the TV set.

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

Solution: Here, R = 8 Ω , I = 15 A and t = 2 hours = 2 × 60 × 60 seconds = 7200 seconds
We know that,

Therefore,the rate at which heat is developed is 1800 W (or 1800 J/s).

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

Answer: Tungsten is used almost exclusively for the filament of electric lamps because it has a very high melting point (around 3380°C), which allows it to remain solid even when heated to high temperatures to emit light. Additionally, it has high resistivity and does not oxidize easily at high temperatures.

(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: The conductors of electric heating devices like bread-toasters and electric irons are made of an alloy rather than a pure metal because alloys have higher resistivity than pure metals. This allows them to generate more heat when current passes through. Additionally, alloys do not oxidize (burn) easily at high temperatures, making them more durable for heating applications.

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

Answer: Series arrangement is not used in domestic circuits because if one appliance fails or is switched off, the entire circuit gets broken and all devices stop working. Also, in series connection, the voltage is divided among appliances, so they do not get the required voltage and do not work properly.

(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 area of cross-section. If the area increases, resistance decreases, and if the area decreases, resistance increases. This is because a thicker wire allows more electrons to pass easily, reducing opposition to current.

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

Answer: Copper and aluminium are used for electricity transmission because they have very low resistivity, so they allow electric current to pass easily with very less energy loss. They are also ductile, so they can be drawn into thin wires, and are relatively cheaper and durable, making them suitable for long-distance power transmission.