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Class 10 Science: Electricity, 8 practice questions with answers

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These are 8 practice questions on NCERT Class 10 Science Chapter 11, Electricity, each with a worked answer, the units, and the textbook page it comes from. Six are numericals (Ohm's law, resistors in series and in parallel, a mixed circuit, heating and electrical energy), one is an assertion-reason question and one is about how resistance depends on length. Try each one on paper before you read the answer.

These questions are our own, drafted with AI help. Every number was re-computed by a test script and every statement was checked against the chapter, but no practising teacher has reviewed them yet. If you find a mistake, please tell us. The ideas behind each formula are in our guide to NCERT Class 10 Science Chapter 11, Electricity.

The formulas you need

Facts from the NCERT textbook (Science, Textbook for Class X, Reprint 2026-27, Chapter 11, printed pages 171 to 194):

Ohm's law, series and parallel

Question 1. A current of 2 A flows through a 5 Ω resistor. What is the voltage across it, in volts?

Answer: 10 V. By Ohm's law, V = I × R = 2 A × 5 Ω = 10 V. (Page 176.)

Question 2. Resistors of 3 Ω, 4 Ω and 5 Ω are connected in series to a 6 V battery. What current flows through the circuit, in amperes?

Answer: 0.5 A. In series the resistances add: R = 3 + 4 + 5 = 12 Ω. By Ohm's law I = V/R = 6 V ÷ 12 Ω = 0.5 A. The same current flows through all three. (Series, page 184.)

Question 3. Two resistors of 6 Ω and 3 Ω are connected in parallel. What is their equivalent resistance, in ohms?

Answer: 2 Ω. 1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2, so R = 2 Ω. This is less than the smaller resistor, as it should be for a parallel combination. (Parallel, page 186.)

Question 4. A 4 Ω resistor is connected in series with a parallel combination of 6 Ω and 3 Ω resistors, across a 12 V battery. What current does the battery supply, in amperes?

Answer: 2 A. The parallel pair is 2 Ω (question 3). In series with 4 Ω the total resistance is 4 + 2 = 6 Ω. Current I = V/R = 12 V ÷ 6 Ω = 2 A. (Pages 184 and 186.)

Heating, power and energy

Question 5. A current of 2 A flows through a resistor of 10 Ω for 60 seconds. How much heat is produced, in joules?

Answer: 2,400 J. By Joule's law of heating, H = I²Rt = 2² × 10 × 60 = 4 × 600 = 2,400 J. (Page 189.)

Question 6. An electric heater rated 1,000 W is used for 3 hours every day. How many units (kW h) of electrical energy does it use in 30 days?

Answer: 90 units. Power = 1,000 W = 1 kW. Energy per day = 1 kW × 3 h = 3 kW h. In 30 days: 3 × 30 = 90 kW h, that is 90 units. (The commercial unit is the kilowatt hour, page 191.)

Reasoning and resistance

Question 7. Assertion (A): Tungsten is used for the filaments of electric bulbs. Reason (R): Tungsten has a very high melting point (about 3,380 °C). Choose the correct option: both true and R explains A; both true but R does not explain A; A true and R false; or A false and R true.

Answer: A and R are both true, and R explains A. The filament must get very hot to give light and must not melt at that temperature, so a strong metal with a high melting point such as tungsten (melting point 3,380 °C) is used. (Pages 179 and 190.)

Question 8. A wire of resistance 12 Ω is cut into three equal pieces, and the three pieces are joined in parallel. What is the resistance of the combination, in ohms?

Answer: 4/3 Ω (about 1.33 Ω). Resistance is proportional to length (R ∝ l, page 178), so each third of the wire has 12 ÷ 3 = 4 Ω. In parallel: 1/R = 1/4 + 1/4 + 1/4 = 3/4, so R = 4/3 Ω. (Page 186.)

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Sources

All opened and read on 21 October 2026. Page numbers are the printed page numbers of the textbook.

The laws and definitions are from the textbook pages above. The questions, worked answers and checks are our own, and every number was re-computed by script. This guide is general information; follow your teacher and the textbook.