Competency Based Questions Class 10 Science: Electricity

Competency based questions in Class 10 Science test whether you can use an idea in a real situation, not just recall it. In this post you will see what these questions look like, a simple method to answer them, and ten practice questions from the chapter Electricity, each solved step by step.

What competency-based questions test

According to CBSE's curriculum documents for 2026-27:

• In Classes 9 and 10, about 50% of exam questions are competency-focused. These include case-based, source-based, integrated, data interpretation, situational and application questions.
• The rest are multiple-choice, short-answer and long-answer questions. Question types also include assertion-reason.
• In Class 10 Science, the paper weights are 50% knowledge and understanding, 30% application, and 20% formulate, analyse, evaluate and create.

So a competency question gives you a situation, a table or a short case, and asks you to apply what you learnt. Electricity is perfect for this, because every formula connects to something in your home: bulbs, heaters, fuses, wiring and electricity units.

How to answer them

Use the same five steps every time:

1. Read the case twice. Underline every number and its unit.
2. Name the idea. Is it Ohm's law, series or parallel, resistivity, heating or power?
3. Write the formula first, for example V = IR or P = VI.
4. Substitute with units and calculate carefully.
5. Answer the question asked, in one clear sentence, with the unit.

The formulas you need from the chapter:

Also remember: 1 kWh = 3.6 × 10⁶ J, and the charge of one electron is 1.6 × 10⁻¹⁹ C.

Questions 1 and 2: circuits at home

Q1. Case: In Meera's house, a bulb and a heater are connected in parallel to the supply. One evening the bulb fuses. (a) Does the heater keep working? (b) Why is house wiring not done in series?

Answer: (a) Yes. In a parallel circuit each appliance has its own path and the same voltage, so if one fails, the others continue to work. (b) In series, if one component fails, the circuit breaks and every other component stops. Also, devices with different current needs, like a bulb and a heater, cannot sensibly be connected in series.

Q2. Application: Resistors of 5 Ω, 8 Ω and 12 Ω are joined in series to a 6 V battery. Find the current and the potential difference across the 12 Ω resistor.

Answer: Rs = 5 Ω + 8 Ω + 12 Ω = 25 Ω. I = V/Rs = 6 V / 25 Ω = 0.24 A. The same current flows through each resistor in series, so V across 12 Ω = I × R = 0.24 A × 12 Ω = 2.88 V.

Check: 0.24 × 5 = 1.2 V and 0.24 × 8 = 1.92 V. Then 1.2 + 1.92 + 2.88 = 6 V, which matches the battery.

Questions 3 and 4: heat and combinations

Q3. Situational: Riya notices that the heating element of a room heater glows red, but the cord carrying current to it stays cool. Explain why.

Answer: The same current flows through both. Heat produced is H = I²Rt, so with the same I and t, heat depends on R. The element has very high resistance, so it produces a lot of heat and glows. The cord has very low resistance, so it produces very little heat.

Q4. Application: A student has four coils of 24 Ω each and a 12 V battery. (a) What is the lowest resistance she can make? (b) What current flows then?

Answer: (a) The lowest resistance comes from connecting all four in parallel. 1/Rp = 1/24 + 1/24 + 1/24 + 1/24 = 4/24 = 1/6, so Rp = 6 Ω. (b) I = V/Rp = 12 V / 6 Ω = 2 A.

Questions 5 to 7: data and materials

Q5. Data interpretation: A student records these readings for one resistor kept at constant temperature.

(a) What is the resistance? (b) Which law do the readings show? (c) Predict the current at 6.0 V.

Answer: (a) R = V/I = 1.0 V / 0.25 A = 4 Ω. Every pair gives the same value, for example 4.0 V / 1.00 A = 4 Ω. (b) V is directly proportional to I, which is Ohm's law. (c) I = V/R = 6.0 V / 4 Ω = 1.5 A.

Q6. Application: A wire of length l and area of cross-section A has a resistance of 4 Ω. Another wire of the same material has length l/2 and area 2A. Find its resistance.

Answer: R = ρ(l/A) = 4 Ω. New R' = ρ(l/2)/(2A) = (1/4) × ρ(l/A) = (1/4) × 4 Ω = 1 Ω. Halving the length halves R, and doubling the area halves it again.

Q7. Case: A company makes toasters and irons. Engineers choose an alloy such as nichrome for the heating element instead of a pure metal. Give two reasons.

Answer: (1) Alloys generally have higher resistivity than the metals they are made from. (2) Alloys do not oxidise (burn) easily at high temperatures. Together, these make alloys like nichrome ideal for heating elements.

Questions 8 to 10: power, energy and charge

Q8. Application: An electric iron uses 880 W from a 220 V supply. Find the current and the resistance of its element.

Answer: P = VI, so I = P/V = 880 W / 220 V = 4 A. R = V/I = 220 V / 4 A = 55 Ω.

Check: P = V²/R = (220 × 220)/55 = 48,400/55 = 880 W.

Q9. Data interpretation: A family uses three appliances each day.

(a) Find the energy used by each in kWh per day. (b) Find the total in kWh and in joules.

Answer: Energy = Power × Time. Fan: 50 W × 10 h = 500 Wh = 0.5 kWh. Bulb: 100 W × 5 h = 500 Wh = 0.5 kWh. Iron: 1000 W × 0.5 h = 500 Wh = 0.5 kWh. Total = 1.5 kWh. In joules: 1.5 × 3.6 × 10⁶ J = 5.4 × 10⁶ J. Notice that a powerful iron used briefly uses the same energy as a small fan used for a long time.

Q10. Application: A filament draws a current of 0.5 A for 10 minutes. (a) How much charge flows? (b) How many electrons is that?

Answer: (a) t = 10 min = 600 s. Q = I × t = 0.5 A × 600 s = 300 C. (b) One coulomb is about 6.25 × 10¹⁸ electrons (1 ÷ 1.6 × 10⁻¹⁹). So n = 300 × 6.25 × 10¹⁸ = 1.875 × 10²¹ electrons.

Common mistakes

• Forgetting to change minutes to seconds before using Q = It or H = I²Rt.
• Adding resistances directly in parallel. In parallel, add the reciprocals.
• Mixing up meters: an ammeter goes in series, a voltmeter goes in parallel.
• Leaving out the unit in the final line.

To practise more questions like these at your own pace, study this chapter with Joy.

Last updated: 7 October 2026. Source: CBSE.