Electricity CBQs — CBSE Class 10 (2027) with step-wise solutions

Sample Electricity CBQs

1. Question 1MCQ1 mark
   Ishaan tests two phone chargers using a meter that records the total charge each one delivers. Charger P delivers 1800 C in 15 minutes, while charger Q delivers 2700 C in 30 minutes. Which statement about the currents supplied by the two chargers is correct?
   • P supplies the larger current: 2.0 A, compared with 1.5 A for Q.
   • Q supplies the larger current, because it delivers more charge in total.
   • P supplies 120 A and Q supplies 90 A.
   • P supplies 0.5 A and Q supplies about 0.67 A.
   Show step-wise solution

   1. [1 mark] I = Q/t with t in seconds: P = 1800 C ÷ 900 s = 2.0 A; Q = 2700 C ÷ 1800 s = 1.5 A. Current is the RATE of flow of charge, so the larger total charge of Q does not mean a larger current; 120 A and 90 A come from forgetting to convert minutes to seconds.

   Answer: P supplies the larger current: 2.0 A, compared with 1.5 A for Q.

2. Question 2Assertion-Reason1 mark
   Assertion (A): In a circuit in which a cell, a key, a torch bulb and an ammeter are all joined in series, the ammeter shows a smaller reading when it is placed after the bulb than when it is placed before the bulb. Reason (R): The current is the same at every point of a series circuit, because charge is not used up as it passes through the bulb.
   • (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
   • (B) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.
   • (C) Assertion is true but Reason is false.
   • (D) Assertion is false but Reason is true.
   Show step-wise solution

   1. [1 mark] (D) A is false and R is true: charge is not consumed by the bulb (only energy is transferred), so the same current flows everywhere in the series loop and the ammeter reads the same before or after the bulb.

   Answer: (D) Assertion is false but Reason is true.

3. Question 3Short answer2 marks
   Priya has a copper wire and an aluminium wire of the same length. Can the electrical resistance of the two wires be the same? Justify your answer.
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   1. [1 mark] Yes, the electrical resistance of the two wires can be the same. (No marks to be awarded if justification is not written.)
   2. [1 mark] if the area of cross-section of the two wires is different OR if the thickness of the two wires is different

   Answer: Yes - the two metals have different resistivities, but a difference in cross-sectional area can compensate.

4. Question 4Case study4 marks
   Read the passage and answer the questions that follow. Deepak, a Class 10 student in Guwahati, noticed that his power bank is labelled '10000 mAh'. His teacher explained that mAh (milliampere-hour) is actually a unit of electric charge, not of energy: 1 mAh is the charge carried by a current of 1 mA flowing for one hour. Deepak copied the labels of two devices at home: Power bank — capacity 10000 mAh, output current 2 A Phone battery — capacity 4000 mAh He wondered how many coulombs these labels really stand for, and how long the power bank could keep supplying its full output current. Take the charge on an electron as 1.6 × 10⁻¹⁹ C and assume that the full capacity of the power bank can be used. (i) Express 1 mAh in coulombs. [1 mark] (ii) Find the charge stored in the power bank, in coulombs. [1 mark] (iii) (a) For how long can the power bank supply its rated output current of 2 A? (b) How many electrons in total flow out of it during this time? [2 marks]
   Show step-wise solution

   1. [1 mark] (i) 1 mAh = (1 × 10⁻³ A) × 3600 s = 3.6 C.
   2. [1 mark] (ii) Q = 10000 × 3.6 C = 36000 C.
   3. [1 mark] (iii)(a) t = Q/I = 36000 C ÷ 2 A = 18000 s = 5 h.
   4. [1 mark] (iii)(b) n = Q/e = 36000 C ÷ (1.6 × 10⁻¹⁹ C) = 2.25 × 10²³ electrons.

   Answer: (i) 3.6 C; (ii) 36000 C; (iii) 5 h (18000 s), 2.25 × 10²³ electrons.

5. Question 5Long answer5 marks
   A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp. Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductor and potential difference across the lamp will take place? Give reason.
   Show step-wise solution

   1. [1 mark] Initial series circuit: $R_{\mathrm{total}}$ = V/I = 10 V / 1 A = 10 Ω.
   2. [1 mark] Resistance of lamp: R_L = $R_{\mathrm{total}}$ − 5 = 10 − 5 = 5 Ω.
   3. [0.5 mark] Now a 10 Ω resistor is connected in parallel with the series combination (lamp + 5 Ω = 10 Ω); the parallel combination is across the same 10 V battery.
   4. [0.5 mark] Potential difference across each branch of a parallel combination equals the battery voltage = 10 V.
   5. [1 mark] Series branch sees 10 V across 10 Ω → current = 1 A (unchanged); PD across lamp = 1 × 5 = 5 V (unchanged).
   6. [1 mark] Reason: in parallel, each branch is independent; adding a parallel branch does not change the voltage or current of an existing branch.

   Answer: $R_{\mathrm{lamp}}$ = 5 Ω. No change — current through 5 Ω stays at 1 A; PD across lamp stays at 5 V.

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