(a) Why does an electric bulb become dim when an electric heater in parallel circuit is switched ON ? (b) How to connect three resistors each of resistance 8 Ω, so that the equivalent resistance of the combination is 12 Ω ? Draw diagram of the combination and justify your answer. (a) Why does an electric bulb become dim when an electric heater in parallel circuit is switched ON ? (b) How to connect three resistors each of resistance 8 Ω, so that the equivalent resistance of the combination is 12 Ω ? Draw diagram of the combination and justify your answer.

ScienceCBSE Class 10 · Board exam 2026, set 31/5/1 · 3 marks · Short answer

CBSE Class 10 Science 2026: question 37

Set in the CBSE Class 10 board exam of 2026, paper 31/5/1, section C, question 37, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q37 · 3 marks · Short answerCompetency-basedElectricity

(a) Why does an electric bulb become dim when an electric heater in parallel circuit is switched ON ?

(b) How to connect three resistors each of resistance $8 Ω$ , so that the equivalent resistance of the combination is $12 Ω$ ? Draw diagram of the combination and justify your answer.

(a)
Why does an electric bulb become dim when an electric heater in parallel circuit is switched ON ?
0 marks
(b)
How to connect three resistors each of resistance $8 Ω$ , so that the equivalent resistance of the combination is $12 Ω$ ? Draw diagram of the combination and justify your answer.
0 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 31/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

A heater has low resistance — what happens to the total current drawn?

Hint 2

12 Ω is 8 Ω plus 4 Ω; how can two 8 Ω resistors give 4 Ω?

Hint 3

Use $1/ R_{p} = 1/ R_{1} + 1/ R_{2}$ and $R_{s} = R_{1} + R_{2}$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(a) A heater has low resistance and draws a large current when switched on. The total current in the supply line rises sharply, so the voltage drop across the connecting wires increases and less voltage/current is available to the bulb in parallel; hence the bulb glows dim.
1 mark
(b) Diagram: Connect two $8 Ω$ resistors in parallel, and connect this parallel pair in series with the third $8 Ω$ resistor between points A and B.
1 mark
Justification: parallel pair $\frac{1}{R _{1}} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} \Rightarrow R_{1} = 4 Ω$ . In series with $8 Ω$ : $R_{e q} = R_{1} + 8 = 4 + 8 = 12 Ω$ .
1 mark

Answer: (a) Heater draws large current, reducing current/voltage for the bulb; (b) two 8 Ω in parallel (4 Ω) in series with third 8 Ω gives 12 Ω

Where the marks are

Reason: heater draws large current so bulb gets less current and glows dim (1 mark)
Correct diagram: two resistors in parallel, in series with third (1 mark)
Calculation $R_{1} = 4 Ω$ , $R_{e q} = 12 Ω$ (1 mark)

Revise the chapter

Chapter: Electricity (notes, formulas, where marks are lost and every past question)
Topic: Resistors in Series and Parallel

Practise this chapterBoard-style questions from Electricity, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 31/5/1: every question, by section.

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CBSE Class 10 Science 2026: question 37

Hints

Step-by-step solution

Revise the chapter

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