ScienceCBSE Class 10 · Sample paper, session 2024-25 · 2 marks · Very short answer

CBSE Class 10 Science sample paper 2024-25: question 25(B)

From the official CBSE sample question paper for 2024-25, section B, question 25(B), for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q25(B) · 2 marks · Very short answerElectricity

The above figure shows two resistors X and Y connected in series to a battery. The power dissipated for this combination is $P_{1}$ . When these resistors are connected in parallel to the same battery then the power dissipated is given by $P_{2}$ . Find out the ratio $\frac{P _{1}}{P _{2}}$ .

(For visually impaired students: A. We have four resistors A, B, C and D of resistance 3 Ω, 6 Ω, 9 Ω and 12 Ω respectively. Find out the lowest resistance which can be obtained by combining these four resistors. OR B. You are given 2 fuse wires A and B with current ratings 2A and 5A respectively. Justify with reason which of the two would you use with a 1000W, 220V room heater?)

Figure in the paper: A battery connected in series with resistor X of resistance R and resistor Y of resistance 2R; current I marked. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

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

Hint 1

The battery is the same in both cases, so use $P = V^{2} / R$ .

Hint 2

Find the equivalent resistance for series and for parallel separately.

Hint 3

Divide the two power expressions; V and R cancel.

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.

Series: $R_{s} = R + 2 R = 3 R$
½ mark
$P_{1} = \frac{V ^{2}}{R _{s}} = \frac{V ^{2}}{3 R}$
½ mark
Parallel: $\frac{1}{R _{p}} = \frac{1}{R} + \frac{1}{2 R} = \frac{3}{2 R}$ , so $R_{p} = \frac{2 R}{3}$
½ mark
$P_{2} = \frac{V ^{2}}{2 R /3} = \frac{3 V ^{2}}{2 R}$ . Hence $\frac{P _{1}}{P _{2}} = \frac{V ^{2}}{3 R} \times \frac{2 R}{3 V ^{2}} = \frac{2}{9}$ , i.e. $P_{1} : P_{2} = 2 : 9$ .
½ mark
(For visually impaired students) A. The lowest resistance comes from joining all four in parallel: $\frac{1}{R} = \frac{1}{3} + \frac{1}{6} + \frac{1}{9} + \frac{1}{12} = \frac{12 + 6 + 4 + 3}{36} = \frac{25}{36}$ , so $R = 1.44 Ω$ . OR B. $I = P / V = 1000/220 \approx 4.54$ A. Use fuse B (5 A): it carries 4.54 A safely, whereas fuse A (2 A) would melt.

Answer: P₁ : P₂ = 2 : 9

Where the marks are

Series resistance 3R (0.5 mark)
$P_{1} = V^{2} /3 R$ (0.5 mark)
Parallel resistance 2R/3 (0.5 mark)
$P_{2} = 3 V^{2} /2 R$ and ratio 2/9 (0.5 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 paperSample paper 2024-25: every question, by section.

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CBSE Class 10 Science sample paper 2024-25: question 25(B)

Hints

Step-by-step solution

Revise the chapter

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