Consider the following electric circuit. Calculate the values of the following: (a) The total resistance of the circuit (b) The total current drawn from the source (c) Potential difference across the parallel combination of 10\ Ω and 15\ Ω resistors

ScienceCBSE Class 10 · Board exam 2025, set 31/1/1 · 3 marks · Short answer

CBSE Class 10 Science 2025: question 32

Set in the CBSE Class 10 board exam of 2025, paper 31/1/1, section C, question 32, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q32 · 3 marks · Short answerElectricity

Consider the following electric circuit. Calculate the values of the following:

(a)
The total resistance of the circuit
0 marks
(b)
The total current drawn from the source
0 marks
(c)
Potential difference across the parallel combination of $10 Ω$ and $15 Ω$ resistors
0 marks

Figure in the paper: Series circuit: a 15 V battery (positive terminal at top), key K, an ammeter A in the right-hand arm, and two parallel groups. The top group has a $10 Ω$ resistor in parallel with a $15 Ω$ resistor. The bottom group has a $60 Ω$ resistor in parallel with a $40 Ω$ resistor. The ammeter's positive terminal faces the top branch. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 31/1/1.

Hints

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

Hint 1

Reduce each parallel pair to a single resistance first.

Hint 2

The two reduced resistances are in series.

Hint 3

For (c), use V = IR with the equivalent resistance of the top pair only.

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) Top parallel group: $\frac{1}{R _{1}} = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{1}{6}$ , so $R_{1} = 6 Ω$ . Bottom group: $\frac{1}{R _{2}} = \frac{1}{60} + \frac{1}{40} = \frac{2 + 3}{120} = \frac{1}{24}$ , so $R_{2} = 24 Ω$ .
½ mark
(a) $R_{1}$ and $R_{2}$ are in series: $R_{t o t a l} = 6 + 24 = 30 Ω$ .
½ mark
(b) By Ohm's law, $I = \frac{V}{R} = \frac{15 V}{30 Ω} = 0.5 A$ .
1 mark
(c) The whole current flows through $R_{1}$ , so the p.d. across the $10 Ω ∥ 15 Ω$ combination is $V_{1} = I R_{1} = 0.5 A \times 6 Ω = 3.0 V$ .
1 mark

Answer: (a) 30 Ω; (b) 0.5 A; (c) 3.0 V

Where the marks are

Finds R₁ = 6 Ω, R₂ = 24 Ω and total 30 Ω (1 mark)
Current I = 15/30 = 0.5 A (1 mark)
p.d. = 0.5 × 6 = 3.0 V (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 paper2025 board paper, set 31/1/1: every question, by section.

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CBSE Class 10 Science 2025: question 32

Hints

Step-by-step solution

Revise the chapter

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