Study the circuit shown in which two resistors X and Y of resistances 3 Ω and 6 Ω respectively are joined in series with a battery of 2 V. (I) Draw a circuit diagram showing the above two resistors X and Y joined in parallel with same battery and same ammeter and voltmeter. (II) In which combination of resistors will the (i) potential difference across X and Y and (ii) current through X and Y, be the same ? (III)(a) Find the current drawn from the battery by the series combination of the two resistors (X and Y). (III)(b) Determine the equivalent resistance of the parallel combination of the two resistors (X and Y).

ScienceCBSE Class 10 · Board exam 2025, set 31/6/1 · 4 marks · Case-based

CBSE Class 10 Science 2025: question 39

Set in the CBSE Class 10 board exam of 2025, paper 31/6/1, section E, question 39, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q39 · 4 marks · Case-basedCompetency-basedElectricity

Study the circuit shown in which two resistors X and Y of resistances 3 $Ω$ and 6 $Ω$ respectively are joined in series with a battery of 2 V.

(I)
Draw a circuit diagram showing the above two resistors X and Y joined in parallel with same battery and same ammeter and voltmeter.
1 mark
(II)
In which combination of resistors will the (i) potential difference across X and Y and (ii) current through X and Y, be the same ?
1 mark
(III)(a)
Find the current drawn from the battery by the series combination of the two resistors (X and Y).
2 marks
OR
(III)(b)
Determine the equivalent resistance of the parallel combination of the two resistors (X and Y).
2 marks

Figure in the paper: Circuit: resistor X (3 Ω) and resistor Y (6 Ω) in series along the top wire, with a voltmeter V connected across X and Y (− on X side, + on Y side). Right side has an ammeter A (− at top, + at bottom). Battery of 2 V at the bottom left with + terminal on the right, and a point-contact switch symbol in the bottom wire; arrow shows current direction. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 31/6/1.

Hints

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

Hint 1

An ammeter always goes in series in the main line, and a voltmeter goes across the part being measured.

Hint 2

In which arrangement does the same charge flow through both resistors one after another? In which are both connected across the same two points?

Hint 3

Use $R_{s} = R_{1} + R_{2}$ with $I = V / R$ , and $1/ R_{p} = 1/ R_{1} + 1/ R_{2}$ .

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.

(I) Circuit diagram (described in words): Connect the 2 V battery, the plug key (switch) and the ammeter A in series in the main line. The ammeter's + terminal goes towards the battery's + terminal.
Join X (3 Ω) and Y (6 Ω) side by side between two common junction points, so that each forms a separate branch.
Connect the voltmeter V across these two junction points, i.e. in parallel with X and Y, with its + terminal on the side joined to the battery's + terminal. Show the direction of current with arrows.
1 mark
(II)(i) The potential difference across X and Y is the same in the parallel combination.
½ mark
(II)(ii) The current through X and Y is the same in the series combination.
½ mark
(III)(a) Equivalent resistance in series: $R_{s} = R_{X} + R_{Y} = 3 Ω + 6 Ω = 9 Ω$
1 mark
(III)(a) By Ohm's law, $I = \frac{V}{R _{s}} = \frac{2 V}{9 Ω} \approx 0.22 A$
1 mark
(III)(b) For parallel combination: $\frac{1}{R _{p}} = \frac{1}{R _{X}} + \frac{1}{R _{Y}}$
½ mark
(III)(b) Substituting: $\frac{1}{R _{p}} = \frac{1}{3} + \frac{1}{6} = \frac{2 + 1}{6} = \frac{3}{6} = \frac{1}{2}$
½ mark
(III)(b) Therefore $R_{p} = 2 Ω$
1 mark

Answer: (I) X and Y in parallel, with the ammeter in series with the battery and the voltmeter across the combination. (II)(i) Parallel; (ii) Series. (III)(a) I ≈ 0.22 A. (III)(b) $R_{p} = 2 Ω$

Where the marks are

(I) Correct parallel circuit with ammeter in series and voltmeter across the combination (1 mark)
(II) (i) parallel (½ mark); (ii) series (½ mark)
(III)(a) Series resistance 9 Ω (1 mark); current 2/9 ≈ 0.22 A (1 mark)
(III)(b) Formula for parallel resistance (½ mark); substitution (½ mark); $R_{p} = 2 Ω$ (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 paper2025 board paper, set 31/6/1: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Science 2025: question 39

Hints

Step-by-step solution

Revise the chapter

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