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Study the following circuit… (Class 10 Science 2024, Q39)

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ScienceCBSE Class 10 · Board exam 2024, set 31/4/1 · 4 marks · Case-based

CBSE Class 10 Science 2024: question 39

Set in the CBSE Class 10 board exam of 2024, paper 31/4/1, 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 following circuit :

On the basis of this circuit, answer the following questions :

  1. (a)

    Find the value of total resistance between the points A and B.

    1 mark
  2. (b)

    Find the resistance between the points B and C.

    1 mark
  3. (c)(i)

    Calculate the current drawn from the battery, when the key is closed.

    2 marks

  4. OR

    (c)(ii)

    In the above circuit, the resistor or the parallel combination of two resistors of , which one of the two will have more potential difference across its two ends ? Justify your answer.

    2 marks

Figure in the paper: Closed circuit with a 6 V battery and a key. Starting from point A on the left, three resistors 4 Ω, 6 Ω and 16 Ω are connected in series along the top wire to point B. Between B and C, two resistors of 8 Ω each are connected in parallel (one above the other). From C the wire returns along the bottom to the key and battery, and back up to A. Arrows show current flowing from A through the series resistors to B, through the parallel pair to C, and back to the battery. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 31/4/1.

Hints

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

Hint 1

Resistors joined end to end in one line simply add up. For two equal resistors in parallel, use 1/Rp = 1/R₁ + 1/R₂.

Hint 2

The AB section and the BC section are in series with each other. Add them, then apply Ohm's law to find the current.

Hint 3

The same current flows through the 16 Ω resistor and through the parallel pair. Compare V = IR for each.

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.

  1. (a) Between A and B, the 4 Ω, 6 Ω and 16 Ω resistors are in series:

    1 mark

  2. (b) Between B and C, the two 8 Ω resistors are in parallel: , so

    1 mark

  3. (c)(i) The series part and the parallel part are in series with each other: total resistance

    1 mark

  4. (c)(i) By Ohm's law,

    ½ mark

  5. (c)(i) A

    ½ mark

  6. (c)(ii) The 16 Ω resistor has the greater potential difference. Both are in series, so the same current flows through them, and by Ohm's law () the larger resistance has the larger potential difference.

    1 mark

  7. (c)(ii) Check:

    • V
    • V

    Since 3.2 V > 0.8 V, the 16 Ω resistor has more potential difference.

    1 mark

Answer: (a) 26 Ω; (b) 4 Ω; (c)(i) 0.2 A; (c)(ii) 16 Ω resistor (3.2 V vs 0.8 V)

Where the marks are

  • R_AB = 26 Ω (1 mark)
  • R_BC = 4 Ω (1 mark)
  • Total resistance 30 Ω (1 mark)
  • I = V/R = 6/30 = 0.2 A (½ + ½ mark)
  • Names the 16 Ω resistor (1 mark)
  • Justification: same current, V = IR gives 3.2 V vs 0.8 V (1 mark)

Revise the chapter

  • Chapter: Electricity (notes, formulas, where marks are lost and every past question)
  • Topic: Ohm's Law
Practise this chapterBoard-style questions from Electricity, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 31/4/1: 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.