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Answer the following. (Science 2023 31/1/1 Q36)

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ScienceCBSE Class 10 · Board exam 2023, set 31/1/1 · 5 marks · Long answer

CBSE Class 10 Science 2023: question 36

Set in the CBSE Class 10 board exam of 2023, paper 31/1/1, section D, question 36, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q36 · 5 marks · Long answerCompetency-basedElectricity

Answer the following.

  1. (a)

    An electric iron consumes energy at a rate of 880 W when heating is at the maximum rate and 330 W when the heating is at the minimum. If the source voltage is 220 V, calculate the current and resistance in each case.

    0 marks
  2. (b)

    What is heating effect of electric current ?

    0 marks
  3. (c)

    Find an expression for the amount of heat produced when a current passes through a resistor for some time.

    0 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 31/1/1.

Hints

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

Hint 1

Power, voltage and current are linked by P = VI.

Hint 2

Once you know I, use Ohm's law to get R for each setting.

Hint 3

For heat, start from the work done in moving charge Q = It through potential difference V.

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) Maximum heating: , . Using ,

    ½ mark

  2. ½ mark

  3. Using Ohm's law,

    ½ mark

  4. ½ mark

  5. Minimum heating: , , so

    ½ mark

  6. ½ mark

  7. (b) When an electric current flows through a resistor, part of the electrical energy is dissipated as heat. The resistor gets heated. This is called the heating effect of electric current (Joule heating).

    1 mark

  8. (c) Let a current flow through a resistor for time , with potential difference across it. Charge passed: . Work done in moving this charge: . This work appears as heat, so . Putting :

    (This is Joule's law of heating.)

    1 mark

Answer: (a) Maximum: I = 4 A, R = 55 Ω; Minimum: I = 1.5 A, R ≈ 146.7 Ω; (b) electrical energy dissipated as heat in a resistor; (c) H = I²Rt (= VIt)

Where the marks are

  • Formula I = P/V for the maximum case (½ mark)
  • I = 4 A (½ mark)
  • Formula R = V/I (½ mark)
  • R = 55 Ω (½ mark)
  • I = 1.5 A for the minimum case (½ mark)
  • R ≈ 146.6–146.7 Ω (½ mark)
  • Explains the heating effect of current (1 mark)
  • Derives/writes H = I²Rt or H = VIt (1 mark)

Revise the chapter

Practise this chapterBoard-style questions from Electricity, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 31/1/1: 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.