In a domestic circuit five LED bulbs are arranged as shown. The source voltage is 220 V and the power rating of each bulb is marked in the circuit diagram. Based on the following circuit diagram, answer the following questions : (a) State what happens when (i) key K_1 is closed. (ii) key K_2 is closed. (b) Find the current drawn by the bulb B when it glows. (c) Calculate (i) the resistance of bulb B, and (ii) total resistance of the combination of four bulbs B, C, D and E. (c) What would happen to the glow of all the bulbs in the circuit when keys K_1 and K_2 both are closed and the bulb C suddenly get fused ? Give reason to justify your answer.

ScienceCBSE Class 10 · Board exam 2024, set 31/3/1 · 4 marks · Case-based

CBSE Class 10 Science 2024: question 39

Set in the CBSE Class 10 board exam of 2024, paper 31/3/1, 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

In a domestic circuit five LED bulbs are arranged as shown. The source voltage is 220 V and the power rating of each bulb is marked in the circuit diagram. Based on the following circuit diagram, answer the following questions :

(a)
State what happens when (i) key $K_{1}$ is closed. (ii) key $K_{2}$ is closed.
1 mark
(b)
Find the current drawn by the bulb B when it glows.
1 mark
(c)
Calculate (i) the resistance of bulb B, and (ii) total resistance of the combination of four bulbs B, C, D and E.
2 marks
OR
(c)
What would happen to the glow of all the bulbs in the circuit when keys $K_{1}$ and $K_{2}$ both are closed and the bulb C suddenly get fused ? Give reason to justify your answer.
2 marks

Figure in the paper: Circuit with a battery (cell stack) on the left, an ammeter at the top, bulb A (22 W, 220 V) on the upper branch, keys K1 and K2 at the top right, and four bulbs B, C, D, E (each 11 W; 55 V) connected in a branch along the bottom in parallel to the lower line. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 31/3/1.

Hints

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

Hint 1

Trace which bulbs each key connects to the source.

Hint 2

Use the rating on bulb B: $P = V I$ , so $I = P / V$ . Then $R = V / I$ .

Hint 3

In series, resistances add and one break stops current in the whole branch. A parallel branch works on its own.

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)(i) **Key $K_{1}$ closed: The branch containing bulb A is completed, so bulb A glows**.
½ mark
(a)(ii) **Key $K_{2}$ closed: The branch containing the series combination of bulbs B, C, D and E is completed, so bulbs B, C, D and E glow**.
½ mark
(b) Formula: For bulb B, power $P = 11$ W and voltage $V = 55$ V. Use $P = V I$ .
½ mark
(b) Current drawn by B:
$I = \frac{P}{V} = \frac{11 W}{55 V} = \frac{1}{5} A = 0.2 A$
½ mark
(c)(a)(i) Resistance of bulb B:
$R = \frac{V}{I} = \frac{55 V}{0.2 A} = 275 Ω$
Check: $R = \frac{V ^{2}}{P} = \frac{5 5 ^{2}}{11} = \frac{3025}{11} = 275 Ω$ .
1 mark
(c)(a)(ii) Total resistance of B, C, D, E: The four identical bulbs are in series, so their resistances add:
$R_{s} = 4 \times 275 Ω = 1100 Ω$
Check: $220 V /1100 Ω = 0.2$ A, which is the rated current of each bulb.
1 mark
(c)(b) OR alternative (marked instead of (c)(a) if chosen) – Bulb A: It keeps glowing with the same brightness. [½ mark]
(c)(b) Bulbs B, D and E: They stop glowing. [½ mark]
(c)(b) Reason: B, D and E are in series with C. When C fuses, that branch becomes an open circuit, so no current flows through B, D or E. Bulb A is on a separate parallel branch. It still gets the full 220 V and the same current, so it is not affected. [1 mark]

Answer: (a) (i) Bulb A glows; (ii) bulbs B, C, D, E glow. (b) 0.2 A. (c)(a) (i) 275 Ω, (ii) 1100 Ω. (c)(b) A glows with the same brightness; B, D, E stop glowing, because they are in series with fused C, while A is on a parallel branch

Where the marks are

(a)(i) Bulb A glows (½ mark)
(a)(ii) Bulbs B, C, D, E glow (½ mark)
(b) Uses P = VI (½ mark) and gets I = 0.2 A (½ mark)
(c)(i) R = 275 Ω using V/I or V²/P (1 mark)
(c)(ii) Series total = 4 × 275 = 1100 Ω (1 mark)
OR (c) A keeps glowing with the same brightness (½ mark); B, D, E stop glowing (½ mark); reason based on series/parallel connection (1 mark)

Revise the chapter

Chapter: Electricity (notes, formulas, where marks are lost and every past question)

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 31/3/1: 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 2024: question 39

Hints

Step-by-step solution

Revise the chapter

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