Science (086) — AI-generated practice question

(i) Joule's Law of Heating:
The heat produced in a resistor is directly proportional to (a) the square of current (I²), (b) the resistance (R), and (c) the time (t) for which current flows.
$$H = I^2 R t$$

Derivation: When charge Q flows through resistance R under potential difference V in time t,
Work done = W = VQ = V × It (since Q = It)
Using V = IR: W = IR × It = I²Rt
Since this work converts to heat: H = I²Rt

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(ii) Given: R = 44 Ω, V = 220 V, t = 5 min = 300 s

(a) Current: $I = \dfrac{V}{R} = \dfrac{220}{44} = \mathbf{5\ A}$

(b) Power: $P = VI = 220 \times 5 = \mathbf{1100\ W}$

(c) Heat generated: $H = I^2Rt = (5)^2 \times 44 \times 300 = 25 \times 44 \times 300 = \mathbf{3,30,000\ J}$

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(iii) Tungsten is used because it has a very high melting point (3380°C), so it can become white-hot and emit light without melting. Copper, though a better conductor, has a much lower melting point and would melt at the high operating temperatures of a bulb filament.

Source: Chapter 11, Section 11.7.1 and 11.8

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• Joule's law derivation must start from W = VIt and substitute V = IR — examiners look for this logical step.
• Part (ii): Use I = V/R first, then P = VI, then H = I²Rt (or P×t). Show substitution clearly for full marks.
• Part (iii): The key words examiners expect are high melting point and the contrast with copper's lower melting point. The source passage explicitly states tungsten's melting point (3380°C) — mention it for full credit.
• Budget your time: derivation ~2 marks, numericals ~2 marks, reason ~1 mark.