Q1. [3] deep thorough-understanding
An electric circuit has a battery, a resistor of resistance R, and a fuse, all in series. The battery maintains a constant potential difference V. (i) Write an expression for the power dissipated in the resistor. (ii) If R is halved (by replacing it with a shorter wire of the same material and cross-section), what happens to the power dissipated? (iii) Explain why this change could cause the fuse to melt.
Generated by claude-sonnet-4-6 · 2026-06-26 01:12 · grounding rag
Model Answer
(i) Power dissipated in the resistor:
$$P = \frac{V^2}{R}$$
(ii) When R is halved (R becomes R/2):
$$P' = \frac{V^2}{R/2} = \frac{2V^2}{R} = 2P$$
The power dissipated doubles.
(iii) Higher power means a larger current flows through the circuit (since $I = V/R$, halving R doubles the current). A fuse melts when current exceeds its rated value. The doubled current produces excess heat in the fuse wire, causing it to melt and break the circuit.
Source: Chapter 11, Section 11.7 Heating Effect of Electric Current
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Explanation
- Examiners award 1 mark each for the three parts. Part (i) needs the correct formula ($V^2/R$); part (ii) needs to show working and state "doubles"; part (iii) needs to link increased current → excess heating → fuse melts.
- The key formula here is $P = V^2/R$ (not $I^2R$), because V is constant in this circuit.
- Always state the conclusion explicitly ("power doubles", "fuse melts") — don't leave the examiner to infer it.