An electric kettle is rated 2 kW at 240 V. Assuming the resistance of the heating element remains constant, calculate the power consumed when it is connected to a 120 V supply. A student argues that halving the voltage will halve the power. Is the student correct? Justify your answer with the appropriate formula.
Generated by claude-sonnet-4-6 · 2026-06-26 01:12 · grounding rag
Model Answer
Step 1: Find the resistance of the kettle.
Using $P = \dfrac{V^2}{R}$:
$$R = \frac{V^2}{P} = \frac{(240)^2}{2000} = \frac{57600}{2000} = 28.8 \ \Omega$$
Step 2: Find power at 120 V.
$$P' = \frac{V'^2}{R} = \frac{(120)^2}{28.8} = \frac{14400}{28.8} = 500 \ \text{W}$$
Is the student correct?
No. The student is wrong. Since $P = \dfrac{V^2}{R}$, power is proportional to the square of voltage. Halving the voltage reduces the power to one-fourth (not one-half): $2000 \times \frac{1}{4} = 500$ W.
Source: Chapter 11 – Electricity, Section 11.8 Electric Power
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Explanation
- Examiners expect you to use $P = V^2/R$ (not $P = VI$) when resistance is constant, because R doesn't change but I does.
- The common error the student makes is assuming $P \propto V$, but the correct relation is $P \propto V^2$. Always state this clearly to earn the justification mark.
- Show all three steps: find R, find new P, then answer the conceptual part. Each step carries roughly 1 mark.