An electric heater is rated 2 kW at 220 V. A fuse rated 5 A is installed in the circuit. Will the fuse operate safely? Justify your answer with a calculation.
Generated by claude-sonnet-4-6 · 2026-06-26 01:09 · grounding rag
Model Answer
Current drawn by the heater:
$$I = \frac{P}{V} = \frac{2000 \text{ W}}{220 \text{ V}} \approx 9.09 \text{ A}$$
Since the current drawn (9.09 A) is greater than the fuse rating (5 A), the fuse will blow (melt). It will NOT operate safely — the fuse will cut the circuit, protecting the appliance.
Source: Chapter 11, Section 11.7.1
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Explanation
- Use $I = P/V$ to find the actual current. 2 kW = 2000 W.
- Compare it with the fuse rating: 9.09 A > 5 A, so the fuse melts and breaks the circuit.
- Note the phrasing: "will the fuse operate safely?" means will it work without blowing — the answer is No, it will blow. That is actually the fuse doing its job of protection, but it means the circuit cannot run safely on a 5 A fuse; a higher-rated fuse (e.g., 10 A) is needed. Examiners want you to state this clearly with the calculation.