Answer the following on Joule's heating and electric energy / resistance.
Generated by claude-sonnet-4-6 · 2026-06-15 06:45 · grounding rag
Model Answer
(A) Two applications of Joule's heating in domestic electric circuits:
- Electric fuse – excess current melts the fuse wire, protecting the circuit.
- Electric heating devices – electric iron, toaster, heater, kettle convert electric energy to heat.
(B) The commercial unit of electric energy is kilowatt hour (kW h).
1 kW h = 1000 W × 3600 s = 3.6 × 10⁶ W s = 3.6 × 10⁶ J
Thus, 1 kW h = 3.6 × 10⁶ J (SI unit).
Source: Chapter 11 – Electricity, Sections 11.7.1 and 11.8
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Explanation
- For (A), the textbook explicitly lists the fuse and heating appliances as key applications of Joule's heating. Either two from: fuse, electric iron, toaster, oven, kettle, heater, electric bulb will earn marks.
- For (B), always show the conversion step-by-step: 1 kW h = 1000 W × 3600 s = 3.6 × 10⁶ J. The examiner looks for the derivation, not just the final value.