An electric kettle, connected to a 240V mains, produces 6.0 x 105J of heat energy to boil a quantity of water in 5 minutes. Find the resistance of the kettle.
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Correct Answer: Option B
Explanation:
Heat energy = Electrical energy
6 x 105 = \(\frac{V^2}{R}\)t
R = \(\frac{240 \times 240 \times 5 \times 60}{6 \times 105}\)
= 28.8\(\Omega\)
Heat energy = Electrical energy
6 x 105 = \(\frac{V^2}{R}\)t
R = \(\frac{240 \times 240 \times 5 \times 60}{6 \times 105}\)
= 28.8\(\Omega\)