A lamp is rated 240\,{V}, 60\,{W}. Determine the resistance of the lamp when lit?

Correct Answer is : 960 Ω960\,960Ω

Correct Answer is : 960 Ω960\,960Ω

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WAEC Physics 2022 Paper

Question : 29

Total: 50

240\,\mathrm{V}, 60\,\mathrm{W}

Solution:

Concept:

The power rating of an electrical device relates its voltage and resistance by

P = \frac{V^2}{R}

Explanation:

The lamp is rated at

P = 60\,\mathrm{W}

and

V = 240\,\mathrm{V}

Using the formula

P = \frac{V^2}{R}

, substitute the given values:

60 = \frac{(240)^2}{R}

(240)^2 = 57600

, so

60 = \frac{57600}{R}

Multiply both sides by

R

60R = 57600

Divide both sides by

60

R = \frac{57600}{60}

Thus,

R = 960\,\Omega

Answer:

The resistance of the lamp when lit is

960\,\Omega

The correct option is D.

960\,\Omega

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