An electric bulb is rated as 220 V and 80 W. When it is operated on 110 V, the power rating would be :
- A.80 W
- B.60 W
- C.40 W
- D.20 W
Correct answer
Explanation
The correct answer is D, 20 W. The bulb's resistance stays the same, and since power varies as the square of voltage, halving the voltage cuts the power to one-fourth.
Power is P = V²/R. From the rating, the resistance of the filament is R = 220²/80 = 48,400/80 = 605 ohm. At 110 V, the power is 110²/605 = 12,100/605 = 20 W. A shortcut: the new voltage is half the rated voltage, so the power becomes (1/2)² = 1/4 of 80 W, which is 20 W. The current also halves, from about 0.36 A to about 0.18 A. This is why a bulb glows dimly when the supply voltage drops.
A is wrong: 80 W is the power only at the rated 220 V. B is wrong: 60 W does not follow from any correct relation between voltage and power. C is wrong: 40 W would be the result if power varied directly with voltage, which it does not when resistance is fixed.
Exam tip: for a fixed resistance, P ∝ V²; half the voltage gives a quarter of the power.
Practice Questions
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