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 — D, 20 W. A bulb's filament resistance is roughly fixed, so power varies as P = V^2/R. Halving the voltage (220 V → 110 V) quarters the power: 80 W ÷ 4 = 20 W.
- (a)80 W — That is the rated power at the rated 220 V; at half the voltage the power drops, it does not stay the same.
- (b)60 W — Not consistent with P ∝ V^2; halving V gives one-quarter power, not three-quarters.
- (c)40 W — That would be half the power — but power falls with the square of voltage, so halving V gives one-quarter, i.e. 20 W, not 40 W.
For a resistive device at (approximately) constant resistance, P = V^2/R. The rated values fix R = V^2/P = 220^2/80. Operating at 110 V gives P' = 110^2/R = (110/220)^2 × 80 = (1/4) × 80 = 20 W. Power scales with the square of the applied voltage.
Use P ∝ V^2 at fixed R. Halving the voltage does not halve the power — it quarters it. The '40 W' trap assumes P ∝ V.
- For fixed resistance, P = V^2/R, so P ∝ V^2.
- Rated 220 V, 80 W → R = 220^2/80 = 605 ohm.
- At 110 V: P = 110^2/605 = 20 W.
- Halving the voltage gives one-quarter of the power.
- Assuming power halves when voltage halves (it quarters).
- Forgetting resistance is treated as constant here.
Asked as the new power of a rated bulb at a different (usually half) operating voltage.
No directly related past PYQ was found.
- practice — not a real PYQ
A 100 W, 200 V bulb is run at 100 V. Its power becomes
- (a)100 W
- (b)50 W
- (c)25 W
- (d)200 W
Answer(c) 25 W (P ∝ V^2).
- practice — not a real PYQ
Electric power of a resistor in terms of voltage and resistance is
- (a)P = V/R
- (b)P = V^2/R
- (c)P = VR
- (d)P = R/V^2
Answer(b) P = V^2/R.