What is the current required to light a 60 W incandescent bulb in a domestic supply of 240 V?
- (a)0·5 A
- (b)0·25 A
- (c)1·0 A
- (d)5·0 A
Correct — B, 0·25 A. Electric power in a resistive device is P = V × I, so the current is I = P/V. Here I = 60 W ÷ 240 V = 0·25 A.
- (a)0·5 A — 0·5 A would mean P = VI = 240 × 0·5 = 120 W, twice the bulb's 60 W rating.
- (c)1·0 A — 1·0 A implies 240 W (240 × 1), four times the actual 60 W power.
- (d)5·0 A — 5·0 A would draw 1200 W, far above a 60 W bulb; this ignores the P = VI relation.
For a device operating on a supply voltage V and consuming power P, the current drawn follows from P = VI. Rearranged, I = P/V. This is the everyday relation used to size household wiring and choose fuses.
Simply divide power by voltage. Wrong options come from multiplying instead of dividing, or misplacing the 240 V. A 60 W bulb on 240 V draws a very small current — a quarter of an ampere.
- Electric power for a resistive load is P = V × I.
- Current drawn here is I = P/V = 60/240 = 0·25 A.
- The bulb's resistance follows from R = V²/P = 240²/60 = 960 Ω.
- Power can also be written P = I²R or P = V²/R for a resistor.
- Multiplying power and voltage instead of dividing.
- Confusing power rating (watt) with energy (watt-hour).
A direct plug-in of I = P/V, or the reverse — find the power given voltage and current.
No directly related past PYQ was found.
- practice — not a real PYQ
A 100 W bulb operates on a 200 V supply. The current through it is
- (a)0.2 A
- (b)0.5 A
- (c)2 A
- (d)5 A
Answer(b) 0.5 A — I = P/V = 100/200 = 0.5 A.
- practice — not a real PYQ
The resistance of a 60 W bulb working at 240 V is
- (a)240 Ω
- (b)480 Ω
- (c)960 Ω
- (d)1440 Ω
Answer(c) 960 Ω — R = V²/P = 240²/60 = 960 Ω.