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
Answer
Why
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.
Why the others are wrong
- (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.
Concept
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.
Key facts
- 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.
Study next
Common traps
- 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.
Related PYQs
An incandescent electric bulb converts 20 % of its power consumption into light, and the remaining power is dissipated as heat. The bulb’s filament has a resistance of 200 Ω and 2 A current flows through it. If the bulb remains ON for 10 h and the rate of electricity charge is ₹5 per unit, then which among the following is the correct amount for the money spent on producing light?
- (a) ₹ 5
- (b) ₹ 6
- (c) ₹ 7
- (d) ₹ 8
Answer(d) ₹ 8
Same concept — electric power in an incandescent bulb. This 2022 item uses P = VI to get the current; the 2024 item extends the same power relations (P = I²R and energy = power × time) to a cost calculation.
Practice
- 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.