JEE Main 2025 · 3 April · evening · Single correct · +4 / −1
An electric bulb rated as 100 W-220 V is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is :
- (a) 0.32 A
- (b) 0.64 A
- (c) 0.45 A
- (d) 2.2 A
Answer and solution
Answer (b) 0.64 A
Solution
Physics sir · 3 steps
The rating gives you the rms current; multiply by root two to reach the peak.
- Rated power is rms. A bulb rated 100 W, 220 V draws 100 W when the rms voltage across it is 220 V. Here the source gives exactly 220 V rms, so the bulb runs at its rated power.
- Find rms current. For a resistive bulb, P = Vᵣₘₛ Iᵣₘₛ, so Iᵣₘₛ = 100/220 = 0.4545 A.
I₀ = √2 Iᵣₘₛ = 1.414 × 0.4545 A = 0.643 A
- So. The peak current through the bulb is about 0.64 A, option (b).
The idea to keep. Ratings on bulbs and appliances are always rms values. Get the rms current from P/V first, then use I₀ = √2 Iᵣₘₛ for the peak.
Answer from the NTA final answer key. Solution by PrepWiser.