Q 12-07-154JEE MainJEE Main 2025 (3 Apr, Shift 2)Easy
An electric bulb rated as $100\ \text{W}$–$220\ \text{V}$ is connected to an ac source of rms voltage $220\ \text{V}$. The peak value of current through the bulb is:
Answer: (A) $0.64\ \text{A}$
The bulb runs at its rated power: $I_{\text{rms}} = \dfrac{P}{V_{\text{rms}}} = \dfrac{100}{220} = 0.4545\ \text{A}$.
$$I_0 = \sqrt2\,I_{\text{rms}} = 1.414\times0.4545 \approx 0.64\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics