Physics · Current Electricity

JEE Main 2025 — 3 April, Evening Shift — Question 57

An electric bulb rated as $100 \mathrm{~W}-220 \mathrm{~V}$ is connected to an ac source of rms

voltage 220 V . The peak value of current through the bulb is

Answer: A

$R_{B}=\frac{V^{2}}{P}=\frac{220 \times 220}{100}=484$

$I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{R}=\frac{220}{484}=\frac{10}{22}=\frac{5}{11}$

$I_{0}=\frac{5 \sqrt{2}}{11} \approx \frac{7.07}{11} \approx 0.64$

Answer key and solution verified before publishing.

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