JEE Main 2025 Apr 3 Shift 2, Physics Q15: Introduction to Alternating Currents and Circuits

JEE Main2025Apr 3, Shift 2Physics

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 :

Solution

Average power on a resistive load: $P = V_{r m s} I_{r m s}$ . Hence

$I_{r m s} = \frac{P}{V _{r m s}} = \frac{100}{220} = \frac{5}{11}$ A.

Peak current: $I_{0} = 2 I_{r m s} = 2 \cdot \frac{5}{11} \approx 0.643$ A.

$∴ I_{0} \approx 0.64$ A.

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