Alternating Current

JEE Main level, single correct, medium

8 of 190

Question 8 of 190

An electric bulb rated as 100 W−220 V100 \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 :

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Answer: (A) 0.64 A0.64\mathrm{~A}

P=vrmsirmsirms=100220i0=2irms=0.64 A\begin{aligned} & \mathrm{P}=\mathrm{v}_{\mathrm{rms}} \mathrm{i}_{\mathrm{rms}} \\ & \mathrm{i}_{\mathrm{rms}}=\frac{100}{220} \\ & \mathrm{i}_0=\sqrt{2} \mathrm{i}_{\mathrm{rms}}=0.64 \mathrm{~A}\end{aligned}

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