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JEE Main 20253 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Evening Shift
Subject
Physics
Chapter
Alternating Current

JEE Main 2025 · 3 April · Evening Shift · Physics · Alternating Current

An electric bulb rated as 100 ~W-220 ~V is connected to an ac...

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 :

Options

  1. A0.64 A0.64\mathrm{~A}
  2. B0.45 A0.45\mathrm{~A}
  3. C2.2 A2.2\mathrm{~A}
  4. D0.32 A0.32\mathrm{~A}

Correct answer

A. 0.64 A0.64\mathrm{~A}

Step-by-step solution

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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