Q51 / 75 · JEE Main 2025 (3 Apr Shift 2) · Physics

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 :

  • Correct answer
    A
    $0.64\mathrm{~A}$
  • B
    $0.45\mathrm{~A}$
  • C
    $2.2\mathrm{~A}$
  • D
    $0.32\mathrm{~A}$

Solution

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

Hence, the correct answer is option: 'A'.

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