A light bulb is rated as 100 W, 230 V and is connected to a 230 V ac source. The value of peak current is

0.61 A

The correct answer is Option (1) → 0.61 A

Given:

Power, $P = 100\,W$

RMS voltage, $V_{rms} = 230\,V$

For an AC circuit with purely resistive load:

$P = V_{rms} I_{rms}$

⟹ $I_{rms} = \frac{P}{V_{rms}} = \frac{100}{230} = 0.435\,A$

Peak current:

$I_0 = \sqrt{2}\, I_{rms} = 1.414 \times 0.435 = 0.615\,A$

Final Answer: $I_0 = 0.62\,A$