JEE MAIN - Physics (2025 - 3rd April Evening Shift - No. 19)
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 :
0.32 A
0.64 A
0.45 A
2.2 A
Explanation
First find the bulb’s rms current and then convert to its peak value:
rms current
$$I_{\text{rms}}=\frac{P}{V}=\frac{100\text{ W}}{220\text{ V}}\approx0.455\text{ A}$$
peak current
$$I_{\text{peak}}=I_{\text{rms}}\sqrt{2}\approx0.455\times1.414\approx0.64\text{ A}$$
So the correct choice is 0.64 A (Option B).
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