JEE MAIN - Physics (2024 - 30th January Evening Shift - No. 4)
An alternating voltage $$V(t)=220 \sin 100 \pi t$$ volt is applied to a purely resistive load of $$50 \Omega$$. The time taken for the current to rise from half of the peak value to the peak value is:
7.2 ms
3.3 ms
5 ms
2.2 ms
Explanation
Rising half to peak
$$\begin{aligned} & \mathrm{t}=\mathrm{T} / 6 \\ & \mathrm{t}=\frac{2 \pi}{6 \omega}=\frac{\pi}{3 \omega}=\frac{\pi}{300 \pi}=\frac{1}{300}=3.33 \mathrm{~ms} \end{aligned}$$
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