Q 12-07-067JEE MainJEE Main 2024 (30 Jan, Shift 2)Medium
An alternating voltage $V(t) = 220\sin100\pi t$ 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:
Answer: (B) $3.3\ \text{ms}$
The current is in phase with the voltage: $i = i_0\sin100\pi t$.
$i = i_0/2$ at $100\pi t_1 = \pi/6$, and $i = i_0$ at $100\pi t_2 = \pi/2$.
$$\Delta t = \frac{\pi/2 - \pi/6}{100\pi} = \frac{1}{300}\ \text{s} \approx 3.3\ \text{ms}$$
Solution by Sreeraj P, M.Sc Physics