Q 12-07-144JEE MainJEE Main 2019 (8 Apr, Shift 1)Easy
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
Answer: (D) $3.33\ \text{ms}$
In a resistor the current is in phase with the voltage: $i = i_0\sin(100\pi t)$.
Half the peak value: $100\pi t_1 = \pi/6$. Peak value: $100\pi t_2 = \pi/2$.
$$\Delta t = \frac{\pi/2 - \pi/6}{100\pi} = \frac{1}{300}\ \text{s} \approx 3.33\ \text{ms}$$
Solution by Sreeraj P, M.Sc Physics