A sinusoidal alternating voltage of time period 36 ms has the maximum value of 250 V. Its value will reach −125 V (half the value of negative maximum) after _______ milliseconds.

Correct option is D

$$\text{The voltage equation for a sinusoidal wave is given by:} \\[4pt]v(t) = V_m \sin(\omega t + \phi) \\[8pt]\text{Where:} \\[4pt]\bullet \; V_m = \text{the maximum value of the voltage} \\[4pt]\bullet \; \omega = \frac{2\pi}{T} \text{ is the angular frequency} \\[4pt]\bullet \; \phi = \text{the phase difference} \\[4pt]\bullet \; T = \text{the time period} \\[8pt]\textbf{Given Data:} \\[4pt]\bullet \; V_m = 250\,\text{V} \\[4pt]\bullet \; T = 36\,\text{ms} \\[4pt]\bullet \; \text{The voltage value to reach: } -125\,\text{V} \\[8pt]\textbf{First, calculate the angular frequency } (\omega): \\[4pt]\omega = \frac{2\pi}{T} = \frac{2\pi}{36 \times 10^{-3}} = 174.532\,\text{rad/s} \\[8pt]-125 = 250 \sin(\omega t) \\[4pt]\sin(\omega t) = \frac{-125}{250} = -0.5 \\[8pt]\omega t = \arcsin(-0.5) \\[4pt]\omega t = -\frac{\pi}{6} \\[8pt]t = \frac{-\pi/6}{174.532} = \frac{-\pi/6}{174.532} \approx 21\,\text{ms}$$​​