MP Board · Class 12 · Physics · Alternating CurrentAn AC voltage given by $V = 200 \sqrt{2} \sin(100 \pi t)$ is applied across a circuit. Find its peak voltage and frequency.

Comparing the given alternating voltage equation $V = 200 \sqrt{2} \sin(100 \pi t)$ with the standard AC voltage equation $V = V_0 \sin(\omega t)$:

1. Peak Voltage ($V_0$): Direct comparison gives the peak voltage $V_0 = 200 \sqrt{2}\text{ V} \approx 282.8\text{ V}$.
2. Frequency ($f$): Angular frequency $\omega = 100 \pi\text{ rad/s}$. Since $\omega = 2\pi f$, we have $2\pi f = 100 \pi$, which yields frequency $f = \frac{100 \pi}{2 \pi} = 50\text{ Hz}$.\nTherefore, the peak voltage is $200 \sqrt{2}\text{ V}$ and the frequency of the AC supply is $50\text{ Hz}$.