For a domestic AC supply of 220 V at 50 tex | vedclass

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Question

(A) The root mean square voltage is given as $V_{rms} = 220 \ V$. The peak voltage $V_0$ is related to $V_{rms}$ by the formula $V_0 = V_{rms} \sqrt{2

Vedclass · Accepted Answer

$V(t) = 220 \sqrt{2} \cos(100 \pi t)$

Vedclass · Answer

(A) The root mean square voltage is given as $V_{rms} = 220 \ V$. The peak voltage $V_0$ is related to $V_{rms}$ by the formula $V_0 = V_{rms} \sqrt{2}$. Therefore, $V_0 = 220 \sqrt{2} \ V$. The angular frequency $\omega$ is given by $\omega = 2 \pi f$, where $f = 50 \ 	ext{Hz}$. Thus, $\omega = 2 \pi 	imes 50 = 100 \pi \ 	ext{rad/s}$. The instantaneous potential difference is given by $V(t) = V_0 \cos(\omega t)$. Substituting the values, we get $V(t) = 220 \sqrt{2} \cos(100 \pi t)$.

For a domestic $AC$ supply of $220 \ V$ at $50 \ \text{cycles per sec}$, the potential difference between the terminals of a two-pin electric outlet in a room is given by

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