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(C) The instantaneous value of an alternating voltage is given by $V(t) = V_{\max} \sin(\omega t)$.To find the average value over a complete cycle (fr

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Zero

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(C) The instantaneous value of an alternating voltage is given by $V(t) = V_{\max} \sin(\omega t)$.To find the average value over a complete cycle (from $t = 0$ to $t = T$),we integrate the function over the period $T$ and divide by the period:$V_{	ext{avg}} = \frac{1}{T} \int_{0}^{T} V_{\max} \sin(\omega t) dt$.Since $\omega = \frac{2\pi}{T}$,the integral becomes:$V_{	ext{avg}} = \frac{V_{\max}}{T} \int_{0}^{T} \sin\left(\frac{2\pi}{T} tight) dt$.The integral of the sine function over a full period is zero because the positive area of the first half-cycle exactly cancels out the negative area of the second half-cycle.Therefore,$V_{	ext{avg}} = 0$.

What is the average value of the $AC$ voltage over a complete cycle?

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