The average value of an A.C. voltage given by | vedclass

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(D) The average value of an $A$.$C$. voltage $V = V_{m} \sin(\omega t)$ over the time interval $t = 0$ to $t = \frac{\pi}{\omega}$ is calculated as fo

Vedclass · Accepted Answer

$\frac{2 V_{m}}{\pi}$

Vedclass · Answer

(D) The average value of an $A$.$C$. voltage $V = V_{m} \sin(\omega t)$ over the time interval $t = 0$ to $t = \frac{\pi}{\omega}$ is calculated as follows:$V_{av} = \frac{\int_{0}^{\frac{\pi}{\omega}} V dt}{\int_{0}^{\frac{\pi}{\omega}} dt}$$V_{av} = \frac{\int_{0}^{\frac{\pi}{\omega}} V_{m} \sin(\omega t) dt}{\frac{\pi}{\omega} - 0}$$V_{av} = \frac{V_{m}}{\frac{\pi}{\omega}} \left[ -\frac{\cos(\omega t)}{\omega} \right]_{0}^{\frac{\pi}{\omega}}$$V_{av} = \frac{V_{m} \omega}{\pi} \left( -\frac{1}{\omega} \right) [\cos(\pi) - \cos(0)]$$V_{av} = -\frac{V_{m}}{\pi} [-1 - 1]$$V_{av} = -\frac{V_{m}}{\pi} [-2] = \frac{2 V_{m}}{\pi}$

The average value of an $A$.$C$. voltage given by $V = V_{m} \sin(\omega t)$ over the time interval $t = 0$ to $t = \frac{\pi}{\omega}$ is:

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