Q131 / 150 · MHT CET 2022 (07 Aug Shift 1) · Physics

The average value of a.c. voltage is given by $\mathrm{V}=\mathrm{V}_{\mathrm{m}} \sin (\omega \mathrm{t})$ over time interval $t=0$ to $=\frac{\pi}{\omega}$ is

  • A
    $\frac{\mathrm{v}_{\mathrm{m}}}{\pi}$
  • B
    $0$
  • C
    $\mathrm{V}_{\mathrm{m}}$
  • Correct answer
    D
    $\frac{2 \mathrm{~V}_{\mathrm{m}}}{\pi}$

Solution

What is the average value of ac voltage $\mathrm{V}=\mathrm{V}_0 \sin \omega \mathrm{t}$ over the time interval $t=0$ to $t=\frac{\pi}{\omega}$ $\begin{aligned} & \mathrm{V}_{\mathrm{av}}=\frac{\int_0^{\frac{\pi}{\omega}} \mathrm{Vdt}}{\int_0^{\frac{\pi}{\omega}} \mathrm{dt}}=\frac{\int_0^{\frac{\pi}{\omega}} \mathrm{V}_0 \sin \omega t \mathrm{dt}}{[\mathrm{t}]_0^{\pi / \omega}} \\ & =\frac{\mathrm{V}_0\left\{-\frac{\cos \omega \mathrm{t}}{\omega}\right\}_0^\omega}{\omega} \\ & =-\frac{\mathrm{V}_0}{\pi}[\cos \pi-\cos 0]\end{aligned}$

Hence, the correct answer is option: 'D'.

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