Question

The voltage over a cycle varies as

$V=V_{0}sin\omega tfor0\le t\le \frac{\pi }{\omega }$ and

$V=-V_{0}sin\omega tfor\frac{\pi }{\omega }\le t\le \frac{2\pi }{\omega }$

Then the average value of voltage is

• A. $\frac{V_0}{\sqrt{2}}$
• B. $\frac{V_0}{2}$
• C. $zero$
• D. $\frac{2V_0}{\pi }$

Answer 1

Answer: (D)

$\frac{2V_0}{\pi }$

The average value of voltage is

$V_{av}=\frac{{\int }_0^{2\pi /\omega }Vdt}{{\int }_0^{2\pi /\omega }dt}$ = $\frac{{\int }_0^{\frac{\pi }{\omega }}{V}_{0}sin\omega tdt+{\int }_{\pi /\omega }^{2\pi /\omega }(-V_{0}sin\omega t)dt}{\frac{2\pi }{\omega }}$

$=\frac{\omega }{2\pi }[{\left|\frac{-V_{0}cos\omega t}{\omega }\right|}_0^{\pi /\omega }+{\left|\frac{V_{0}cos\omega t}{\omega }\right|}_{\pi /\omega }^{2\pi /\omega }]=\frac{2V_0}{\pi }$