Question

The voltage over a cycle varies as
$V=V_{0}sin\omega t$ for $0\le t\le \frac{\pi }{\omega }$
$=-V_{0}sin\omega t$ for $\frac{\pi }{\omega }\le t\le \frac{2\pi }{\omega }$
The average value of the voltage for one cycle is

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

Answer 1

Answer: (D)

The average value of the voltage is
$V_{av}=\frac{\underset{0}{\overset{2\pi /\omega }{\int }}Vdt}{\underset{0}{\overset{2\pi /\omega }{\int }}dt}$$=\frac{\underset{0}{\overset{\pi /\omega }{\int }}{V}_{0}sin\omega tdt+\underset{\pi /\omega }{\overset{2\pi /\omega }{\int }}(-V_{0}sin\omega t)dt}{\frac{2\pi }{\omega }}$
$=\frac{\omega }{2\pi }\left[{\left|\frac{-V_{0}cos\omega t}{\omega }\right|}_{{}_{{}_0}}^{{}^{\pi /\omega }}+{\left|\frac{V_{0}cos\omega t}{\omega }\right|}_{{}_{{}_{\pi /\omega }}}^{{}^{2\pi /\omega }}\right]$
$=\frac{2V_0}{\pi }$