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{\int\limits^{2\pi/\omega}_{{0}}\,V\,dt}{\int\limits^{2\pi/\omega}_{{0}} \,dt}$$=\frac{\int\limits^{\pi/\omega}_{{0}}\,V_0\,sin\,\omega tdt +\int\limits^{2\pi/\omega}_{{\pi/\omega}} (-V_0\,sin\,\omega t)dt}{\frac{2\pi}{\omega}}$
$=\frac{\omega}{2\pi}\left[\left|\frac{-V_{0}\,cos\,\omega t}{\omega}\right|^{^{\pi/\omega}}_{_{_0}}+\left|\frac{V_{0}\,cos\,\omega t}{\omega}\right|^{^{2\pi/\omega}}_{_{_{\pi/\omega}}}\right]$
$=\frac{2V_{0}}{\pi}$