Question

For a given A.C voltage $V=V_0\sin \omega t$, if $V_1$ is average voltage of half cycle and $V_2$ is rms voltage, then $\frac{V_1}{V_2}$ is

• A. $0$
• B. $\sqrt{2}$
• C. $\frac{2\sqrt{2}}{\pi }$
• D. $\frac{\sqrt{2}}{\pi }$

Answer 1

The correct option is C $\frac{2\sqrt{2}}{\pi }$

$V_1=\frac{{\int }_0^{\frac{T}{2}}Vdt}{{\int }_0^{\frac{T}{2}}dt}=\frac{V_0}{\frac{T}{2}}{\int }_0^{\frac{T}{2}}\sin \omega tdt=\frac{2V_0}{\pi }$
$V_2=\frac{V_0}{\sqrt{2}}\Rightarrow \frac{V_1}{V_2}=\frac{2\sqrt{2}}{\pi }$