Question

A half wave diode rectifier is shown in figure. The load contains pure inductor of inductance $L=10mH$. Assume the diode is ideal. The ratio of rms value of output current to the average value of output current is___________

• A. 1.22

Answer 1

The correct option is A 1.22
$\begin{array}{ccc}\text{By applying KVL during positive half cycle of source}& & \\ V_m\sin \omega t& =L\frac{di}{dt}& \\ i\left(t\right)& =\frac{-V_m}{\omega L}\cos \omega t+A& \cdots \left(i\right)\\ At\omega t=0,i\left(t\right)& =0& \\ 0& =\frac{-V_m}{\omega L}\cos 0+A& \\ A& =\frac{V_m}{\omega L}& \\ \text{Substituting the value of A in equation (i),}& & \\ i\left(t\right)& =\frac{-V_m}{\omega L}\cos \omega t+\frac{V_m}{\omega L}=\frac{V_m}{\omega L}(1-\cos \omega t)& \\ I_{0avg}& =\frac{1}{2\pi }{\int }_0^{2\pi }\frac{V_m}{\omega L}(1-\cos \omega t)d\left(\omega t\right)& \\ I_{0avg}& =\frac{V_m}{\omega L}& \\ I_{0rms}& =\sqrt{{\left(\frac{V_m}{\omega L}\right)}^2+{\left(\frac{\frac{V_m}{\omega L}}{\sqrt{2}}\right)}^2}=\sqrt{\frac{3}{2}}\times \frac{V_m}{\omega L}& \end{array}$

$\begin{array}{cc}\therefore \text{The ratio of rms value of output current to the average value of output current is,}\frac{I_{0rms}}{I_{0avg}}& =\frac{\sqrt{\frac{3}{2}}\frac{V_m}{\omega L}}{\frac{V_m}{\omega L}}=\sqrt{\frac{3}{2}}=1.22\end{array}$