Question

A single phase VSI is feeding a purely inductive load of 0.5H. If the source voltage is 400 V and output frequency is 50 Hz, then the peak value of inductor current $i_0$ will be (Assume the load current does not have any dc component)

Answer 1

From $0<\omega t<\pi$, the load current tends to peak value. At $t_1$, peak value of output current is $I_p.$

$t_1=\frac{\pi }{\omega }=\frac{\pi }{2\pi \times 50}=\frac{1}{100}s$

At t = 0,
$i_0=-I_p$

$V_0=L\frac{di}{dt}$

$400=0.5\left(\frac{I_p-(-I_p)}{t_1-0}\right)$

$400=0.5\times \frac{2I_p}{t_1}$

$\frac{400}{0.5\times 2}\times \frac{1}{100}=I_P=4A$