Question

A single phase VSI is feeding a purely inductive load of 0.5 H. 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)

• A. 5 A
• B. 6 A
• C. 4 A
• D. 7 A

Answer 1

The correct option is C 4 A

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

$i_0=-I_p$

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

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

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

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

$Alternatesolution:i_{0max}=\frac{V_{dc}}{4fL}$