Question

In a single phase VSI bridge inverter, the load current is $I_0$ = 50 sin ($\omega t-30°$) A. If the supply voltage is 200 V, then the power drawn from the supply is

• A. 11.02 kW
• B. 17.30 kW
• C. 12.24 kW
• D. 5.51 kW

Answer 1

The correct option is D 5.51 kW

$I_0$ = 50 sin ($\omega t-30°$)A

The given current have only fundamental component so only fundamental voltage component will be responsible for power.

$V_{01}=\frac{4V_s}{\sqrt{2}\pi }=\frac{4\times 200}{\sqrt{2}\pi }$

$=180.06V$
Taking voltage as reference,

$\varphi =30{}^{\circ }$

Active power, $P=V_{01}{I}_{01}\cos \varphi$

$=\frac{180.06\times 50}{\sqrt{2}}\times \cos 30{}^{\circ }=5513.19W\simeq 5.51kW$

$P=5.51kW$