Question

A single phase full bridge inverter is connected to an RL load of 20 $\Omega$ and 0.1 H. For a dc source voltage of 240 V and frequency of 50 Hz, the maximum value of load current at the end of first half cycle is _______ A. (Assume the circuit is initially relaxed)

• A. 10.37

Answer 1

The correct option is A 10.37
During the positive half cycle the circuit is,

$i\left(t\right)=\frac{V_s}{R}(1-e^{-t/\tau })=\frac{240}{20}(1-e^{\frac{-0.01\times 20}{0.1}})$

$i_{\left(0.01\right)}=10.37A$