Question

A single phase half wave rectifier with a freewheeling diode has R- L load with resistance of value R = 8 $\Omega$ and very large inductance. If source is ac supply with 240 V and 50 Hz frequency , then the rms value of source current will be_____A.

• A. 9.55

Answer 1

The correct option is A 9.55
$\text{Output voltage,}{V}_0=\frac{1}{2\pi }{\int }_0^{\pi }{V}_m\sin \omega td\left(\omega t\right)=\frac{V_m}{\pi }$
$\text{Output current,}{I}_0=\frac{V_m}{\pi R}\frac{240\sqrt{2}}{\pi \times 8}=13.5A$
$I_{0,rms}=13.5A$
$\text{Power in the resistor},P=I_{rms}^{2}R=(13.5{)}^2\times 8=1458W$
$\text{Source current},I_{srms}=\frac{I_{0rms}}{\sqrt{2}}=\frac{13.5}{\sqrt{2}}=9.55A$