Question

A single phase half bridge inverter has a resistive load of R = 10 $\Omega$ and the dc input voltage of 48 volts. The harmonic factor of the lowest order harmonic would be equal to _____ %.

• A. 33.33

Answer 1

The correct option is A 33.33
Fourier expression of output voltage is,
$V_{0,n}=\sum _{n=1,3,5}^{\infty }\frac{2V_{dc}}{n\pi }\sin n\omega tV$
Harmonic factor for 3rd harmonic

$H.F.=\frac{V_3,rms}{V_1,rms}$

$H.F.=\frac{V_3,rms}{V_1,rms}=\frac{\frac{2\times 48}{\sqrt{2}\times 3\pi }}{\frac{2\times 48}{\sqrt{2}\pi }}\times 100$

$=\frac{1}{3}\times 100=33.33\%$