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. 29.21%
• B. 30.24%
• C. 33.33%
• D. 25.51%

Answer 1

The correct option is C 33.33%

Fourier expression of output voltage is,
$V_{0,n}=\sum _{n=1,3,5}^{\infty }\frac{2V_{dc}}{n\pi }sinn\omega tV$

Harmonic factor for $3^{rd}harmonic$
$H.F.=\frac{V_{3,rms}}{V_{1,rms}}$
$H.F.=\frac{V_{3,rms}}{V_{1,rms}}=\frac{\frac{2\times 48}{\sqrt{2}\times 3\pi }}{\frac{2\times 48}{\sqrt{2}\pi }}\times 100$

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