Question

A single phase full bridge inverter is shown below. Supply frequency is 60 Hz.



$R=4\Omega ,L=35mH$ and $C=155\mu F$

Then the rms value of $3rd$ harmonic output current is

Answer 1

$V_{03\left(rms\right)}=\frac{4V_s}{3\pi \sqrt{2}}==\frac{4\times 230}{3\pi \sqrt{2}}=69.024V$
$|Z_{03}|=\sqrt{R^2+{(n\omega L-\frac{1}{n\omega C})}^2}$
$=\sqrt{4^2+{(3\times 2\pi \times 60\times 35\times {10}^{-3}-\frac{1}{3\times 2\pi \times 60\times 155\times {10}^{-6}})}^2}$
$=\sqrt{4^2+(33.8795{)}^2}$
$|Z_{03}|=34.1148\Omega$
The rms value of $3rd$ harmonic output current is,
$I_{03\left(rms\right)}=\frac{V_{03\left(rms\right)}}{|Z_{03}|}=\frac{69.024}{34.1148}=2.02A$