Question

The output voltages $e_1$ and $e_2$ of two full bridge inverters are added using output transformers. In order to eliminate the fifth harmonic from the output voltage, the phase angle between $e_1$ and $e_2$ should be

Answer 1

$e_1=\frac{4V_s}{n\pi }sinn\omega t$
$e_2=\frac{4V_s}{n\pi }sinn(\omega t+\varphi )$
$V_0=e_1+e_2$
$=\frac{4V_s}{n\pi }[sinn\omega t+sin(n\omega t+n\varphi \left)\right]$
$V_0=\frac{4V_s}{n\pi }[sinn\omega t+sinn\omega tcosn\varphi +cosn\omega tsinn\varphi ]$
In order to eliminate fifth harmonic
$V_{05}=0$
i.e., $sin5\omega t+sin5\omega tcos5\varphi +cos5\omega tsin5\varphi =0$
To safisfy the above equation,
$cos5\varphi =-1$ and $sin5\varphi =0$
$cos5\varphi =cos\pi$ $sin5\varphi =sin\pi$
$\varphi =\frac{\pi }{5};$ $\varphi =\frac{\pi }{5}$
$\therefore$ The phase angle between $e_1$ and $e_2$ should be $\varphi =\frac{\pi }{5}$ to eliminate fifth harmonic in the output.