Question

A single -phase full bridge inverter has RLC load of $R=6\Omega$, L = 30 mH and $C=150\mu F$. If the input voltage is 230 V dc and output frequency is 50 Hz, then the magnitude of instantaneous value of fundamental load current at t = 20 msec will be

Answer 1

$V_{on}=\sum _{n=1,3,\mathrm{5...}}^{\infty }\frac{4V_s}{n\pi }sinn\omega tV$

$Z_{01}=6+(j\omega L-\frac{j}{\omega C})$

$=6+(j\times 2\pi \times 50\times 30\times {10}^{-3}-\frac{j}{2\pi \times 50\times 150\times {10}^{-6}})$

$Z_{01}=13.237\angle -{63.047}^{\circ }\Omega$

$I_{01}=\frac{4V_s}{\pi \left|Z_{01}\right|}sin(\omega t+{63.047}^{\circ })$

$=\frac{4\times 230}{\pi \times 13.237}sin(\omega t+{63.047}^{\circ })$

$=22.123sin(\omega t+{63.047}^{\circ })$

At, t = 20 msec

Magnitude, $I_{01}{|}_{20msec}=22.123\sin (2\pi \times 50\times 20\times {10}^{-3}\times \frac{{180}^{\circ }}{\pi }+{63.047}^{\circ })$

$=22.123sin({360}^{\circ }+{63.047}^{\circ })=19.72A$