Square-wave inverter ±100 V: fundamental load current

The output is a square wave of $\pm100$ V with $T = 0.02$ s, so $f = 50$ Hz and $\omega = 2\pi f = 314.16$ rad/s. Its fundamental is found from the Fourier series $v_0 = \sum_{n\ \text{odd}}\frac{4V}{n\pi}\sin n\omega t$.

Load: $R = 10\ \Omega$, $L = 20$ mH in series.

Fundamental component of output voltage

$$\begin{aligned} V_{1m} &= \frac{4V}{\pi} = \frac{4\times100}{\pi} = 127.32\ \text{V} \\ V_{1,rms} &= \frac{127.32}{\sqrt2} = 90.03\ \text{V} \end{aligned}$$

Phase: the square wave starts its positive half at $t = 0$, so the fundamental is in phase with it (0°):

$$v_1(t) = 127.32\sin(314.16t)\ \text{V}$$

Fundamental load current

$$\begin{aligned} X_1 &= \omega L = 314.16\times0.02 = 6.283\ \Omega \\ Z_1 &= \sqrt{10^2 + 6.283^2} = 11.810\ \Omega \\ \theta_1 &= \tan^{-1}\frac{6.283}{10} = 32.14^\circ \ (\text{lagging}) \\ I_{1m} &= \frac{127.32}{11.810} = 10.78\ \text{A}, \qquad I_{1,rms} = 7.62\ \text{A} \end{aligned}$$ $$i_1(t) = 10.78\sin(314.16t - 32.14^\circ)\ \text{A}$$

Answer: fundamental voltage $= 127.32$ V peak (90.03 V rms) at 0°; $i_1(t) = 10.78\sin(314.16t - 32.14^\circ)$ A. Fundamental power delivered $= 7.62^2\times10 = 581$ W.