Square-wave inverter RL load current up to 5th order

The output is a square wave of $\pm100$ V with period 0.02 s, so $f = 50$ Hz and $\omega = 314.16$ rad/s (the top level is taken as +100 V; the figure's sign is a misprint). Its Fourier series is $v_0 = \sum_{n\ \text{odd}}\frac{4V}{n\pi}\sin n\omega t$ with $V = 100$ V.

Load: $R = 5\ \Omega$, $L = 10$ mH in series.

Fundamental component of output voltage

$$\begin{aligned} V_{1m} &= \frac{4\times100}{\pi} = 127.32\ \text{V (peak)}, \qquad V_{1,rms} = 90.03\ \text{V} \\ v_1 &= 127.32\sin(314.16t)\ \text{V} \quad (\text{phase } 0^\circ) \end{aligned}$$

Harmonic currents

$$Z_n = \sqrt{R^2 + (n\omega L)^2}, \quad \theta_n = \tan^{-1}\frac{n\omega L}{R}, \quad I_{nm} = \frac{4V}{n\pi Z_n}$$

Time-domain load current (up to 5th order)

$$i_0(t) = 21.56\sin(314.16t - 32.14^\circ) + 3.98\sin(942.48t - 62.05^\circ) + 1.54\sin(1570.8t - 72.34^\circ)\ \text{A}$$

RMS of these components: $I_1 = 15.25$ A, $I_3 = 2.81$ A, $I_5 = 1.09$ A, so $I_0 \approx \sqrt{15.25^2 + 2.81^2 + 1.09^2} = 15.54$ A.

Answer: fundamental voltage $= 127.32$ V peak (90.03 V rms) at 0°; $i_0$ as above, with the fundamental current $21.56$ A peak lagging by $32.14^\circ$.