Full-bridge square-wave inverter, 400 V, RL load

The square-wave output of a full-bridge inverter has the Fourier series $v_0 = \sum_{n\ \text{odd}}\frac{4V_s}{n\pi}\sin n\omega t$. Taking the voltage as the reference (phase 0°), each harmonic current lags its voltage by the load angle $\theta_n = \tan^{-1}(n\omega L/R)$.

Given: $V_s = 400$ V, $f = 50$ Hz ($\omega = 314.16$ rad/s), $R = 10\ \Omega$, $L = 50$ mH.

Fundamental component ($n = 1$)

$$\begin{aligned} V_{1m} &= \frac{4V_s}{\pi} = \frac{4\times400}{\pi} = 509.30\ \text{V}, \qquad V_{1,rms} = 360.13\ \text{V} \\ X_1 &= \omega L = 314.16\times0.05 = 15.708\ \Omega \\ Z_1 &= \sqrt{10^2 + 15.708^2} = 18.621\ \Omega, \quad \theta_1 = \tan^{-1}\frac{15.708}{10} = 57.52^\circ \\ I_{1m} &= \frac{509.30}{18.621} = 27.35\ \text{A}, \qquad I_{1,rms} = 19.34\ \text{A} \end{aligned}$$

Third harmonic ($n = 3$)

$$\begin{aligned} V_{3m} &= \frac{509.30}{3} = 169.77\ \text{V}, \qquad V_{3,rms} = 120.04\ \text{V} \\ X_3 &= 3\omega L = 47.124\ \Omega \\ Z_3 &= \sqrt{10^2 + 47.124^2} = 48.173\ \Omega, \quad \theta_3 = \tan^{-1}\frac{47.124}{10} = 78.02^\circ \\ I_{3m} &= \frac{169.77}{48.173} = 3.524\ \text{A}, \qquad I_{3,rms} = 2.492\ \text{A} \end{aligned}$$

Results

$$\begin{aligned} v_1 &= 509.30\sin\omega t, & i_1 &= 27.35\sin(\omega t - 57.52^\circ) \\ v_3 &= 169.77\sin3\omega t, & i_3 &= 3.524\sin(3\omega t - 78.02^\circ) \end{aligned}$$

The third harmonic voltage is one third of the fundamental, but its current is only about 13% of the fundamental current, because the inductive reactance triples at 150 Hz.