Full-bridge inverter RLC load to fifth harmonic

For a square-wave full-bridge inverter, the output voltage contains only odd harmonics, $v_0 = \sum \frac{4V_s}{n\pi}\sin n\omega t$. Each harmonic current is found by dividing by the load impedance at that harmonic, and the currents are added (superposition).

Given: $R = 4\ \Omega$, $L = 35$ mH, $C = 155\ \mu$F, $V_s = 230$ V, $f = 50$ Hz, $\omega = 314.16$ rad/s.

i) Load current up to the fifth harmonic

$$\begin{aligned} V_{nm} &= \frac{4V_s}{n\pi} = \frac{292.85}{n}\ \text{V} \\ Z_n &= \sqrt{R^2 + \Big(n\omega L - \frac{1}{n\omega C}\Big)^2}, \qquad \theta_n = \tan^{-1}\frac{n\omega L - 1/(n\omega C)}{R} \end{aligned}$$

The current is $i_n = (V_{nm}/Z_n)\sin(n\omega t - \theta_n)$. The fundamental sees a capacitive load, so it leads:

$$i_0(t) = 28.31\sin(314.16t + 67.25^\circ) + 3.69\sin(942.48t - 81.30^\circ) + 1.15\sin(1570.8t - 85.50^\circ)\ \text{A}$$

ii) RMS value of fundamental load current

$$I_{1} = \frac{28.31}{\sqrt2} = 20.02\ \text{A}$$

iii) Power absorbed by the load and fundamental power

RMS load current (harmonics up to the 5th):

$$\begin{aligned} I_3 &= \frac{3.691}{\sqrt2} = 2.61\ \text{A}, \qquad I_5 = \frac{1.148}{\sqrt2} = 0.81\ \text{A} \\ I_0 &= \sqrt{20.02^2 + 2.61^2 + 0.81^2} = 20.20\ \text{A} \end{aligned}$$ $$\begin{aligned} P_0 &= I_0^2R = 20.20^2\times4 = 1632.5\ \text{W} \\ P_1 &= I_1^2R = 20.02^2\times4 = 1602.6\ \text{W} \end{aligned}$$

(Including all higher harmonics changes $P_0$ only slightly, to about 1633.6 W.)

Answer: $I_{1,rms} = 20.02$ A; load power $\approx 1632.5$ W; fundamental power $= 1602.6$ W.