Single-phase full-wave inverter with 5 Ω load

A single-phase full-wave (full-bridge) inverter with square-wave switching gives $v_0 = \pm V_s$ across the load; with a resistive load the current is a square wave of $\pm V_s/R$.

Given: $R = 5\ \Omega$, $V_s = 30$ V (unit taken as volts).

(i) RMS output voltage

$$V_{0,rms} = \left[\frac{2}{T}\int_0^{T/2}V_s^2\,dt\right]^{1/2} = V_s = 30\ \text{V}$$

(ii) RMS of fundamental component

From $v_0 = \sum_{n\ \text{odd}}\frac{4V_s}{n\pi}\sin n\omega t$:

$$V_{1,rms} = \frac{4V_s}{\sqrt2\,\pi} = 0.9003\times30 = 27.01\ \text{V}$$

(iii) Output power

$$P_0 = \frac{V_{0,rms}^2}{R} = \frac{30^2}{5} = 180\ \text{W}$$

(Fundamental power alone is $27.01^2/5 = 145.9$ W.)

(iv) Peak current in each thyristor

Each pair (T1, T2 or T3, T4) carries the full load current for its half cycle:

$$I_{T,peak} = \frac{V_s}{R} = \frac{30}{5} = 6\ \text{A}$$

Also, average thyristor current $= 6\times0.5 = 3$ A and rms thyristor current $= 6/\sqrt2 = 4.24$ A.

Answer: $V_{0,rms} = 30$ V; $V_{1,rms} = 27.01$ V; $P_0 = 180$ W; $I_{T,peak} = 6$ A.