AC voltage controller, R = 10 Ω, α = 45°

In a full-wave AC voltage controller with resistive load, each of the two anti-parallel thyristors conducts from $\alpha$ to $\pi$ in its half cycle.

Given: $V_s = 230$ V, 50 Hz, $R = 10\ \Omega$, $\alpha = 45° = \pi/4$, $V_m = \sqrt{2}\times 230 = 325.27$ V.

RMS output voltage

$$\begin{aligned} V_o &= V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)} \\ &= 230\sqrt{\frac{1}{\pi}\left(\pi - \frac{\pi}{4} + \frac{\sin 90°}{2}\right)} \\ &= 230\sqrt{\frac{2.3562 + 0.5}{\pi}} = 230\sqrt{0.9092} \\ &= 230 \times 0.9535 = 219.30\ \text{V} \end{aligned}$$

RMS load current

$$I_o = \frac{V_o}{R} = \frac{219.30}{10} = 21.93\ \text{A}$$

Input power factor

$$\begin{aligned} P_o &= I_o^2 R = 21.93^2 \times 10 = 4809.4\ \text{W} \\ pf &= \frac{P_o}{V_s I_o} = \frac{V_o}{V_s} = \frac{219.30}{230} = 0.9535\ \text{(lagging)} \end{aligned}$$

Average thyristor current

Each thyristor conducts only in its own half cycle:

$$\begin{aligned} I_{T,avg} &= \frac{1}{2\pi}\int_{\alpha}^{\pi}\frac{V_m}{R}\sin\omega t\; d(\omega t) = \frac{V_m}{2\pi R}(1 + \cos\alpha) \\ &= \frac{325.27}{2\pi \times 10}(1 + 0.7071) = 8.84\ \text{A} \end{aligned}$$

(For reference, rms thyristor current $= I_o/\sqrt{2} = 15.51$ A.)

 vo   (alpha = 45 deg)
        .--.
       |    \
 ------|-----\-----.------
 0    45    180    |    / 360
                   |  /
                   '-'
                  225

Answer: $V_o = 219.30$ V, $I_o = 21.93$ A, pf = 0.9535 lagging, $I_{T,avg} = 8.84$ A.