AC voltage controller, R = 20 Ω, α = 45°

In a single-phase full-wave AC voltage controller with resistive load, T1 conducts from $\alpha$ to $\pi$ and T2 from $\pi + \alpha$ to $2\pi$.

Given: $V_s = 230$ V, 50 Hz, $R = 20\ \Omega$, $\alpha = 45° = \pi/4$, $V_m = \sqrt{2}\times 230 = 325.27$ V.

i) RMS output voltage

$$\begin{aligned} V_o &= V_s\sqrt{\frac{1}{\pi}\left(\pi - \alpha + \frac{\sin 2\alpha}{2}\right)} \\ &= 230\sqrt{\frac{1}{\pi}\left(\frac{3\pi}{4} + \frac{1}{2}\right)} = 230\sqrt{0.9092} \\ &= 230 \times 0.9535 = 219.30\ \text{V} \end{aligned}$$

ii) Load power and input power factor

$$\begin{aligned} I_o &= \frac{V_o}{R} = \frac{219.30}{20} = 10.97\ \text{A} \\ P_o &= \frac{V_o^2}{R} = \frac{219.30^2}{20} = 2404.7\ \text{W} \\ pf &= \frac{P_o}{V_s I_o} = \frac{2404.7}{230 \times 10.965} = \frac{V_o}{V_s} = 0.9535\ \text{(lagging)} \end{aligned}$$

iii) Thyristor currents

Each thyristor carries the load current only in its own half cycle.

$$\begin{aligned} I_{T,avg} &= \frac{V_m}{2\pi R}(1 + \cos\alpha) = \frac{325.27}{2\pi \times 20}(1 + 0.7071) = 4.42\ \text{A} \\ I_{T,rms} &= \frac{I_o}{\sqrt{2}} = \frac{10.965}{\sqrt{2}} = 7.75\ \text{A} \end{aligned}$$

The rms thyristor current formula follows from

$$I_{T,rms} = \left[\frac{1}{2\pi}\int_{\alpha}^{\pi}\left(\frac{V_m}{R}\sin\omega t\right)^2 d(\omega t)\right]^{1/2} = \frac{I_o}{\sqrt{2}}$$

Answer: $V_o = 219.30$ V; $P_o = 2404.7$ W; pf = 0.9535; $I_{T,avg} = 4.42$ A, $I_{T,rms} = 7.75$ A.