Topic: Power Electronics
A single-phase full-wave AC phase controller feeds power to a resistive load of 100$\Omega$ from a 220 V, 50 Hz supply. What will be the R.M.S. output voltage at delay angles $\alpha_1=\alpha_2=\alpha=\pi/2$ of both transistor?
- A. $\frac{220}{\sqrt{2}}\text{ V}$
- B. $\sqrt{2 \times 220}\text{ V}$
- C. $\sqrt{2} \times 110\text{ V}$
- D. $\sqrt{2} \times 220\text{ V}$
Answer: A. $\frac{220}{\sqrt{2}}\text{ V}$
Explanation: For a single-phase full-wave AC phase controller, the RMS output voltage is $V_{o(rms)} = \frac{V_m}{\sqrt{2}} \sqrt{\frac{1}{\pi} [(\pi - \alpha) + \frac{\sin 2\alpha}{2}]}$. Given $V_s = 220\text{ V}$ RMS (so $V_m = 220\sqrt{2}$) and $\alpha = \pi/2$. Substituting these values: $V_{o(rms)} = 220 \sqrt{\frac{1}{\pi} (\pi - \pi/2)} = 220 \sqrt{1/2} = \frac{220}{\sqrt{2}}\text{ V}$.