--- title: "AC voltage controller, R = 20 Ω, α = 45°" subject: "Power Electronics" chapter: 6 key: "semester_7/power_electronics:acvc-r20-230v-45deg" source: https://www.bctengineer.com/q/semester_7/power_electronics/acvc-r20-230v-45deg site: BCT Engineering --- # AC voltage controller, R = 20 Ω, α = 45° Power Electronics · Chapter 6: AC voltage controller - Asked in: 2071 Shrawan, 8 marks - Video: [Single phase full wave AC voltage controller with R-load](https://www.youtube.com/watch?v=h3UAP09Hk1Q) — Anet Jose - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/acvc-r20-230v-45deg - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-6 ## Question A single-phase full wave ac voltage controller feeds a load of resistance 20 ohm with an input voltage of 230 V, 50 Hz. If the firing angle for both thyristor is 45°, calculate: i) rms value of output voltage ii) Load power and input power factors iii) Average and rms current of thyristors ## Short answer RMS output voltage = 219.30 V, load power = Vo²/R = 2404.7 W. Input power factor = 0.9535 lagging. Average thyristor current = 4.42 A, RMS thyristor current = Io/√2 = 7.75 A. ## Answer 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the full-wave AC voltage controller, over which intervals do T1 and T2 conduct? A. T1: 0 to alpha, T2: pi to pi+alpha B. T1: alpha to pi, T2: pi+alpha to 2pi C. T1: alpha to 2pi, T2: pi to 2pi D. T1: 0 to pi, T2: pi to 2pi Answer: B. Each thyristor is fired at its delay angle and turns off naturally at the zero crossing, so T1 conducts alpha to pi and T2 pi+alpha to 2pi. (Key point: Conduction intervals) 2. When R is changed from 10 ohm to 20 ohm with the same alpha = 45°, what happens to the rms output voltage? A. It doubles B. It halves C. It stays 219.3 V D. It drops to 109.7 V Answer: C. For a resistive load Vo depends only on Vs and alpha, so it is still 219.3 V; only the current changes. (Key point: Effect of R on Vo) 3. What is the load power for Vo = 219.3 V and R = 20 ohm? A. 4809 W B. 2404.7 W C. 1202 W D. 10.97 W Answer: B. Po = Vo^2/R = 219.3^2/20 = 2404.7 W. (Key point: Load power) 4. How is the rms thyristor current related to the rms load current Io? A. Io B. Io/2 C. Io/sqrt(2) D. Io*sqrt(2) Answer: C. Each thyristor carries current for half of the cycles, so its mean-square current is half of the load's, giving Io/sqrt(2) = 7.75 A. (Key point: Thyristor rms current) ## Close variants set in other exams - [AC voltage controller, R = 10 Ω, α = 45°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Aacvc-r10-230v-45deg) — Power Electronics, chapter 6, asked 2× ## Related in Power Electronics - [Full-wave thyristor rectifier, 220 V, 50 Ω, 45°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-wave-thyristor-220v-50ohm-45) — Power Electronics, chapter 2 - [AC voltage controller RL load extinction angle](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Aacvc-rl-extinction-angle) — Power Electronics, chapter 6 - [Single-phase AC voltage controller with R load](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Aacvc-r-load-rms) — Power Electronics, chapter 6, asked 6× - [AC voltage controller rms at α = 90°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Aacvc-r-load-220v-90deg) — Power Electronics, chapter 6, asked 2× - [GTO full converter, extinction angle: RMS and input current](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Agto-extinction-angle-rms) — Power Electronics, chapter 2, asked 2× - [Single-phase full-wave inverter with 5 Ω load](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-wave-inverter-5ohm) — Power Electronics, chapter 5 ## Related in other subjects - [3-phase thyristor bridge: no-load speed and firing angle](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Afull-bridge-motor-no-load-speed) — Utilization of Electrical Energy, chapter 3, asked 3× - [3-phase half controlled bridge: no-load speed, firing angle](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Ahalf-bridge-motor-no-load-speed) — Utilization of Electrical Energy, chapter 3, asked 3× - [Full wave bridge rectifier load power](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Abridge-rectifier-power-5ohm) — Electronics Devices and Circuit, chapter 6