--- title: "AC voltage controller, R = 10 Ω, α = 45°" subject: "Power Electronics" chapter: 6 key: "semester_7/power_electronics:acvc-r10-230v-45deg" source: https://www.bctengineer.com/q/semester_7/power_electronics/acvc-r10-230v-45deg site: BCT Engineering --- # AC voltage controller, R = 10 Ω, α = 45° Power Electronics · Chapter 6: AC voltage controller - Asked in: 2073 Chaitra, 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-r10-230v-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 R = 10Ω with an ac input voltage of 230 V, 50Hz. Firing angle for both the thyristors is 45°. Calculate rms value of output voltage, load current, input power factor and average value of current of thyristors. ## Short answer RMS output voltage = 219.30 V, RMS load current = 21.93 A. Input power factor = 0.9535 lagging. Average thyristor current = (Vm/(2πR))(1+cos 45°) = 8.84 A. ## Answer 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.) ```text 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a single-phase full-wave AC voltage controller with R load, which formula gives the rms output voltage? A. Vs*sqrt((1/pi)(pi - alpha + sin2alpha/2)) B. Vs*sqrt((1/pi)(pi - alpha - sin2alpha/2)) C. Vs*(1 + cos alpha)/2 D. Vm*sqrt((1/pi)(pi - alpha + sin2alpha/2)) Answer: A. Integrating Vm^2 sin^2(wt) from alpha to pi gives Vs*sqrt((1/pi)(pi - alpha + sin2alpha/2)), with Vs the rms supply voltage. (Key point: RMS output voltage formula) 2. With Vs = 230 V and alpha = 45°, what is the rms output voltage? A. 230 V B. 162.6 V C. 219.3 V D. 325.3 V Answer: C. Vo = 230*sqrt(0.9092) = 230*0.9535 = 219.3 V. (Key point: Computing Vo at 45°) 3. For the resistive-load controller, the input power factor equals which ratio? A. cos alpha B. Vo/Vs C. Vs/Vo D. Io/Vs Answer: B. pf = Po/(Vs*Io) = Io^2R/(Vs*Io) = Vo/Vs = 0.9535 lagging. (Key point: Input power factor) 4. Why is the average thyristor current Vm/(2 pi R)*(1+cos alpha) and not Vm/(pi R)*(1+cos alpha)? A. Each thyristor conducts only in its own half cycle, so the average is taken over 2 pi B. The load resistance doubles C. The thyristor conducts from 0 to alpha D. Both thyristors share the current equally in every half cycle Answer: A. Each SCR conducts only from alpha to pi of the full 2 pi period, so the integral is divided by 2 pi, giving 8.84 A here. (Key point: Average thyristor current) ## Close variants set in other exams - [AC voltage controller, R = 20 Ω, α = 45°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Aacvc-r20-230v-45deg) — Power Electronics, chapter 6, asked 2× ## Related in Power Electronics - [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× - [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× - [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× - [Single-phase full converter, 10 A, α = 30°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Afull-converter-10a-30deg) — Power Electronics, chapter 2, asked 4× - [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 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× - [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× - [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