---
title: "AC voltage controller on-off cycle control numerical"
subject: "Power Electronics"
chapter: 6
key: "semester_7/power_electronics:integral-cycle-control-numerical"
source: https://www.bctengineer.com/q/semester_7/power_electronics/integral-cycle-control-numerical
site: BCT Engineering
---

# AC voltage controller on-off cycle control numerical

Power Electronics · Chapter 6: AC voltage controller

- Asked in: 2082 Baishakh, 8 marks
- Video: [34. Principle of On Off Control for AC Voltage Controllers](https://www.youtube.com/watch?v=Nrm7tGdvbdU) — Electronic Waves
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/integral-cycle-control-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-6

## Question

A single-phase AC voltage control has input voltage of 230 V, 50 Hz and a load of Resistor of 15 Ω. For 6 Cycles ON and 4 cycles OFF, calculate: (i) RMS Output voltage (ii) Input pf (iii) Average and RMS thyristor currents (iv) Draw input and output waveform

## Short answer

For duty ratio 0.6, RMS output voltage is 178.16 V, input power factor is 0.775. Average thyristor current is 4.14 A and RMS thyristor current is 8.40 A, using peak current 21.68 A.

## Answer

In integral cycle (on-off) control, the thyristors are fired at the zero crossings so that complete cycles are passed to the load for $n$ cycles and blocked for $m$ cycles. The duty ratio is

$$
k = \frac{n}{n+m} = \frac{6}{6+4} = 0.6
$$

**Given:** $V_s = 230$ V, $f = 50$ Hz, $R = 15\ \Omega$, $V_m = \sqrt{2}\times 230 = 325.27$ V.

#### (i) RMS output voltage

$$
V_o = V_s\sqrt{k} = 230\sqrt{0.6} = 230 \times 0.7746 = 178.16\ \text{V}
$$

Load current $I_o = V_o/R = 178.16/15 = 11.88$ A; load power $P = V_o^2/R = 2116$ W.

#### (ii) Input power factor

$$
pf = \frac{P}{V_s I_o} = \frac{V_o^2/R}{V_s\,V_o/R} = \frac{V_o}{V_s} = \sqrt{k} = \sqrt{0.6} = 0.775
$$

#### (iii) Thyristor currents

Peak current $I_m = V_m/R = 325.27/15 = 21.68$ A. Each thyristor conducts one half-cycle in each ON cycle.

$$
\begin{aligned}
I_{T,avg} &= \frac{k\,I_m}{\pi} = \frac{0.6 \times 21.68}{\pi} = 4.14\ \text{A} \\
I_{T,rms} &= \frac{I_m}{2}\sqrt{k} = \frac{21.68}{2}\times 0.7746 = 8.40\ \text{A}
\end{aligned}
$$

#### (iv) Waveforms

```text
 vs  /\  /\  /\  /\  /\  /\  /\  /\  /\  /\
      \/  \/  \/  \/  \/  \/  \/  \/  \/  \/
     |<-------- 10 cycles = 200 ms -------->|

 vo  /\  /\  /\  /\  /\  /\
      \/  \/  \/  \/  \/  \/ ________________
     |<----- 6 ON (120 ms) ->|<- 4 OFF 80ms->|
```

- Input current has the same shape as $v_o$ (R load), flowing only in the 6 ON cycles.
- Thyristors switch at zero voltage, so little EMI; suited to heaters with large thermal time constants.

**Answer: $V_o = 178.16$ V, pf = 0.775, $I_{T,avg} = 4.14$ A, $I_{T,rms} = 8.40$ A.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For 6 cycles ON and 4 cycles OFF, what is the duty ratio k?
   A. 0.4
   B. 1.5
   C. 0.6
   D. 0.67
   Answer: C. k = n/(n+m) = 6/10 = 0.6. (Key point: Duty ratio)

2. What is the rms output voltage for Vs = 230 V and k = 0.6?
   A. 138 V
   B. 178.16 V
   C. 230 V
   D. 206 V
   Answer: B. Vo = Vs sqrt(k) = 230 x 0.7746 = 178.16 V. (Key point: RMS output voltage)

3. What is the input power factor with R load in integral cycle control?
   A. sqrt(k) = 0.775
   B. k = 0.6
   C. 1.0
   D. cos(alpha)
   Answer: A. pf = P/(Vs Io) = Vo/Vs = sqrt(k) = 0.775. (Key point: Input power factor)

4. Which pair gives the thyristor average and rms currents (Im = 21.68 A)?
   A. Average k Im/pi = 4.14 A, rms (Im/2) sqrt(k) = 8.40 A
   B. Average Im/pi = 6.9 A, rms Im/2 = 10.84 A
   C. Average k Im = 13 A, rms Im sqrt(k) = 16.8 A
   D. Average 4.14 A, rms 11.88 A
   Answer: A. Each thyristor carries one half-cycle in each ON cycle, so the average is k Im/pi and the rms is (Im/2) sqrt(k). (Key point: Thyristor currents)

## Related in Power Electronics

- [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 RL load extinction angle](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Aacvc-rl-extinction-angle) — Power Electronics, chapter 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×
- [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×
- [AC voltage controller rms and ELC use](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Aacvc-rms-elc-mhp) — Power Electronics, chapter 6
- [AC voltage controller RL load at α = 90°](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Aacvc-rl-230v-90deg) — Power Electronics, chapter 6

## Related in other subjects

- [Input current, pf and torque of single phase IM](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Asingle-phase-im-input-torque) — Electric Machines II, chapter 3
- [Speed, input current and output of 2/3 HP motor](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Atwo-thirds-hp-motor-performance) — Electric Machines II, chapter 3
- [Arc furnace: kW input, arc voltage, pf, kVA](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aarc-furnace-arc-voltage-kva) — Utilization of Electrical Energy, chapter 5
- [Design: 500 kVA single phase transformer with tubes](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-500kva-single-phase-tubes) — Electric Machine Design, chapter 3, asked 2×