---
title: "AC voltage controller RL load at α = 90°"
subject: "Power Electronics"
chapter: 6
key: "semester_7/power_electronics:acvc-rl-230v-90deg"
source: https://www.bctengineer.com/q/semester_7/power_electronics/acvc-rl-230v-90deg
site: BCT Engineering
---

# AC voltage controller RL load at α = 90°

Power Electronics · Chapter 6: AC voltage controller

- Asked in: 2074 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/acvc-rl-230v-90deg
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-6

## Question

The single phase ac controller has input voltage of 230V, 50Hz and the positive and negative thyristors are triggered at an angle of 90° and π+90° respectively. The series RL load has R = 20 Ω, L = 50mH then determine the rms value of load voltage and load current.

## Short answer

Load angle is 38.15°, so conduction is discontinuous. Extinction angle β is found numerically as 215.37°. RMS output voltage is 169.99 V, RMS load current is 5.575 A.

## Answer

With an R-L load, current lags; since $\alpha$ is greater than the load angle $\phi$, conduction is discontinuous and each SCR conducts from $\alpha$ to the extinction angle $\beta$.

**Given:** $V_s = 230$ V, 50 Hz, $R = 20\ \Omega$, $L = 50$ mH, $\alpha = 90°$.

#### Load parameters

$$
\begin{aligned}
X_L &= 2\pi f L = 2\pi(50)(0.05) = 15.708\ \Omega \\
Z &= \sqrt{20^2 + 15.708^2} = 25.431\ \Omega \\
\phi &= \tan^{-1}\frac{15.708}{20} = 38.15°, \quad \tan\phi = 0.7854
\end{aligned}
$$

Since $\alpha = 90° > \phi = 38.15°$, conduction is discontinuous.

#### Extinction angle

$$
i(\omega t) = \frac{V_m}{Z}\left[\sin(\omega t - \phi) - \sin(\alpha - \phi)\,e^{-(\omega t - \alpha)/\tan\phi}\right]
$$

Setting $i(\beta) = 0$:

$$
\sin(\beta - 38.15°) = \sin(51.85°)\,e^{-(\beta - 90°)\pi/(180°\times 0.7854)}
$$

Solving numerically (trial and error):

$$
\beta = 215.37°, \qquad \gamma = \beta - \alpha = 125.37°
$$

Check: $\beta < 180° + \alpha = 270°$, so T1 stops before T2 is fired (discontinuous), as expected.

#### RMS load voltage

$$
\begin{aligned}
V_o &= V_s\sqrt{\frac{1}{\pi}\left[\beta - \alpha + \frac{\sin 2\alpha - \sin 2\beta}{2}\right]} \\
&= 230\sqrt{\frac{1}{\pi}\left[2.1880 + \frac{0 - 0.9440}{2}\right]} = 230\sqrt{\frac{1.7160}{\pi}} \\
&= 230 \times 0.7391 = 169.99\ \text{V}
\end{aligned}
$$

(Here $\beta - \alpha = 125.37° = 2.1880$ rad, $\sin 2\alpha = \sin 180° = 0$ and $\sin 2\beta = \sin 430.73° = 0.9440$.)

#### RMS load current

$$
I_o = \left[\frac{1}{\pi}\int_{\alpha}^{\beta} i^2(\omega t)\, d(\omega t)\right]^{1/2}
$$

Evaluating with $V_m/Z = 325.27/25.431 = 12.79$ A gives

$$
I_o = 5.575\ \text{A}
$$

Check by power balance: $P = I_o^2 R = 5.575^2 \times 20 = 621.6$ W, equal to the average of $v_o i_o$.

**Answer: $\beta = 215.37°$, $V_{o,rms} = 169.99$ V, $I_{o,rms} = 5.575$ A.**

## Check yourself

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

1. For R = 20 ohm and L = 50 mH at 50 Hz, what are XL and phi?
   A. XL = 15.708 ohm, phi = 38.15 degrees
   B. XL = 50 ohm, phi = 68 degrees
   C. XL = 5 ohm, phi = 14 degrees
   D. XL = 15.708 ohm, phi = 51.85 degrees
   Answer: A. XL = 2 pi f L = 15.708 ohm and phi = tan^-1(15.708/20) = 38.15 degrees. (Key point: Load parameters)

2. Why is conduction discontinuous here?
   A. Because alpha = 90 degrees is greater than phi = 38.15 degrees
   B. Because alpha is less than phi
   C. Because the load is purely resistive
   D. Because L is very large
   Answer: A. When alpha exceeds the load angle, the current reaches zero at beta before the other SCR is fired. (Key point: Conduction mode)

3. How is beta found, and what is its value?
   A. Setting i(beta) = 0 in the load current equation, solved numerically; beta = 215.37 degrees
   B. From beta = alpha + 180 degrees = 270 degrees
   C. From beta = phi + 90 degrees
   D. From beta = 180 degrees always
   Answer: A. The transcendental equation sin(beta - phi) = sin(alpha - phi) e^-(beta - alpha)/tan(phi) is solved by trial and error. (Key point: Extinction angle)

4. What are the rms load voltage and current?
   A. Vo = 230 V, Io = 9.05 A
   B. Vo = 169.99 V, Io = 5.575 A
   C. Vo = 155.56 V, Io = 7.78 A
   D. Vo = 169.99 V, Io = 12.79 A
   Answer: B. The rms voltage formula with beta gives 169.99 V, and the rms of the current expression gives 5.575 A (power check I^2 R = 621.6 W). (Key point: Results)

## 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
- [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 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, 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×
- [Three-phase AC-DC with three thyristors, inductive load](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athree-phase-three-thyristor-waveforms) — Power Electronics, chapter 3
- [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×

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