---
title: "Power angle and series compensation"
subject: "Power System Analysis II"
chapter: 6
key: "semester_5/power_system_analysis_ii:power-angle-series-compensation"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/power-angle-series-compensation
site: BCT Engineering
---

# Power angle and series compensation

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2069 Chaitra, 5+5 marks
- Video: [Series Compensation - Part 1](https://www.youtube.com/watch?v=B7s9ms6BPNE) — Dr Nimal Madhu
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/power-angle-series-compensation
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

A transmission line connecting a generator to an infinite bus has a series reactance of 0.8 p.u. Assuming the sending and receiving end voltage are at 1 p.u. each compute the following: i. Power angle if the line delivers a 1 p.u. power to infinite bus ii. Series compensation required to bring the power angle to 30°.

## Short answer

The power angle is 53.13°. To reduce it to 30°, a series capacitor of XC = 0.3 pu (37.5% compensation) is required.

## Answer

For a lossless line, $P = \dfrac{V_SV_R}{X}\sin\delta$. Series capacitors reduce the net reactance $X$. For the same power, this reduces $\delta$ and raises the stability margin.

**Data:** $X_L = 0.8$ pu, $V_S = V_R = 1$ pu, $P = 1$ pu.

#### (i) Power angle without compensation

$$
\begin{aligned}
\sin\delta &= \frac{PX_L}{V_SV_R} = \frac{1\times0.8}{1\times1} = 0.8 \\
\delta &= \sin^{-1}0.8 = 53.13^\circ
\end{aligned}
$$

#### (ii) Series compensation for $\delta = 30^\circ$

The net reactance needed is

$$
X_{net} = \frac{V_SV_R\sin30^\circ}{P} = \frac{1\times1\times0.5}{1} = 0.5\ \text{pu}
$$

So the series capacitor must have

$$
X_C = X_L - X_{net} = 0.8 - 0.5 = 0.3\ \text{pu}
$$

$$
\text{Degree of compensation} = \frac{X_C}{X_L}\times100 = \frac{0.3}{0.8}\times100 = 37.5\%
$$

#### Effect on stability

| Quantity | Uncompensated | Compensated |
|---|---|---|
| Net $X$ (pu) | 0.8 | 0.5 |
| $P_{max}$ (pu) | 1.25 | 2.0 |
| $\delta$ for 1 pu | 53.13° | 30° |
| Steady-state margin | 20% | 50% |

**Answer: (i) $\delta = 53.13^\circ$; (ii) series capacitive reactance $X_C = 0.3$ pu, i.e. 37.5% series compensation.**

## Check yourself

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

1. For P = 1 pu and X = 0.8 pu, how is delta found?
   A. cos(delta) = 0.8
   B. delta = sin^-1(1/0.8)
   C. delta = 30 degrees
   D. sin(delta) = P X/(Vs Vr) = 0.8, so delta = 53.13 degrees
   Answer: D. P = (VsVr/X) sin(delta), so sin(delta) = PX/(VsVr). (Key point: Power angle formula)

2. What net reactance is needed for delta = 30 degrees?
   A. 0.5 pu
   B. 0.8 pu
   C. 0.3 pu
   D. 1.6 pu
   Answer: A. X = VsVr sin30/P = 0.5 pu. (Key point: Net reactance for 30 degrees)

3. What is the series capacitor reactance and degree of compensation?
   A. Xc = 0.5 pu, 62.5%
   B. Xc = 0.3 pu, 37.5%
   C. Xc = 0.3 pu, 62.5%
   D. Xc = 1.3 pu, 37.5%
   Answer: B. Xc = 0.8 - 0.5 = 0.3 pu; 0.3/0.8 = 37.5%. (Key point: Capacitor and compensation)

## Related in Power System Analysis II

- [Series capacitor to limit power angle to 30°](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aseries-capacitor-30-degree) — Power System Analysis II, chapter 6
- [Critical clearing for temporary sending-end fault](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-temporary-fault-new) — Power System Analysis II, chapter 6
- [Critical clearing angle/time, fault at infinite bus](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-infinite-bus-h5) — Power System Analysis II, chapter 6
- [Rotor angle and swing equation, infinite bus](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Arotor-angle-swing-equation) — Power System Analysis II, chapter 6
- [Fault at receiving end of 132 kV lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-132kv-receiving-end-2071) — Power System Analysis II, chapter 3
- [Critical clearing angle, breakers 1 and 2](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-angle-breakers-1-2) — Power System Analysis II, chapter 6, asked 2×

## Related in other subjects

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