---
title: "Series capacitor to limit power angle to 30°"
subject: "Power System Analysis II"
chapter: 6
key: "semester_5/power_system_analysis_ii:series-capacitor-30-degree"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/series-capacitor-30-degree
site: BCT Engineering
---

# Series capacitor to limit power angle to 30°

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2073 Shrawan, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/series-capacitor-30-degree
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

For a synchronous generator connected to infinite bus as shown in figure below, compute the value of series capacitor so that the power angle at steady state could be limited to 30° for generator supplying power of 1 p.u. [Figure: generator 6.6 kV, 50 MVA, 0.3 p.u (Y grounded) - transformer 6.6/132 kV, 50 MVA, 0.1 p.u - line j0.3 - series capacitor −jXc - line j0.3 - 132 kV infinite bus]

## Short answer

With E = V = 1 pu, the total reactance for 1 pu power at 30° must be 0.5 pu. The uncompensated reactance is 1.0 pu, so a series capacitor of 0.5 pu (174.24 Ω, ~18.27 μF per phase at 50 Hz) is needed.

## Answer

**Assumptions:** the generator internal emf is not given, so $|E| = 1.0$ pu is assumed, with infinite bus voltage $|V| = 1.0$ pu. The base is 50 MVA, 132 kV on the line side.

**Data (pu on 50 MVA):** $X_g = 0.3$, $X_T = 0.1$, line sections $0.3 + 0.3$, series capacitor $-jX_C$.

#### Step 1: Required total reactance

$$
\begin{aligned}
P &= \frac{EV}{X}\sin\delta \\
1.0 &= \frac{1\times1}{X}\sin30^\circ \;\Rightarrow\; X = 0.5\ \text{pu}
\end{aligned}
$$

#### Step 2: Capacitor reactance

Uncompensated reactance: $0.3 + 0.1 + 0.3 + 0.3 = 1.0$ pu. So

$$
X_C = 1.0 - 0.5 = 0.5\ \text{pu}
$$

Without the capacitor, $\sin\delta = 1.0$, giving $\delta = 90^\circ$, which is right at the limit. The capacitor is therefore essential.

#### Step 3: Actual value

$$
\begin{aligned}
Z_{base} &= \frac{(132)^2}{50} = 348.48\ \Omega \\
X_C &= 0.5\times 348.48 = 174.24\ \Omega \\
C &= \frac{1}{2\pi f X_C} = \frac{1}{2\pi\times50\times174.24} = 18.27\ \mu\text{F}
\end{aligned}
$$

**Answer: $X_C = 0.5$ pu = 174.24 Ω per phase, i.e. $C \approx 18.27\ \mu$F per phase (50% compensation of the total reactance).**

## Check yourself

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

1. For P = 1 pu, E = V = 1 pu and delta = 30 degrees, what total reactance is needed?
   A. 0.5 pu
   B. 1.0 pu
   C. 0.25 pu
   D. 2.0 pu
   Answer: A. P = (EV/X) sin30 gives X = 0.5 / 1 = 0.5 pu. (Key point: Required total reactance)

2. The uncompensated reactance is 0.3 + 0.1 + 0.3 + 0.3 = 1.0 pu. What is Xc?
   A. 1.5 pu, since Xc = 1.0 + 0.5
   B. 0.5 pu, since Xc = 1.0 - 0.5
   C. 0.3 pu
   D. 1.0 pu
   Answer: B. The capacitor reactance subtracts from the series inductive reactance, so Xc = 1.0 - 0.5 = 0.5 pu. (Key point: Capacitor reactance)

3. How is Xc = 0.5 pu converted to ohms?
   A. Multiply by 6.6^2/50 ohm
   B. Divide by 348.48
   C. Multiply by Zbase = 132^2/50 = 348.48 ohm, giving 174.24 ohm
   D. Multiply by 132 ohm
   Answer: C. The capacitor is on the 132 kV line side, so the 132 kV, 50 MVA base is used; C = 1/(2 pi f Xc) = 18.27 uF. (Key point: Converting to ohms and farads)

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