---
title: "Initial rotor angle and critical clearing angle"
subject: "Power System Analysis II"
chapter: 6
key: "semester_5/power_system_analysis_ii:rotor-angle-critical-clearing-2076"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/rotor-angle-critical-clearing-2076
site: BCT Engineering
---

# Initial rotor angle and critical clearing angle

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2076 Chaitra, 10 marks
- Video: [Power system Stability analysis | Solved Problem -1  | Critical Clearing Angle | PSA](https://www.youtube.com/watch?v=ne0so8Xe52k) — Simple n Easy Learning
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/rotor-angle-critical-clearing-2076
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

In the power system network shown below, find the rotor angle before occurrence of fault at bus 3 when the rotor was running at synchronous speed when the mechanical input to the generator is 1 p.u. Also compute the critical clearing angle. [Figure: generator with terminal voltage |Vt| = 1.08 pu and Xd' = 0.2 pu - transformer j0.1 - bus 1; two parallel lines j0.5 and j0.4 from bus 1 to bus 3; bus 3 connected directly to bus 2, the infinite bus 1∠0° pu]

## Short answer

Pre-fault rotor angle delta0 = 26.45° (E' = 1.173 pu). Critical clearing angle delta_cr = 84.69°, obtained from equal area with zero power during fault.

## Answer

**Data:** $|V_t| = 1.08$ pu, $X_d' = 0.2$, $X_T = 0.1$, lines $j0.5 \parallel j0.4$ between bus 1 and bus 3, bus 3 tied directly to the infinite bus $1\angle0^\circ$, $P_m = 1$ pu.

**Assumption:** the fault at bus 3 is a temporary fault. After it is cleared, the network is unchanged.

#### Step 1: Reactances

$$
\begin{aligned}
X_{lines} &= \frac{0.5\times 0.4}{0.5+0.4} = 0.2222\ \text{pu} \\
X_{t\text{-}\infty} &= 0.1 + 0.2222 = 0.3222\ \text{pu}\ \ \text{(terminal to infinite bus)}
\end{aligned}
$$

#### Step 2: Terminal voltage angle

$$
\begin{aligned}
P &= \frac{|V_t||V|}{X}\sin\theta \;\Rightarrow\; \sin\theta = \frac{1\times 0.3222}{1.08\times 1} = 0.2984 \\
\theta &= 17.36^\circ, \quad V_t = 1.08\angle17.36^\circ = 1.0308 + j0.3222
\end{aligned}
$$

#### Step 3: Current and internal emf

$$
\begin{aligned}
I &= \frac{V_t - V}{jX} = \frac{0.0308 + j0.3222}{j0.3222} = 1.0 - j0.0956 = 1.0044\angle{-5.46^\circ}\ \text{pu} \\
E' &= V_t + jX_d' I = (1.0308 + j0.3222) + j0.2(1.0 - j0.0956) \\
&= 1.0499 + j0.5222 = 1.1726\angle 26.45^\circ\ \text{pu}
\end{aligned}
$$

**Rotor angle before the fault: $\delta_0 = 26.45^\circ$.**

#### Step 4: Power-angle curves

$$
\begin{aligned}
X_{total} &= 0.2 + 0.3222 = 0.5222\ \text{pu} \\
P_{max} &= \frac{1.1726\times 1}{0.5222} = 2.2455\ \text{pu}
\end{aligned}
$$

Check: $2.2455\sin 26.45^\circ = 1.0$ pu.

During the fault, bus 3 is shorted, and it is tied directly to the infinite bus, so $P_e = 0$. After clearing, $P_{max3} = P_{max1} = 2.2455$ pu. Then $\delta_{max} = 180^\circ - 26.45^\circ = 153.55^\circ$.

#### Step 5: Critical clearing angle

$$
\begin{aligned}
\cos\delta_{cr} &= (\pi - 2\delta_0)\sin\delta_0 - \cos\delta_0 \\
&= (3.1416 - 0.9231)(0.4453) - 0.8954 \\
&= 0.0926 \\
\delta_{cr} &= 84.69^\circ
\end{aligned}
$$

**Answer: Initial rotor angle $\delta_0 = 26.45^\circ$ ($E' = 1.173$ pu); critical clearing angle $\delta_{cr} \approx 84.69^\circ$.**

## Check yourself

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

1. The terminal-to-infinite-bus reactance is (lines 0.5 || 0.4, transformer 0.1):
   A. 0.9 pu
   B. 0.2222 pu
   C. 0.5222 pu
   D. 0.3222 pu
   Answer: D. 0.5||0.4 = 0.2222 and adding the transformer's 0.1 gives 0.3222 pu. (Key point: Equivalent reactance)

2. How is the rotor angle delta0 found?
   A. Compute theta from P = VtV sin(theta)/X, then current I, then E' = Vt + jXd' I; delta0 is the angle of E'
   B. Take delta0 = theta directly
   C. Take delta0 = asin(Pm)
   D. Take delta0 = 90 degrees minus theta
   Answer: A. Terminal voltage angle 17.36 degrees gives I, then E' = 1.1726 at 26.45 degrees. (Key point: Finding delta0)

3. Why is Pe = 0 during the fault at bus 3?
   A. Bus 3 is shorted and tied directly to the infinite bus
   B. The lines are open
   C. Pm is zero
   D. E' is zero
   Answer: A. Bus 3 is directly connected to the infinite bus, so a fault there leaves no voltage to transfer power. (Key point: Fault at bus 3)

## Related in Power System Analysis II

- [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
- [Critical clearing time and angle, fault at P](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-time-point-p-h5) — Power System Analysis II, chapter 6
- [Critical clearing, fault at infinite bus, H = 6](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-fault-infinite-bus-h6) — Power System Analysis II, chapter 6
- [Critical clearing angle, Pmax 2, 0.5, 1.5](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-angle-pmax-two) — Power System Analysis II, chapter 6
- [Critical clearing angle and time, H = 2](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-h2-parallel-lines) — Power System Analysis II, chapter 6
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