---
title: "Critical clearing angle, Pm = 1.2 pu"
subject: "Power System Analysis II"
chapter: 6
key: "semester_5/power_system_analysis_ii:critical-clearing-line2-fault-pm1-2"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/critical-clearing-line2-fault-pm1-2
site: BCT Engineering
---

# Critical clearing angle, Pm = 1.2 pu

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2079 Bhadra, 8 marks
- Video: [POWER SYSTEM - Critical clearing angle (using equal area criterion)](https://www.youtube.com/watch?v=-bJ_X-5Lyfk) — Tulsiram shinde
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/critical-clearing-line2-fault-pm1-2
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

For the system shown in figure, the per unit values of different quantities are E = 1.1, V = 1, Xd' = 0.15, X1 = X2 = 0.4. The system is operating in equilibrium with Pm = 1.2 pu. Find the critical clearing angle if a 3-phase fault occurs on line-2 close to the generator. [Figure: generator E∠δ behind Xd' feeding two parallel lines X1 and X2 (each with circuit breakers at both ends) to an infinite bus V∠0; fault F at the generator end of line 2]

## Short answer

Pre-fault Pmax=3.143 pu, delta0=22.45°. During fault Pe=0. Post-fault Pmax=2 pu, delta_max=143.13°. Equal area gives delta_cr = 62.37°.

## Answer

**Data:** $E = 1.1$, $V = 1$, $X_d' = 0.15$, $X_1 = X_2 = 0.4$ pu, $P_m = 1.2$ pu. The fault is on line 2 close to the generator and is cleared by opening line 2.

#### Step 1: Pre-fault (both lines in service)

$$
\begin{aligned}
X_I &= 0.15 + \frac{0.4}{2} = 0.35\ \text{pu} \\
P_{max1} &= \frac{1.1\times 1}{0.35} = 3.143\ \text{pu} \\
\delta_0 &= \sin^{-1}\frac{1.2}{3.143} = 22.45^\circ = 0.3918\ \text{rad}
\end{aligned}
$$

#### Step 2: During the fault

The fault is at the generator end of line 2, which is the generator-side bus. That bus is at zero voltage, so no power reaches the infinite bus: $P_{max2} = 0$.

#### Step 3: Post-fault (line 2 open)

$$
\begin{aligned}
X_{III} &= 0.15 + 0.4 = 0.55\ \text{pu} \\
P_{max3} &= \frac{1.1}{0.55} = 2.0\ \text{pu} \\
\delta_{max} &= 180^\circ - \sin^{-1}\frac{1.2}{2.0} = 180^\circ - 36.87^\circ = 143.13^\circ = 2.4981\ \text{rad}
\end{aligned}
$$

#### Step 4: Critical clearing angle

From $A_1 = A_2$:

$$
\cos\delta_{cr} = \frac{P_m(\delta_{max}-\delta_0) + P_{max3}\cos\delta_{max} - P_{max2}\cos\delta_0}{P_{max3}-P_{max2}}
$$

$$
\begin{aligned}
\cos\delta_{cr} &= \frac{1.2(2.4981-0.3918) + 2.0\cos 143.13^\circ - 0}{2.0 - 0} \\
&= \frac{2.5276 - 1.6000}{2.0} = 0.4638 \\
\delta_{cr} &= 62.37^\circ
\end{aligned}
$$

**Answer: Critical clearing angle $\delta_{cr} \approx 62.37^\circ$.**

## Check yourself

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

1. With Xd' = 0.15 and two 0.4 pu lines in parallel, the pre-fault Pmax is (E = 1.1, V = 1):
   A. 2.0 pu
   B. 1.1/0.55 pu
   C. 3.143 pu
   D. 2.75 pu
   Answer: C. X = 0.15 + 0.2 = 0.35, so Pmax1 = 1.1/0.35 = 3.143 pu. (Key point: Pre-fault Pmax)

2. Why is Pmax2 = 0 for a fault on line 2 close to the generator?
   A. Line 2 is open
   B. Pm is zero
   C. The infinite bus is shorted
   D. The generator-side bus is at zero voltage, so no power gets to the infinite bus
   Answer: D. A three-phase fault at the generator end puts that bus at zero voltage and blocks power transfer. (Key point: During-fault power)

3. After line 2 is cleared, Pmax3 is:
   A. 2.0 pu, from X = 0.15 + 0.4 = 0.55
   B. 3.143 pu
   C. 1.1 pu
   D. 0 pu
   Answer: A. Only line 1 remains, so X = 0.55 and Pmax3 = 1.1/0.55 = 2.0 pu, giving delta_max = 143.13 degrees. (Key point: Post-fault network)

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