---
title: "EAC from given power-angle curves"
subject: "Power System Analysis II"
chapter: 6
key: "semester_5/power_system_analysis_ii:power-angle-curves-eac"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/power-angle-curves-eac
site: BCT Engineering
---

# EAC from given power-angle curves

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2072 Chaitra, 12 marks
- Video: [Power System -ll || equal area criterion of stability part -1 ||](https://www.youtube.com/watch?v=fJpPnOJLIds) — Technical Subham Edu.
- 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-curves-eac
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

A generator is operating at rated voltage and is connected to infinite bus at 50 Hz with power transfer of 1.0 pu. A three-phase fault occurs and it is cleared after some time. Power angle (Pe-δ) curves for the pre-fault condition, during fault and after the fault is cleared is shown in figure below. Use equal area criterion to check whether the system is stable or not. If the system is stable, determine critical clearing angle. [Figure: Pe-δ curves: Curve-I (pre-fault) with peak 1.8 pu; Curve-III (post-fault) with peak 1.3 pu; Curve-II (during fault) with peak 0.4 pu; Pm = 1.0 pu]

## Short answer

The system is stable if cleared before the critical clearing angle. Using equal area criterion, δcr = 55.35°.

## Answer

**Data from the curves:** $P_m = 1.0$ pu, pre-fault $P_{max1} = 1.8$ pu, during fault $P_{max2} = 0.4$ pu, post-fault $P_{max3} = 1.3$ pu.

#### Step 1: Can the system be stable?

$$
\begin{aligned}
\delta_0 &= \sin^{-1}\frac{1.0}{1.8} = 33.75^\circ = 0.5890\ \text{rad} \\
\delta_{max} &= 180^\circ - \sin^{-1}\frac{1.0}{1.3} = 180^\circ - 50.28^\circ = 129.72^\circ = 2.2640\ \text{rad}
\end{aligned}
$$

During the fault, $P_{max2} = 0.4 < P_m$, so the rotor accelerates. After clearing, $P_{max3} = 1.3 > P_m$, so a decelerating area exists. The system is **stable if the fault is cleared before the critical clearing angle**, where $A_1 = A_2$.

#### Step 2: Critical clearing angle

Equating the accelerating area $A_1$ (on curve II, from $\delta_0$ to $\delta_{cr}$) with the decelerating area $A_2$ (on curve III, from $\delta_{cr}$ to $\delta_{max}$):

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

$$
\begin{aligned}
\cos\delta_{cr} &= \frac{1.0(2.2640-0.5890) + 1.3\cos129.72^\circ - 0.4\cos33.75^\circ}{1.3-0.4} \\
&= \frac{1.6750 - 0.8307 - 0.3326}{0.9} \\
&= 0.5685 \\
\delta_{cr} &= 55.35^\circ
\end{aligned}
$$

```text
 Pe
 1.8 |      ____ I
 1.3 |    /  __ \__ III
 1.0 |---*--|----\---\---- Pm
 0.4 |  / A1| A2  \   \ II
     +--d0--dcr---dmax---- delta
       33.75 55.35 129.72
```

**Answer: The system is stable if the fault is cleared at an angle below the critical clearing angle $\delta_{cr} \approx 55.35^\circ$; if clearing is later, it loses synchronism.**

## Check yourself

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

1. Why does the rotor accelerate during the fault?
   A. Pmax2 is above Pm
   B. Pe is greater than Pm
   C. Because curve III is lower than curve I
   D. Pmax2 = 0.4 pu is below Pm = 1.0, so Pm exceeds Pe
   Answer: D. On the fault curve Pe is below the input Pm, so there is net accelerating power. (Key point: Fault accelerates the rotor)

2. Why can the system be stable after clearing?
   A. Pmax3 = 1.3 pu exceeds Pm, so Pe > Pm gives a decelerating area
   B. Pmax3 is less than Pm
   C. Because the fault is cleared at delta_0
   D. Because Pmax1 is the biggest
   Answer: A. Post-fault curve crosses above Pm, so decelerating power exists between the crossing points. (Key point: Post-fault decelerating area)

3. Which curve does delta_max come from?
   A. Pre-fault curve I
   B. Post-fault curve III: 180 degrees - sin^-1(1.0/1.3) = 129.72 degrees
   C. Fault curve II
   D. It is fixed at 90 degrees
   Answer: B. The decelerating area lies on the post-fault curve, so its second crossing with Pm sets delta_max. (Key point: Which angle uses which curve)

4. What does delta_cr = 55.35 degrees mean?
   A. It is stable only if cleared after this angle
   B. It is the post-fault equilibrium
   C. The system is stable if the fault is cleared before this angle, unstable if later
   D. It is the angle at which the fault starts
   Answer: C. At delta_cr, A1 = A2; later clearing gives A1 > A2,max and synchronism is lost. (Key point: Meaning of critical clearing angle)

## 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
- [Fault cleared at 30°, Pe = 2.1 sin δ](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-cleared-thirty-degrees) — 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 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
- [Fault cleared at 45°: stability and maximum swing](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-cleared-45-max-swing) — Power System Analysis II, chapter 6
- [Mid-line fault, cleared at 60°](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Amid-line-fault-cleared-60) — Power System Analysis II, chapter 6

## Related in other subjects

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- [Generator on infinite bus with 15% excitation rise](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Agenerator-infinite-bus-excitation-15) — Electric Machines II, chapter 1
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