---
title: "Critical clearing angle, Pmax 2, 0.5, 1.5"
subject: "Power System Analysis II"
chapter: 6
key: "semester_5/power_system_analysis_ii:critical-clearing-angle-pmax-two"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/critical-clearing-angle-pmax-two
site: BCT Engineering
---

# Critical clearing angle, Pmax 2, 0.5, 1.5

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2078 Kartik, 10 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-angle-pmax-two
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

A 50 Hz generator with inertia constant 5 MJ/MVA is delivering 1 pu power to an infinite bus. When a fault occurs, the maximum power transferable reduces to 0.5 pu. The maximum power transferable before the occurrence of fault was 2 pu. The maximum power after clearance of fault is 1.5 pu. Compute the critical fault clearing angle.

## Short answer

Pre-fault Pmax=2 pu gives delta0=30°, post-fault Pmax=1.5 pu gives delta_max=138.19°. Using equal area, delta_cr = 70.3°.

## Answer

**Data:** $P_m = 1$ pu, $P_{max1} = 2$ pu (pre-fault), $P_{max2} = 0.5$ pu (during fault), $P_{max3} = 1.5$ pu (post-fault), $H = 5$ MJ/MVA, $f = 50$ Hz. $H$ is not needed for the angle.

#### Step 1: Initial and maximum angles

$$
\begin{aligned}
\delta_0 &= \sin^{-1}\frac{P_m}{P_{max1}} = \sin^{-1}\frac{1}{2} = 30^\circ = 0.5236\ \text{rad} \\
\delta_{max} &= 180^\circ - \sin^{-1}\frac{P_m}{P_{max3}} = 180^\circ - 41.81^\circ = 138.19^\circ = 2.4119\ \text{rad}
\end{aligned}
$$

#### Step 2: Equal area condition

Accelerating area (from $\delta_0$ to $\delta_{cr}$ on curve 2) = decelerating area (from $\delta_{cr}$ to $\delta_{max}$ on curve 3):

$$
\int_{\delta_0}^{\delta_{cr}}(P_m - P_{max2}\sin\delta)\,d\delta = \int_{\delta_{cr}}^{\delta_{max}}(P_{max3}\sin\delta - P_m)\,d\delta
$$

This gives

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

#### Step 3: Substitute

$$
\begin{aligned}
\cos\delta_{cr} &= \frac{1(2.4119-0.5236) + 1.5\cos138.19^\circ - 0.5\cos30^\circ}{1.5-0.5} \\
&= \frac{1.8883 - 1.1180 - 0.4330}{1.0} \\
&= 0.3372 \\
\delta_{cr} &= 70.29^\circ
\end{aligned}
$$

**Answer: Critical clearing angle $\delta_{cr} \approx 70.3^\circ$.**

## Check yourself

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

1. With Pm = 1, Pmax1 = 2 and Pmax3 = 1.5, delta0 and delta_max are:
   A. 30 degrees and 138.19 degrees
   B. 30 degrees and 150 degrees
   C. 41.81 degrees and 138.19 degrees
   D. 60 degrees and 120 degrees
   Answer: A. delta0 = asin(1/2) = 30 degrees; delta_max = 180 - asin(1/1.5) = 138.19 degrees. (Key point: Initial and maximum angle)

2. In the equal area method, the accelerating area is taken:
   A. On curve 3 from d0 to dcr
   B. On curve 1 from d0 to dmax
   C. On curve 2 (fault) from d0 to dcr, and the decelerating area on curve 3 from dcr to dmax
   D. On curve 2 from dcr to dmax
   Answer: C. During the fault Pe follows Pmax2 sin(delta); after clearing it follows Pmax3 sin(delta). (Key point: Areas)

3. Is H needed to find the critical clearing angle?
   A. Yes, it sets the area
   B. Yes, it sets delta0
   C. Yes, it sets Pmax3
   D. No, the angle depends only on the power-angle curves; H is only needed for time
   Answer: D. The equal area equation has no H; H converts the angle into the critical clearing time. (Key point: Role of H)

## Related in Power System Analysis II

- [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
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- [Critical clearing angle, Pmax 1.5, 0.5, 1.2](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-angle-reactive-line) — Power System Analysis II, chapter 6
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- [Critical clearing angle, Pm = 1.2 pu](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-line2-fault-pm1-2) — Power System Analysis II, chapter 6

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