---
title: "Synchronizing power coefficient and application"
subject: "Power System Analysis II"
chapter: 6
key: "semester_5/power_system_analysis_ii:synchronizing-power-coefficient"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/synchronizing-power-coefficient
site: BCT Engineering
---

# Synchronizing power coefficient and application

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2076 Chaitra, 6 marks
- Video: [SM32 Synchronizing power Coefficient](https://www.youtube.com/watch?v=ERoR6UzgRz4) — Lectures in Electrical Engineering
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/synchronizing-power-coefficient
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

What is synchronizing power coefficient? Where it finds application?

## Short answer

Synchronizing power coefficient P_s is the derivative of electrical power with respect to rotor angle at the operating point: P_s = (EV/X) cos delta = P_max cos delta. It represents the restoring power per unit angle change. Applications: steady state stability check (requires P_s > 0), stability margin, natural frequency of rotor oscillation, synchronising of parallel alternators, and hunting analysis.

## Answer

**Synchronizing power coefficient** $P_s$ is the rate of change of electrical power output with rotor angle at the operating point:
$$
P_s = \left.\frac{\partial P_e}{\partial\delta}\right|_{\delta_0} = \frac{EV}{X}\cos\delta_0 = P_{max}\cos\delta_0\ \ \text{(pu power/rad)}
$$
It gives the change in power that tends to pull the rotor back to synchronism for a small change in rotor angle: $\Delta P_e = P_s\,\Delta\delta$. It is also called the stiffness of the coupling between the machine and the system.

#### Derivation
For a machine connected to an infinite bus, $P_e = P_{max}\sin\delta$. If the rotor angle increases from $\delta_0$ by $\Delta\delta$:
$$
\Delta P_e = P_{max}\sin(\delta_0 + \Delta\delta) - P_{max}\sin\delta_0 \approx (P_{max}\cos\delta_0)\Delta\delta = P_s\Delta\delta
$$
With $P_m$ constant, an increase of $\Delta\delta$ makes $P_e > P_m$ when $P_s > 0$; the rotor decelerates and returns to $\delta_0$. So $P_s$ acts like the spring constant of the electrical coupling.

```text
 Ps = Pmax cos(d0)
  ^
 Pmax.
  |    '.
  |      '.
  +--------'.------> d0
  0          90 deg (Ps = 0, limit)
```

#### Applications
1. **Steady state stability check:** the operating point is stable only if $P_s > 0$, i.e. $\delta_0 < 90^\circ$; the limit is $P_s = 0$.
2. **Measure of stability margin/stiffness:** larger $P_s$ (lower loading, lower reactance, higher excitation) means a more stable machine.
3. **Natural frequency of rotor oscillation:** from the linearised swing equation $M\frac{d^2\Delta\delta}{dt^2} + P_s\Delta\delta = 0$,
$$
f_n = \frac{1}{2\pi}\sqrt{\frac{P_s}{M}}, \quad M = \frac{H}{\pi f}
$$
used for small-signal (dynamic) stability studies and design of damper windings and power system stabilisers.
4. **Parallel operation and synchronising of alternators:** $P_s$ gives the synchronising power that keeps paralleled machines in step and shares load changes.
5. **Hunting analysis:** used to find the period of oscillation of a machine after a small disturbance.

**Example:** $P_{max} = 2$ pu, $P_m = 1$ pu → $\delta_0 = 30^\circ$, $P_s = 2\cos30^\circ = 1.732$ pu/rad.

## Check yourself

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

1. What does the synchronizing power coefficient Ps represent?
   A. dPe/ddelta at the operating point, the restoring power per radian of angle change
   B. The mechanical power Pm
   C. The inertia constant
   D. The total power at 90°
   Answer: A. Ps = Pmax cos(delta_0) gives delta Pe = Ps delta(delta), the restoring power. (Key point: Definition)

2. With Pmax = 2 pu and Pm = 1 pu, what is Ps?
   A. 2 pu/rad
   B. 1 pu/rad
   C. 0.866 pu/rad
   D. 1.732 pu/rad
   Answer: D. delta_0 = sin^-1(0.5) = 30°, so Ps = 2 cos30° = 1.732 pu/rad. (Key point: Numerical example)

3. The natural frequency of rotor oscillation is:
   A. f_n = (1/2pi) sqrt(M/Ps)
   B. f_n = (1/2pi) sqrt(Ps/M), M = H/(pi f)
   C. f_n = Ps x M
   D. f_n = 2 pi sqrt(Ps/M)
   Answer: B. From M d2(delta)/dt2 + Ps delta = 0, omega_n = sqrt(Ps/M). (Key point: Natural frequency)

4. At what operating angle does Ps become zero, marking the stability limit?
   A. 0°
   B. 180°
   C. 45°
   D. 90°
   Answer: D. cos(90°) = 0, so Ps = 0 at the steady state limit. (Key point: Stability condition)

## Related in Power System Analysis II

- [Positive synchronizing power coefficient](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Asynchronizing-coefficient-positive) — Power System Analysis II, chapter 6
- [Steady state stability and synchronizing coefficient](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Asteady-state-stability-sync-coefficient) — Power System Analysis II, chapter 6