---
title: "Rotor angle and swing equation, infinite bus"
subject: "Power System Analysis II"
chapter: 6
key: "semester_5/power_system_analysis_ii:rotor-angle-swing-equation"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/rotor-angle-swing-equation
site: BCT Engineering
---

# Rotor angle and swing equation, infinite bus

Power System Analysis II · Chapter 6: Power System Stability

- Asked in: 2073 Shrawan, 6 marks
- Video: [Swing Equation derivation | Power system Stability | Power system Analysis](https://www.youtube.com/watch?v=ZpfAQ3G_8DE) — 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-swing-equation
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-6

## Question

What do you mean by rotor angle? Derive the swing equation for a single synchronous generator connected to infinite bus.

## Short answer

Rotor angle δ is the angular displacement between internal emf E and terminal or infinite-bus voltage V, with transferred power Pe = (EV/X) sinδ. For a generator on an infinite bus, the swing equation is (2H/ωs)d2δ/dt2 = Pm - Pe, with Pe = (EV/X) sinδ, derived from Newton's law and inertia constant H.

## Answer

#### Rotor angle
The **rotor angle** (power angle or load angle) $\delta$ is the angular displacement between the rotor axis (direction of the internal emf $E$) and a synchronously rotating reference axis (direction of the terminal or infinite-bus voltage $V$). In phasor terms it is the angle between $E$ and $V$:
```text
        E
       /
      /  delta
     /_________ V (reference)
```
For a generator connected to an infinite bus through reactance $X$, the power transferred depends on it: $P_e = \dfrac{EV}{X}\sin\delta$. Under steady operation $\delta$ is constant; after a disturbance the rotor speeds up or slows down and $\delta$ changes ("swings").

#### Swing equation for a generator connected to an infinite bus
```text
 Turbine -> [G] E<d --- jX --- V<0 (infinite bus)
   Pm          Pe
```
#### Derivation
Consider the rotor of a synchronous generator. Let
- $J$ = moment of inertia of rotor and turbine (kg·m²)
- $\theta_m$ = angular position of the rotor with respect to a stationary axis (mech. rad)
- $T_m$ = mechanical (shaft) torque, $T_e$ = electromagnetic torque (N·m)

**Step 1: Newton's law of rotation.** Accelerating torque = moment of inertia × angular acceleration:
$$
J\frac{d^2\theta_m}{dt^2} = T_a = T_m - T_e
$$

**Step 2: Measure angle from a synchronously rotating frame.** The rotor angle $\delta_m$ is the angle of the rotor relative to a reference axis rotating at synchronous speed $\omega_{sm}$:
$$
\theta_m = \omega_{sm}t + \delta_m \;\Rightarrow\; \frac{d\theta_m}{dt} = \omega_{sm} + \frac{d\delta_m}{dt}, \quad \frac{d^2\theta_m}{dt^2} = \frac{d^2\delta_m}{dt^2}
$$
so
$$
J\frac{d^2\delta_m}{dt^2} = T_m - T_e
$$

**Step 3: Convert torque to power.** Multiply by the rotor speed $\omega_m$ (power = torque × speed):
$$
J\omega_m\frac{d^2\delta_m}{dt^2} = P_m - P_e
$$
$M = J\omega_m$ is the **angular momentum** (inertia constant). Since the speed changes very little during a swing, $\omega_m \approx \omega_{sm}$ and $M$ is taken as constant:
$$
M\frac{d^2\delta_m}{dt^2} = P_m - P_e
$$

**Step 4: Introduce the inertia constant $H$.** $H$ is the stored kinetic energy at synchronous speed per MVA rating $S$:
$$
H = \frac{\tfrac12 J\omega_{sm}^2}{S}\ \ \text{(MJ/MVA or s)} \;\Rightarrow\; M = J\omega_{sm} = \frac{2HS}{\omega_{sm}}
$$
Substituting and dividing by $S$:
$$
\frac{2H}{\omega_{sm}}\frac{d^2\delta_m}{dt^2} = \frac{P_m - P_e}{S} = P_m - P_e\ \ \text{(pu)}
$$

**Step 5: Use electrical angle.** For a machine with $p$ poles, $\delta = \frac{p}{2}\delta_m$ and $\omega_s = \frac{p}{2}\omega_{sm}$, so $\delta/\omega_s = \delta_m/\omega_{sm}$:
$$
\boxed{\frac{2H}{\omega_s}\frac{d^2\delta}{dt^2} = P_m - P_e} \qquad \text{or} \qquad \frac{H}{\pi f}\frac{d^2\delta}{dt^2} = P_m - P_e
$$
with $\delta$ in electrical radians and $\omega_s = 2\pi f$. If $\delta$ is in electrical degrees, $\dfrac{H}{180f}\dfrac{d^2\delta}{dt^2} = P_m - P_e$.

For a machine connected to an infinite bus through reactance $X$, $P_e = \dfrac{EV}{X}\sin\delta = P_{max}\sin\delta$, so
$$
\frac{H}{\pi f}\frac{d^2\delta}{dt^2} = P_m - P_{max}\sin\delta
$$
This is the **swing equation**. It is a second-order non-linear differential equation; its solution $\delta(t)$ describes how the rotor swings after a disturbance. If damping is included, a term $D\frac{d\delta}{dt}$ is added on the left.

## Check yourself

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

1. The rotor angle delta of a generator on an infinite bus is the angle between:
   A. Internal emf E and bus voltage V
   B. Terminal voltage and current
   C. Stator and rotor currents
   D. Field flux and armature current
   Answer: A. Delta is the angular displacement between E (rotor axis) and the synchronously rotating reference V. (Key point: Rotor angle definition)

2. In the derivation, why can M = J*w_m be taken as constant?
   A. Because J changes with load
   B. Because the speed changes very little during a swing, so w_m is about w_sm
   C. Because the damping is zero
   D. Because torque equals power
   Answer: B. Speed deviations are small during a swing, so w_m is approximately w_sm and M is constant. (Key point: Why speed treated as constant)

3. For a generator connected to an infinite bus through X, the swing equation becomes:
   A. (H/pi f) d2delta/dt2 = Pmax - Pm
   B. (H/pi f) d2delta/dt2 = Pm - Pmax sin(delta), with Pmax = EV/X
   C. (H/pi f) d delta/dt = Pm - Pmax cos(delta)
   D. H d2delta/dt2 = Pm + Pmax sin(delta)
   Answer: B. Accelerating power is Pm - Pe, with Pe = (EV/X) sin(delta). (Key point: Final swing equation with infinite bus)

## Related in Power System Analysis II

- [Swing equation of synchronous machine](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aswing-equation-derivation) — Power System Analysis II, chapter 6
- [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
- [Rotor dynamics for incremental angle change](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aincremental-rotor-dynamics) — Power System Analysis II, chapter 6
- [Initial rotor angle and critical clearing angle](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Arotor-angle-critical-clearing-2076) — Power System Analysis II, chapter 6
- [Line opened: stability and rotor swing](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Abreaker-a-opens-rotor-swing) — Power System Analysis II, chapter 6
- [Stability definitions and swing equation from dynamics](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astability-define-swing-from-dynamics) — Power System Analysis II, chapter 6

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