---
title: "Power angle depends on real power"
subject: "Power System Analysis II"
chapter: 1
key: "semester_5/power_system_analysis_ii:power-angle-real-power"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/power-angle-real-power
site: BCT Engineering
---

# Power angle depends on real power

Power System Analysis II · Chapter 1: Interconnected Power System

- Asked in: 2083 Baishakh (new course), 5 marks
- Video: [Power Angle Curve ](https://www.youtube.com/watch?v=l4iCuwIjYlU) — Chandan Bobby
- 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-real-power
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-1

## Question

Using appropriate mathematical derivations, [show that] the phase angle (power/loading angle) δ between Ef and Vt is due solely to the real power delivered by the synchronous generator.

## Short answer

From the complex power expression, P = (Ef Vt / Xs) sin δ and Q = (Ef Vt cos δ – Vt^2)/Xs. The angle δ arises from the quadrature component of Ef, which is proportional to P/Xs; at zero real power, δ = 0 regardless of reactive load, showing δ is due solely to real power.

## Answer

Consider a cylindrical-rotor synchronous generator with internal emf $E_f\angle\delta$, terminal voltage $V_t\angle 0^\circ$, synchronous reactance $X_s$ and negligible armature resistance.

#### Derivation
Armature current:

$$
I_a = \frac{E_f\angle\delta - V_t\angle 0}{jX_s}
$$

Complex power delivered at the terminal:

$$
\begin{aligned}
S &= V_tI_a^* = V_t\left[\frac{E_f\angle -\delta - V_t}{-jX_s}\right]
= \frac{jV_t\left(E_f\cos\delta - jE_f\sin\delta - V_t\right)}{X_s} \\
&= \frac{V_tE_f\sin\delta}{X_s} + j\,\frac{V_tE_f\cos\delta - V_t^2}{X_s}
\end{aligned}
$$

So

$$
P = \frac{E_fV_t}{X_s}\sin\delta,\qquad
Q = \frac{V_t(E_f\cos\delta - V_t)}{X_s}
$$

#### Showing δ is due to real power alone
Write $I_a = I_a\cos\phi - jI_a\sin\phi$ (lagging). Then

$$
E_f = V_t + jX_sI_a = (V_t + X_sI_a\sin\phi) + j\,X_sI_a\cos\phi
$$

$$
\tan\delta = \frac{X_sI_a\cos\phi}{V_t + X_sI_a\sin\phi}
$$

- The **imaginary (quadrature) part** of $E_f$, which creates the angle $\delta$, is $X_sI_a\cos\phi = X_sP/V_t$, i.e. proportional only to **real power**.
- The reactive current $I_a\sin\phi$ adds only to the **in-phase part** of $E_f$, changing its magnitude, not producing any angle by itself.

Two limiting cases make this clear:
1. **P = 0** (pure reactive load, $\cos\phi = 0$): $E_f\sin\delta = 0$, so $\delta = 0$ however large Q is; $E_f$ and $V_t$ are in phase and differ only in magnitude.
2. **Q = 0** (unity pf): $E_f = V_t + jX_sI_a$, and $\delta = \tan^{-1}(X_sI_a/V_t)$, a non-zero angle set by P.

Hence the power angle $\delta$ between $E_f$ and $V_t$ exists only because real power is delivered ($\delta = 0 \Leftrightarrow P = 0$), while Q affects mainly the magnitude of $E_f$. This is the basis of P–δ (P–f) and Q–V decoupling.

## Check yourself

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

1. The part of Ef that creates delta is:
   A. Xs Ia cos(phi) = Xs P / Vt
   B. Xs Ia sin(phi)
   C. Vt
   D. Xs Q
   Answer: A. Quadrature component is proportional to real power. (Key point: Quadrature part)

2. The reactive current Ia sin(phi) affects Ef by:
   A. creating angle delta
   B. changing its magnitude only
   C. reducing P
   D. nothing at all
   Answer: B. It adds to the in-phase part. (Key point: Reactive current)

3. If P = 0 (pure reactive load):
   A. delta = 90 deg
   B. delta depends on Q
   C. delta = 0 and Ef, Vt differ only in magnitude
   D. Ef = 0
   Answer: C. Ef sin(delta) = 0. (Key point: P = 0 case)

4. The real power delivered is:
   A. P = Ef Vt cos(delta)/Xs
   B. P = (Ef cos(delta) - Vt)Vt/Xs
   C. P = Vt^2/Xs
   D. P = Ef Vt sin(delta) / Xs
   Answer: D. Real part of S = Vt Ia*. (Key point: Power equation)

## Related in Power System Analysis II

- [Series capacitor to limit power angle to 30°](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aseries-capacitor-30-degree) — Power System Analysis II, chapter 6
- [Maximum steady state power with Vt held](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Amax-steady-state-power-terminal-voltage) — Power System Analysis II, chapter 6
- [EAC from given power-angle curves](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Apower-angle-curves-eac) — Power System Analysis II, chapter 6
- [Interconnection advantages and real power-frequency balance](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ainterconnection-advantages-frequency) — Power System Analysis II, chapter 1, asked 3×
- [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
- [Real power/frequency control in interconnected system](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Areal-power-frequency-control) — Power System Analysis II, chapter 1

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