---
title: "Sending end active power from ABCD, 132/130.5 kV"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:sending-power-132kv-130-5kv"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/sending-power-132kv-130-5kv
site: BCT Engineering
---

# Sending end active power from ABCD, 132/130.5 kV

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2077 Chaitra, 4 marks
- Video: [Power Flow Equations in Transmission Lines | Derivation | GATE/ESE/ISRO/BARC](https://www.youtube.com/watch?v=WfD3GfzYG4I) — ELECTRICAL ENGINEERING CONCEPTS
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/sending-power-132kv-130-5kv
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

A 3-phase overhead line has the following parameters: VS = 132∠0° kV, VR = 130.5∠10° kV, A = 0.97∠0.5°, B = 100∠75° ohms/ph. Determine the sending end active power.

## Short answer

Sending end active power is −27.63 MW, meaning 27.63 MW flows into the sending-end bus. Result from power-circle equation with VS=132 kV, VR=130.5∠10° kV, A=0.97∠0.5°, B=100∠75° Ω.

## Answer

Use the sending-end power-circle equation (taking $D = A$). With line voltages in kV, the result is in three-phase MW:
$$
P_S = \frac{|A||V_S|^2}{|B|}\cos(\beta-\alpha) - \frac{|V_S||V_R|}{|B|}\cos(\beta+\delta)
$$
Here $\delta$ is the angle by which $V_S$ leads $V_R$. Since $V_S = 132\angle0^\circ$ and $V_R = 130.5\angle10^\circ$,
$$
\delta = 0^\circ - 10^\circ = -10^\circ
$$

Data: $|A| = 0.97$, $\alpha = 0.5^\circ$, $|B| = 100\ \Omega$, $\beta = 75^\circ$.

#### Substitution
$$
\begin{aligned}
\frac{|A||V_S|^2}{|B|} &= \frac{0.97\times132^2}{100} = 169.01 \\
\frac{|V_S||V_R|}{|B|} &= \frac{132\times130.5}{100} = 172.26 \\
\beta-\alpha &= 74.5^\circ,\quad \beta+\delta = 65^\circ
\end{aligned}
$$
$$
\begin{aligned}
P_S &= 169.01\cos74.5^\circ - 172.26\cos65^\circ \\
&= 169.01(0.2672) - 172.26(0.4226) \\
&= 45.17 - 72.80 = -27.63\ \text{MW}
\end{aligned}
$$

#### Meaning
$P_S$ is negative because the receiving-end voltage **leads** the sending-end voltage by $10^\circ$. Active power flows from the leading to the lagging end, so about 27.6 MW actually flows **out of the line at the sending end** (from the "receiving" bus towards the "sending" bus).

**Answer: $P_S$ ≈ −27.63 MW (27.63 MW flows into the sending-end bus).**

## Check yourself

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

1. Which formula gives sending-end active power?
   A. P_S = |A||Vs|^2/|B| cos(beta - alpha) - |Vs||Vr|/|B| cos(beta + delta)
   B. P_S = |A||Vs|^2/|B| cos(beta + alpha) - ...cos(beta - delta)
   C. P_S = |Vs||Vr| sin(delta)/|B| only
   D. P_S = |Vr|^2/|B| cos(beta)
   Answer: A. Sending-end power equation, with D = A. (Key point: Sending end formula)

2. V_S = 132 at 0 deg and V_R = 130.5 at 10 deg. What is delta?
   A. +10 deg
   B. -10 deg, since delta is the angle by which V_S leads V_R
   C. 0 deg
   D. 130.5 deg
   Answer: B. Delta = 0 - 10 = -10 deg, so beta + delta = 65 deg. (Key point: Sign of delta)

3. Why is P_S negative (-27.63 MW)?
   A. Because the line has losses
   B. Because |V_R| < |V_S|
   C. V_R leads V_S, so active power flows from the receiving bus toward the sending bus
   D. Because A is below 1
   Answer: C. Power flows from the leading to the lagging end. (Key point: Negative P_S)

## Related in Power System Analysis I

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