---
title: "Receiving end power, 140.2∠18.2° kV and 132 kV"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:receiving-power-140-2kv"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/receiving-power-140-2kv
site: BCT Engineering
---

# Receiving end power, 140.2∠18.2° kV and 132 kV

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2068 Magh, 5 marks
- 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/receiving-power-140-2kv
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

For a 3-phase line the following data is available at a particular instant: VS (line) = 140.2∠18.2° kV, VR (line) = 132 kV∠0°, A = 0.94∠0.4°, B = 64.2∠78.8° Ohms. Compute the receiving end active power.

## Short answer

Using P_R = (|V_S||V_R|/|B|) cos(β−δ) − (|A||V_R|²/|B|) cos(β−α) with given ABCD parameters yields receiving-end active power of 90.2 MW.

## Answer

Use the receiving-end power-circle equation. With line kV, the result is three-phase MW:
$$
P_R = \frac{|V_S||V_R|}{|B|}\cos(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha)
$$

Data: $|V_S| = 140.2$ kV, $\delta = 18.2^\circ$, $|V_R| = 132$ kV, $|A| = 0.94$, $\alpha = 0.4^\circ$, $|B| = 64.2\ \Omega$, $\beta = 78.8^\circ$.

#### Substitution
$$
\begin{aligned}
\frac{|V_S||V_R|}{|B|} &= \frac{140.2\times132}{64.2} = 288.26 \\
\frac{|A||V_R|^2}{|B|} &= \frac{0.94\times132^2}{64.2} = 255.12 \\
\beta-\delta &= 60.6^\circ,\quad \beta-\alpha = 78.4^\circ
\end{aligned}
$$
$$
\begin{aligned}
P_R &= 288.26\cos60.6^\circ - 255.12\cos78.4^\circ \\
&= 141.51 - 51.30 = 90.21\ \text{MW}
\end{aligned}
$$

**Answer: Receiving-end active power ≈ 90.2 MW.**

## Check yourself

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

1. Which equation gives the receiving-end active power?
   A. P_R = |Vs||Vr|cos(beta-delta)/|B| - |A||Vr|^2 cos(beta-alpha)/|B|
   B. P_R = |Vs||Vr|sin(delta)/|B| only
   C. P_R = |A||Vs|^2 cos(beta-alpha)/|B| - |Vs||Vr|cos(beta+delta)/|B|
   D. P_R = |Vs||Vr|cos(beta+delta)/|B|
   Answer: A. This is the receiving-end power-circle equation, giving 90.21 MW. (Key point: Power equation)

2. What angles are used in the two cosine terms?
   A. beta + delta = 97 and beta + alpha = 79.2
   B. delta = 18.2 and alpha = 0.4
   C. beta - delta = 60.6 and beta - alpha = 78.4
   D. beta - delta = 78.8 and beta - alpha = 0.4
   Answer: C. beta = 78.8, so beta - delta = 60.6 degrees and beta - alpha = 78.4 degrees. (Key point: Angles used)

3. With line voltages in kV and B in ohm, what is the unit and nature of the result?
   A. Three-phase MW
   B. Per-phase kW
   C. Amps
   D. Three-phase MVAr
   Answer: A. Line kV squared over ohm gives three-phase MVA, so the cosine form gives MW (90.2). (Key point: Units)

## Related in Power System Analysis I

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