---
title: "Receiving end power in ABCD terms and maximum power"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:receiving-power-abcd-max"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/receiving-power-abcd-max
site: BCT Engineering
---

# Receiving end power in ABCD terms and maximum power

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2078 Chaitra, 4+2 marks
- Video: [PS34 Receiving End Power Circle Diagram](https://www.youtube.com/watch?v=kGee2B5Pmeg) — Lectures in Electrical Engineering
- 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-abcd-max
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

Starting from a suitable point, find the expression of power at the receiving end of a line in terms of sending and receiving end voltages and line ABCD constants. Also, find the expression for maximum active power at the receiving end.

## Short answer

From V_S = A V_R + B I_R, receiving power is P_R = (|V_S||V_R|/|B|) cos(β−δ) − (|A||V_R|^2/|B|) cos(β−α). Maximum active power occurs when δ = β, giving P_Rmax = |V_S||V_R|/|B| − |A||V_R|^2 cos(β−α)/|B|.

## Answer

Start from the ABCD relation of a two-port line:
$$
V_S = A V_R + B I_R
$$
Take $V_R$ as reference and let
$$
V_R = |V_R|\angle0^\circ,\quad V_S = |V_S|\angle\delta,\quad A = |A|\angle\alpha,\quad B = |B|\angle\beta
$$

#### Receiving-end current
$$
I_R = \frac{V_S - A V_R}{B} = \frac{|V_S|}{|B|}\angle(\delta-\beta) - \frac{|A||V_R|}{|B|}\angle(\alpha-\beta)
$$

#### Receiving-end complex power
$$
\begin{aligned}
S_R &= P_R + jQ_R = V_R I_R^* \\
&= \frac{|V_S||V_R|}{|B|}\angle(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\angle(\beta-\alpha)
\end{aligned}
$$
Separating real and imaginary parts:
$$
\begin{aligned}
P_R &= \frac{|V_S||V_R|}{|B|}\cos(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha) \\
Q_R &= \frac{|V_S||V_R|}{|B|}\sin(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha)
\end{aligned}
$$
With phase voltages these give power per phase; with line voltages they give total three-phase power.

#### Maximum receiving-end power
For fixed $|V_S|$, $|V_R|$ and line constants, only $\delta$ can vary. The second term is constant, so $P_R$ is maximum when $\cos(\beta-\delta) = 1$, i.e.
$$
\delta = \beta
$$
$$
P_{R,max} = \frac{|V_S||V_R|}{|B|} - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha)
$$
At this point the reactive power at the receiving end is
$$
Q_R = -\frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha)
$$
which is negative, so the load must supply (leading) reactive power to receive maximum active power.

For a short line ($A = 1\angle0^\circ$, $B = Z\angle\theta$):
$$
P_{R,max} = \frac{|V_S||V_R|}{|Z|} - \frac{|V_R|^2 R}{|Z|^2}
$$

## Check yourself

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

1. Starting from V_S = A V_R + B I_R, the receiving current is
   A. I_R = (V_S - A V_R)/B
   B. I_R = (V_S + A V_R)/B
   C. I_R = (V_S - B V_R)/A
   D. I_R = V_S/(A+B)
   Answer: A. Rearranging the ABCD equation. (Key point: Receiving current)

2. How is S_R obtained?
   A. S_R = V_R I_R without conjugate
   B. S_R = V_R I_R* (conjugate of current), giving |VsVr/B| at angle (beta - delta) minus |A Vr^2/B| at angle (beta - alpha)
   C. S_R = V_S I_R*
   D. S_R = |V_R|^2 / B
   Answer: B. Conjugation reverses angles, giving beta - delta and beta - alpha. (Key point: Complex power)

3. P_R is maximum when
   A. delta = 90 deg always
   B. delta = alpha
   C. delta = beta, so cos(beta - delta) = 1
   D. delta = 0
   Answer: C. Only the first term varies with delta. (Key point: Condition for maximum)

4. At maximum receiving power, Q_R is
   A. Positive, lagging load
   B. Zero
   C. Equal to P_R
   D. -|A||V_R|^2 sin(beta - alpha)/|B|, negative, so load must supply leading Q
   Answer: D. sin(beta - delta) = 0 at delta = beta, leaving only the negative second term. (Key point: Q at maximum)

## Related in Power System Analysis I

- [Sending end active power from ABCD, 132/130.5 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Asending-power-132kv-130-5kv) — Power System Analysis I, chapter 6
- [Receiving and sending end powers from ABCD, 132/134 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aabcd-power-132kv-134kv) — Power System Analysis I, chapter 6
- [150 mile, 60 Hz line: ABCD and sending end](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A150-mile-60hz-line-abcd) — Power System Analysis I, chapter 5
- [220 kV line, 200 MW: sending end from ABCD](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aabcd-220kv-200mw) — Power System Analysis I, chapter 6
- [Long line sending end voltage derivation](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Along-line-sending-voltage-derivation) — Power System Analysis I, chapter 5
- [Receiving end complex power and power circle](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areceiving-power-circle) — Power System Analysis I, chapter 6

## Related in other subjects

- [Expression for receiving-end shunt compensation](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Ashunt-compensation-xp-expression) — High Voltage Engineering, chapter 3
- [Receiving-end voltage infinite at 1500 km](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Alossless-line-1500km-infinite-voltage) — High Voltage Engineering, chapter 3
- [Fault at receiving end of 220 kV lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-220kv-two-generators) — Power System Analysis II, chapter 3
- [Critical clearing for temporary sending-end fault](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Acritical-clearing-temporary-fault-new) — Power System Analysis II, chapter 6