---
title: "Maximum power over 20 + j50 Ω line"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:max-power-20-j50-33kv"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/max-power-20-j50-33kv
site: BCT Engineering
---

# Maximum power over 20 + j50 Ω line

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2074 Bhadra, 2+3+3 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/max-power-20-j50-33kv
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

A 3-phase short overhead line has a series impedance of 20 + j50 ohms per phase. Find the maximum active power which can be delivered at the receiving end voltage of 31.2 kV if the sending end voltage is 33 kV. Also, determine the receiving end power factor and line efficiency under this condition.

## Short answer

Maximum active power ≈12.41 MW; receiving-end power factor ≈0.594 leading; line efficiency ≈58.1%.

## Answer

Short line: $Z = 20 + j50 = 53.85\angle68.20^\circ\ \Omega$, $V_S$ = 33 kV, $V_R$ = 31.2 kV (line values, so powers are three-phase).

#### Maximum active power
For a short line, $P_R$ is maximum when $\delta = \theta$:
$$
\begin{aligned}
P_{R,max} &= \frac{|V_S||V_R|}{|Z|} - \frac{|V_R|^2R}{|Z|^2} \\
&= \frac{33\times31.2}{53.85} - \frac{31.2^2\times20}{53.85^2} \\
&= 19.12 - 6.71 = 12.41\ \text{MW}
\end{aligned}
$$

#### Receiving-end power factor
At $\delta = \theta$:
$$
Q_R = -\frac{|V_R|^2X}{|Z|^2} = -\frac{31.2^2\times50}{2900} = -16.78\ \text{MVAr}
$$
$$
\text{pf}_R = \frac{12.41}{\sqrt{12.41^2 + 16.78^2}} = 0.594\ \text{leading}
$$
(Negative $Q_R$: the load must supply reactive power, i.e. it is leading.)

#### Line efficiency
$$
\begin{aligned}
I &= \frac{\sqrt{12.41^2 + 16.78^2}}{\sqrt3\times31.2} = 0.3862\ \text{kA} = 386.2\ \text{A} \\
P_{loss} &= 3I^2R = 3(386.2)^2(20) = 8.95\ \text{MW} \\
\eta &= \frac{12.41}{12.41 + 8.95}\times100 = 58.1\%
\end{aligned}
$$
(Check: the sending-end power $P_S = \frac{|V_S|^2R}{|Z|^2} - \frac{|V_S||V_R|}{|Z|}\cos 2\theta = 7.51 + 13.84 = 21.35$ MW, so the loss is $21.35 - 12.41 = 8.94$ MW.)

**Answer: $P_{max}$ ≈ 12.41 MW, receiving-end pf ≈ 0.594 leading, efficiency ≈ 58.1 %.**

## Check yourself

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

1. For this short line, P_R is maximum when:
   A. delta = 0
   B. delta = theta (the impedance angle)
   C. delta = 90 degrees
   D. delta = 45 degrees
   Answer: B. P_R = V_S V_R cos(theta - delta)/|Z| - V_R^2 R/|Z|^2 is largest when delta = theta. (Key point: Condition for maximum power)

2. Which expression and result give P_R,max?
   A. V_S V_R/|Z| + V_R^2 R/|Z|^2 = 25.8 MW
   B. V_S V_R/|Z| - V_R^2 R/|Z|^2 = 12.41 MW
   C. V_S^2/|Z| = 20.2 MW
   D. V_R^2/|Z| = 18.1 MW
   Answer: B. 19.12 - 6.71 = 12.41 MW with |Z| = 53.85 ohm. (Key point: Maximum power value)

3. Why is the receiving-end power factor leading?
   A. Efficiency is below 100 %
   B. Q_R = -V_R^2 X/|Z|^2 is negative, so the load must supply reactive power
   C. P_R is negative
   D. The line is long
   Answer: B. At delta = theta, Q_R = -16.78 MVAr; negative Q_R means the receiving end must supply VAr, i.e. leading pf of 0.594. (Key point: Receiving pf)

## Related in Power System Analysis I

- [Power output with 46.85 kV sending, 33 kV receiving](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Apower-output-46-85kv-33kv) — Power System Analysis I, chapter 6
- [Maximum power over 20 + j30 Ω line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Amax-power-20-j30-33kv) — Power System Analysis I, chapter 6
- [Power delivered from sending end data, nominal-π](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-pi-power-delivered) — Power System Analysis I, chapter 5
- [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
- [8 km line receiving 15000 kVA at 33 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A8km-line-15000kva) — Power System Analysis I, chapter 6

## Related in other subjects

- [Line with 50+j50 load, first maximum, power](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Aline-50-j50-load-first-maximum) — Electromagnetism, chapter 5
- [Fault at far end of 33 kV line](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-far-end-33kv-line) — Power System Analysis II, chapter 3
- [Fault at load end of overhead line](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-load-end-33kv-line) — Power System Analysis II, 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