---
title: "Maximum power over 20 + j30 Ω line"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:max-power-20-j30-33kv"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/max-power-20-j30-33kv
site: BCT Engineering
---

# Maximum power over 20 + j30 Ω line

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2069 Poush, 6 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-j30-33kv
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

A 3-phase overhead line has a series impedance of 20 + j30 Ohm per phase. Find the maximum power that can be delivered at the receiving end keeping the receiving end voltage 31.2 kV while the sending end voltage is 33 kV. What will be the efficiency of the line under this condition?

## Short answer

Maximum power occurs when load angle equals impedance angle (56.31°). Using P_R,max = (|V_S||V_R|/|Z|) − (|V_R|²R/|Z|²) gives 13.58 MW. Efficiency is about 49.0% with receiving-end power factor 0.517 leading.

## Answer

Short line: $Z = 20 + j30 = 36.06\angle56.31^\circ\ \Omega$, $|V_S|$ = 33 kV, $|V_R|$ = 31.2 kV (line kV give three-phase MW/MVAr).

#### Maximum power
Maximum $P_R$ occurs when $\delta = \theta = 56.31^\circ$:
$$
\begin{aligned}
P_{R,max} &= \frac{|V_S||V_R|}{|Z|} - \frac{|V_R|^2R}{|Z|^2} \\
&= \frac{33\times31.2}{36.06} - \frac{31.2^2\times20}{1300} \\
&= 28.56 - 14.98 = 13.58\ \text{MW}
\end{aligned}
$$

#### Reactive power at this condition
$$
Q_R = -\frac{|V_R|^2X}{|Z|^2} = -\frac{31.2^2\times30}{1300} = -22.46\ \text{MVAr}
$$
So the receiving end must supply 22.46 MVAr (pf $= 0.517$ leading).

#### Efficiency
$$
\begin{aligned}
I &= \frac{\sqrt{13.58^2 + 22.46^2}}{\sqrt3\times31.2} = 0.4857\ \text{kA} \\
P_{loss} &= 3I^2R = 3(485.7)^2(20) = 14.16\ \text{MW} \\
\eta &= \frac{13.58}{13.58 + 14.16}\times100 = 48.96\%
\end{aligned}
$$
(Check: $P_S = \frac{|V_S|^2R}{|Z|^2} - \frac{|V_S||V_R|}{|Z|}\cos(\theta+\delta) = 16.75 - 28.56\cos112.62^\circ = 27.74$ MW.)

**Answer: Maximum power ≈ 13.58 MW; efficiency ≈ 49.0 % (receiving pf 0.517 leading).**

## Check yourself

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

1. At what load angle is P_R maximum for this short line?
   A. delta = 90 degrees
   B. delta = theta = 56.31 degrees (impedance angle)
   C. delta = 0
   D. delta = 33 degrees
   Answer: B. Maximum power occurs when delta equals the impedance angle. (Key point: Condition for maximum power)

2. Which formula gives P_R,max?
   A. |Vs||Vr|/|Z| + |Vr|^2 R/|Z|^2
   B. |Vs||Vr|/|Z| - |Vr|^2 X/|Z|^2
   C. |Vs||Vr|/|Z| - |Vr|^2 R/|Z|^2
   D. |Vr|^2 R/|Z|^2 only
   Answer: C. P_R,max = 28.56 - 14.98 = 13.58 MW; the subtracted term involves R. (Key point: Maximum power formula)

3. What is the nature of the reactive power at maximum power?
   A. The receiving end absorbs 22.46 MVAr
   B. The receiving end supplies 22.46 MVAr (leading pf 0.517)
   C. Q is zero
   D. The line absorbs nothing and the load is lagging
   Answer: B. Q_R = -|Vr|^2 X/|Z|^2 = -22.46 MVAr, so the load must be leading. (Key point: Reactive power at maximum power)

## Related in Power System Analysis I

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