---
title: "12 km line at 32.5 kV: regulation and capacitor"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:12km-32-5kv-regulation-capacitor"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/12km-32-5kv-regulation-capacitor
site: BCT Engineering
---

# 12 km line at 32.5 kV: regulation and capacitor

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2068 Bhadra, 8 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/12km-32-5kv-regulation-capacitor
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

A 12 km long, 50 Hz, 3-phase line is supplying a load at 32.5 kV∠-20° w.r.t. VS. The line has a resistance of 2.5 × 10⁻³ Ohms/m and inductance of 1.3 × 10⁻⁶ H/m. The sending end voltage is maintained at 34.1 kV∠0°. Calculate the active power flow through the line and voltage regulation. Compute the capacitance per phase to be connected at the receiving end to reduce the voltage regulation by 50%.

## Short answer

Line impedance is 3 + j4.90 Ω. Current from phasor difference yields sending-end power 67.3 MW, receiving-end 54.9 MW. Voltage regulation is 4.92%. To halve regulation, a shunt capacitor of about 17.4 µF per phase (6.06 MVAr) is needed.

## Answer

**Assumption:** the printed resistance $2.5\times10^{-3}\ \Omega$/m gives $R$ = 30 Ω (2.5 Ω/km), far above any overhead conductor; with it the capacitor part has no solution. It is taken as $2.5\times10^{-4}\ \Omega$/m (0.25 Ω/km), a likely misprint.

#### Line constants
$$
\begin{aligned}
R &= 2.5\times10^{-4}\times12\,000 = 3\ \Omega \\
X &= 2\pi(50)(1.3\times10^{-6}\times12\,000) = 4.90\ \Omega \\
Z &= 3 + j4.90 = 5.746\angle58.53^\circ\ \Omega
\end{aligned}
$$

#### Current and power flow
Phase values: $V_S = 19.688\angle0^\circ$ kV, $V_R = 18.764\angle-20^\circ$ kV.
$$
\begin{aligned}
I &= \frac{V_S - V_R}{Z} = \frac{6.739\angle72.24^\circ}{5.746\angle58.53^\circ} = 1.1727\angle13.71^\circ\ \text{kA} \\
S_S &= 3V_SI^* = 67.29 - j16.42\ \text{MVA} \\
S_R &= 3V_RI^* = 54.91 - j36.64\ \text{MVA}
\end{aligned}
$$
Active power sent = 67.29 MW, received = 54.91 MW (loss 12.38 MW). $Q_R$ is negative, so the load is leading (pf 0.832 lead).

#### Voltage regulation
For a short line, the no-load receiving voltage equals $V_S$:
$$
\%VR = \frac{34.1 - 32.5}{32.5}\times100 = 4.92\%
$$

#### Capacitor to halve the regulation
New regulation $= 2.46\%$, so the receiving voltage must rise to
$$
V_R' = \frac{34.1}{1.0246} = 33.28\ \text{kV}
$$
With $P$ = 54.91 MW fixed, find $Q$ from $|V_S|^2|V_R'|^2 = (V_R'^2 + RP + XQ)^2 + (XP - RQ)^2$:
$$
Q = -42.70\ \text{MVAr}
$$
Extra leading VAr needed:
$$
Q_C = 42.70 - 36.64 = 6.06\ \text{MVAr}
$$
$$
C = \frac{Q_C}{\omega V_R'^2} = \frac{6.06\times10^6}{314.16\times(33\,280)^2} = 17.41\ \mu\text{F per phase (star)}
$$

(With the printed $R$ = 30 Ω, the power sent would be 5.95 MW and received 1.53 MW, and no shunt capacitor could raise $V_R$ to 33.28 kV.)

**Answer: Power sent ≈ 67.3 MW (54.9 MW received); regulation 4.92 %; C ≈ 17.4 µF per phase (≈ 6.06 MVAr) to halve the regulation.**

## Check yourself

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

1. How is the line reactance X obtained from the given inductance per metre?
   A. X = 2 pi f L total, with L total = 1.3e-6 x 12000 H
   B. X = L total / (2 pi f)
   C. X = 1.3e-6 x 12 ohm
   D. X = R tan(delta)
   Answer: A. X = 2 pi (50)(0.0156) = 4.90 ohm. (Key point: Line constants)

2. How is the line current found?
   A. I = (V_S + V_R)/Z
   B. I = V_R / Z
   C. I = P / V_R
   D. I = (V_S - V_R)/Z using phase voltages as phasors
   Answer: D. For a short line the current is the phasor voltage difference over Z, giving 1.1727 kA at 13.71 degrees. (Key point: Current from the two end voltages)

3. Why is regulation (V_S - V_R)/V_R = 4.92%?
   A. No-load receiving voltage of a short line equals V_S, so VR = (34.1 - 32.5)/32.5
   B. Because of the Ferranti effect
   C. Because VR is the difference of phase angles
   D. Because the capacitor is already connected
   Answer: A. With no shunt branch, V_R at no load equals V_S = 34.1 kV. (Key point: Regulation of a short line)

4. What is the approach to find the capacitor to halve the regulation?
   A. Halve the line resistance
   B. Find the V_R' = 34.1/1.0246, hold P constant, solve for the new Q, and take Q_C as the extra leading VAr
   C. Halve the load current
   D. Make C = 1/(omega X)
   Answer: B. V_R' = 33.28 kV, new Q = -42.70 MVAr versus existing -36.64, so Q_C = 6.06 MVAr and C = Q_C/(omega V^2) = 17.41 microfarad. (Key point: Capacitor to halve regulation)

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