---
title: "12 km 33 kV line: losses, zero-regulation capacitor"
subject: "Power System Analysis I"
chapter: 6
key: "semester_4/power_system_analysis_i:12km-33kv-losses-zero-regulation"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/12km-33kv-losses-zero-regulation
site: BCT Engineering
---

# 12 km 33 kV line: losses, zero-regulation capacitor

Power System Analysis I · Chapter 6: Performance Analysis

- Asked in: 2068 Magh, 6+4 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-33kv-losses-zero-regulation
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6

## Question

A 12 km long 3-phase line is supplying a load at 33 kV. The line has a resistance of 2.5 × 10⁻³ Ohms/m and inductance of 1.3 × 10⁻⁶ H/m. If the sending end voltage is maintained at 34 kV∠20°, compute the active and reactive power losses in the line. Compute the capacitance per phase that needs to be connected at the end of the line to make the voltage regulation zero.

## Short answer

With corrected R=3 Ω, X=4.90 Ω, line current is 1.173 kA. Active loss is 12.39 MW, reactive loss 20.24 MVAr. For zero regulation, a capacitor of about 22.3 µF per phase (8.1 MVAr) is required.

## Answer

**Assumption:** the printed $2.5\times10^{-3}\ \Omega$/m gives $R$ = 30 Ω for 12 km (2.5 Ω/km), unrealistic and larger than $X$; with it zero regulation is impossible by any shunt capacitor. It is taken as $2.5\times10^{-4}\ \Omega$/m (likely misprint).

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

#### Line current
Phase values: $V_S = 19.630\angle20^\circ$ kV, $V_R = 19.053\angle0^\circ$ kV.
$$
\begin{aligned}
V_S - V_R &= 6.741\angle95.16^\circ\ \text{kV} \\
I &= \frac{6.741\angle95.16^\circ}{5.746\angle58.53^\circ} = 1.1732\angle36.63^\circ\ \text{kA}
\end{aligned}
$$

#### Losses
$$
\begin{aligned}
P_{loss} &= 3I^2R = 3(1173.2)^2(3) = 12.39\ \text{MW} \\
Q_{loss} &= 3I^2X = 3(1173.2)^2(4.90) = 20.24\ \text{MVAr}
\end{aligned}
$$
(Check: $S_S = 66.20 - j19.78$ MVA, $S_R = 53.81 - j40.01$ MVA; differences 12.39 MW and 20.24 MVAr.)

#### Capacitance for zero regulation
Zero regulation means $|V_R| = |V_S| = 34$ kV at full load. Keep the load $P$ = 53.81 MW and find the net $Q$:
$$
34^4 = (34^2 + 3P + 4.90Q)^2 + (4.90P - 3Q)^2 \Rightarrow Q = -48.12\ \text{MVAr}
$$
The load already supplies 40.01 MVAr (leading), so the capacitor must supply
$$
Q_C = 48.12 - 40.01 = 8.11\ \text{MVAr}
$$
$$
C = \frac{Q_C}{\omega V_L^2} = \frac{8.11\times10^6}{314.16\times34\,000^2} = 22.34\ \mu\text{F per phase (star)}
$$

(With the printed $R$ = 30 Ω the losses would be 4.43 MW and 0.72 MVAr.)

**Answer: Losses ≈ 12.39 MW and 20.24 MVAr; C ≈ 22.3 µF per phase (≈ 8.1 MVAr) for zero regulation.**

## Check yourself

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

1. How is the active power loss calculated?
   A. 3 I^2 X
   B. 3 I^2 R
   C. I^2 R
   D. V_S I cos(phi)
   Answer: B. Loss P = 3 I^2 R = 12.39 MW; the reactive loss uses X (20.24 MVAr). (Key point: Active power loss)

2. How is the reactive power loss in the line obtained?
   A. 3 I^2 X from the same current
   B. 3 I^2 R
   C. Q_S + Q_R
   D. V^2/X
   Answer: A. The series reactance absorbs 3 I^2 X = 20.24 MVAr. (Key point: Reactive loss)

3. How can the loss values be verified?
   A. They cannot be verified
   B. By dividing S_S by S_R
   C. By comparing the difference of sending-end and receiving-end S, e.g. 66.20 - 53.81 = 12.39 MW
   D. By adding P_S and P_R
   Answer: C. The loss is the difference S_S - S_R, which matches 12.39 MW and 20.24 MVAr. (Key point: Check using end powers)

4. With |V_R| = 34 kV, the net Q is -48.12 MVAr and the load already supplies 40.01 MVAr leading. What is Q_C?
   A. 48.12 + 40.01 MVAr
   B. 48.12 - 40.01 = 8.11 MVAr
   C. 40.01 MVAr
   D. 48.12 MVAr
   Answer: B. The load already provides part of the leading VAr, so the capacitor supplies only the difference, which gives about 22.3 microfarad per phase. (Key point: Zero regulation capacitor)

## Related in Power System Analysis I

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