---
title: "132 kV triangular line: L, C, charging MVAR"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:triangular-132kv-18mm-parameters"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/triangular-132kv-18mm-parameters
site: BCT Engineering
---

# 132 kV triangular line: L, C, charging MVAR

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2072 Asoj, 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/triangular-132kv-18mm-parameters
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

A 132 kV, 50 Hz, 3-phase line has its conductors, each of diameter 18 mm, in a triangular configuration. The conductors are transposed at regular intervals. The distances between phase conductors are 4.0 m, 5.0 m and 4.5 m. Find the inductance, capacitance, charging current and charging MVAR per km.

## Short answer

With equivalent spacing 4.4814 m, per km: inductance per phase 1.292 mH/km, capacitance 8.958 nF/km, charging current 0.2145 A/km, total three-phase charging reactive power 0.0490 MVAR/km.

## Answer

For a transposed line with unsymmetrical spacing, use the equivalent spacing $D_{eq} = \sqrt[3]{D_{ab}D_{bc}D_{ca}}$.

**Data:** diameter 18 mm, $r = 0.009$ m, $r' = 0.7788 \times 0.009 = 0.0070092$ m; spacings 4.0, 5.0, 4.5 m; 132 kV, 50 Hz.

$$
D_{eq} = \sqrt[3]{4 \times 5 \times 4.5} = \sqrt[3]{90} = 4.4814\ \text{m}
$$

#### Inductance per phase

$$
\begin{aligned}
L &= 2\times10^{-7}\ln\frac{D_{eq}}{r'} = 2\times10^{-7}\ln\frac{4.4814}{0.0070092} \\
&= 2\times10^{-7}\times6.4605 = 1.2921\times10^{-6}\ \text{H/m} = 1.292\ \text{mH/km}
\end{aligned}
$$

#### Capacitance per phase

$$
\begin{aligned}
C &= \frac{2\pi\varepsilon_0}{\ln(D_{eq}/r)} = \frac{55.63\times10^{-12}}{\ln(4.4814/0.009)} = \frac{55.63\times10^{-12}}{6.2105} \\
&= 8.958\times10^{-12}\ \text{F/m} = 8.958\ \text{nF/km}
\end{aligned}
$$

#### Charging current per km

$$
\begin{aligned}
X_C &= \frac{1}{2\pi \times 50 \times 8.958\times10^{-9}} = 3.553\times10^5\ \Omega\text{-km} \\
V_p &= 132/\sqrt3 = 76.21\ \text{kV} \\
I_c &= \frac{76\,210}{355\,349} = 0.2145\ \text{A/km}
\end{aligned}
$$

#### Charging MVAR per km (three phase)

$$
Q_c = \frac{V_L^2}{X_C} = \frac{(132\times10^3)^2}{355\,349} = 49\,033\ \text{VAR} = 0.0490\ \text{MVAR/km}
$$

**Answer:** $L = 1.292$ mH/km, $C = 8.958$ nF/km, $I_c = 0.2145$ A/km, $Q_c = 0.0490$ MVAR/km (49.0 kVAR/km).

## Check yourself

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

1. For spacings 4.0, 5.0 and 4.5 m, D_eq is:
   A. Cube root of 90 = 4.481 m
   B. 4.5 m (arithmetic mean)
   C. 5 m
   D. Square root of 20
   Answer: A. D_eq = (4 x 5 x 4.5)^(1/3) = 90^(1/3) = 4.4814 m. (Key point: Equivalent spacing)

2. For a conductor of 18 mm diameter, which radii are used?
   A. r' = 7.009 mm for inductance and r = 9 mm for capacitance
   B. 18 mm for both
   C. 9 mm for inductance and 7.009 mm for capacitance
   D. r' for both
   Answer: A. r = 9 mm; inductance uses r' = 0.7788r, capacitance uses r. (Key point: Radius in L and C)

3. With C = 8.958 nF/km, how is I_c per km found?
   A. V_p / X_C with V_p = 76.21 kV and X_C = 3.553e5 ohm-km, giving 0.2145 A/km
   B. 132 kV / X_C giving 0.37 A/km
   C. X_C / V_p
   D. V_p x 2*pi f / C
   Answer: A. I_c = 76,210/355,349 = 0.2145 A/km. (Key point: Charging current)

4. The total three-phase charging MVAR per km uses:
   A. V_L^2 / X_C = 0.0490 MVAR/km
   B. V_p^2 / X_C = 0.0163 MVAR/km
   C. V_L x I_c
   D. V_p x X_C
   Answer: A. Q = V_L^2/X_C = (132e3)^2/355,349 = 49 kVAR/km (= 3 V_p I_c). (Key point: Three-phase MVAR)

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