---
title: "132 kV horizontal line: susceptance and charging"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:horizontal-132kv-4-5m-susceptance"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/horizontal-132kv-4-5m-susceptance
site: BCT Engineering
---

# 132 kV horizontal line: susceptance and charging

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2071 Bhadra, 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/horizontal-132kv-4-5m-susceptance
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

A 120 km long, 50 Hz, 3 phase, 132 kV overhead line has flat horizontal configuration with distance between adjacent phases equal to 4.5 m and distance between extreme end phases equalling 9 m. Compute the shunt capacitive susceptance per phase, charging current per phase per km and the reactive power generated by the line.

## Short answer

Assuming conductor radius 12 mm and transposition, Deq = 5.67 m. Capacitance = 9.03 nF/km, giving shunt susceptance 2.84 µS/km per phase (341 µS for 120 km), charging current 0.216 A/km, and total reactive power 5.93 MVAR.

## Answer

The conductor radius is not given; it is **assumed to be 12 mm** (a common value for 132 kV lines in IOE problems). For a transposed line, $C_n = 2\pi\varepsilon_0/\ln(D_{eq}/r)$ and the shunt susceptance is $B = \omega C$.

```text
   a o----4.5 m----o b----4.5 m----o c
     |<----------- 9 m ----------->|
```

**Data:** $D_{ab} = D_{bc} = 4.5$ m, $D_{ca} = 9$ m, $r = 0.012$ m, 132 kV, 50 Hz, 120 km.

#### Equivalent spacing

$$
D_{eq} = \sqrt[3]{4.5 \times 4.5 \times 9} = 5.6696\ \text{m}
$$

#### Capacitance per phase

$$
\begin{aligned}
C &= \frac{2\pi\varepsilon_0}{\ln(D_{eq}/r)} = \frac{55.63\times10^{-12}}{\ln(5.6696/0.012)} = \frac{55.63\times10^{-12}}{6.1580} \\
&= 9.034\times10^{-12}\ \text{F/m} = 9.034\ \text{nF/km}
\end{aligned}
$$

#### Shunt susceptance per phase

$$
\begin{aligned}
B &= \omega C = 2\pi \times 50 \times 9.034\times10^{-9} = 2.838\times10^{-6}\ \text{S/km} \\
B_{120} &= 2.838\times10^{-6} \times 120 = 3.406\times10^{-4}\ \text{S}
\end{aligned}
$$

#### Charging current per phase per km

$$
I_c = V_p B = \frac{132\,000}{\sqrt3} \times 2.838\times10^{-6} = 76\,210 \times 2.838\times10^{-6} = 0.2163\ \text{A/km}
$$

#### Reactive power generated by the line (120 km, 3 phase)

$$
Q = V_L^2 B_{120} = (132\times10^3)^2 \times 3.406\times10^{-4} = 5.934\ \text{MVAR}
$$

**Answer (with $r = 12$ mm):** $B = 2.838\ \mu$S/km per phase (340.6 µS for 120 km), $I_c = 0.216$ A/km, $Q = 5.93$ MVAR.

## Check yourself

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

1. For a flat line with spacings 4.5 m, 4.5 m and 9 m (transposed), what is D_eq?
   A. cube root of (4.5 x 4.5 x 9) = 5.67 m
   B. Average (4.5+4.5+9)/3 = 6 m
   C. 9 m, the largest spacing
   D. 4.5 m, the adjacent spacing
   Answer: A. For a transposed line D_eq is the geometric mean of the three spacings, cube root of 182.25 = 5.67 m. (Key point: Equivalent spacing)

2. Which expression gives the capacitance to neutral per phase used here?
   A. C = 2*pi*eps0 / ln(D_eq / r')  with r' = 0.7788r
   B. C = 2*pi*eps0 / ln(D_eq / r)
   C. C = pi*eps0 / ln(D_eq / r)
   D. C = 2*pi*eps0 * ln(D_eq / r)
   Answer: B. Capacitance uses the actual radius r (no internal flux), and C is inversely proportional to ln(D_eq/r). (Key point: Capacitance formula)

3. How is the shunt susceptance obtained from the capacitance?
   A. B = 1/(omega*C)
   B. B = C / (2*pi*f)
   C. B = omega*C = 2*pi*f*C
   D. B = omega*C^2
   Answer: C. Susceptance is the reciprocal of capacitive reactance, so B = 2*pi*f*C (about 2.84 microS/km here). (Key point: Susceptance)

4. How is the 3-phase reactive power generated by the line found?
   A. Q = V_L x B
   B. Q = V_L^2 / B
   C. Q = sqrt(3) x V_L x B
   D. Q = V_L^2 x B (line-to-line voltage squared times total susceptance)
   Answer: D. Charging Q = V_L^2 * omega*C, which with B = 3.406e-4 S for 120 km gives about 5.93 MVAR. (Key point: Reactive power generated)

## Related in Power System Analysis I

- [Charging current of 60 Hz horizontal 132 kV line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ahorizontal-60hz-132kv-charging) — Power System Analysis I, chapter 4
- [330 kV quad-bundle line: inductance and charging](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aquad-bundle-330kv-6m) — Power System Analysis I, chapter 4
- [Inductive reactance of vertical 132 kV line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Avertical-132kv-12mm-5m) — Power System Analysis I, chapter 4
- [Nominal-T 100 km line, 50 MVA at 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-50mva) — Power System Analysis I, chapter 6
- [132 kV triangular line: L, C, charging MVAR](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atriangular-132kv-18mm-parameters) — Power System Analysis I, chapter 4
- [Charging VAR of 132 kV equilateral line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Acharging-var-132kv-6m-1-1cm) — Power System Analysis I, chapter 4

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