---
title: "Charging VAR of 132 kV equilateral line"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:charging-var-132kv-6m-1-1cm"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/charging-var-132kv-6m-1-1cm
site: BCT Engineering
---

# Charging VAR of 132 kV equilateral line

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2068 Magh, 5 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/charging-var-132kv-6m-1-1cm
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

A 50 Hz, 132 kV, 3-phase overhead transmission line has symmetrical phase spacing of 6 m. The effective radius of the phase conductor is 1.1 cm. Compute the charging VAR per unit length generated by the line.

## Short answer

With symmetrical spacing 6 m and r=1.1 cm, per-phase capacitance = 8.828 nF/km. Three-phase charging VAR = V_L² ω C_n = 48.3 kVAR/km (16.1 kVAR/km per phase).

## Answer

With symmetrical spacing, $C_n = 2\pi\varepsilon_0/\ln(D/r)$ per phase, and the three-phase charging VAR is $Q = V_L^2\,\omega C_n$.

**Data:** $D = 6$ m, $r = 1.1$ cm $= 0.011$ m, 132 kV, 50 Hz.

#### Capacitance per phase

$$
\begin{aligned}
C_n &= \frac{55.63\times10^{-12}}{\ln(6/0.011)} = \frac{55.63\times10^{-12}}{6.3016} \\
&= 8.828\times10^{-12}\ \text{F/m} = 8.828\ \text{nF/km}
\end{aligned}
$$

#### Capacitive reactance

$$
X_C = \frac{1}{2\pi \times 50 \times 8.828\times10^{-9}} = 3.606\times10^5\ \Omega\text{-km}
$$

#### Charging current and VAR

$$
\begin{aligned}
I_c &= \frac{132\,000/\sqrt3}{360\,564} = 0.2114\ \text{A/km} \\
Q_{3\phi} &= \frac{V_L^2}{X_C} = \frac{(132\times10^3)^2}{360\,564} = 48\,324\ \text{VAR/km}
\end{aligned}
$$

Per phase: $48.32/3 = 16.11$ kVAR/km.

**Answer:** charging VAR $= 48.3$ kVAR per km for the three-phase line (16.1 kVAR/km per phase).

## Check yourself

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

1. Since the effective radius 1.1 cm is given, it is used as:
   A. The r in ln(D/r) directly
   B. r' after multiplying by 0.7788
   C. The diameter
   D. r/2
   Answer: A. The effective radius is the given value for capacitance; no extra correction. (Key point: Effective radius)

2. Three-phase charging VAR is:
   A. Q = V_p^2 omega C_n
   B. Q = V_L^2 omega C_n
   C. Q = V_L omega C_n
   D. Q = V_L^2 / (omega C_n)
   Answer: B. Q = V_L^2 / X_C = V_L^2 omega C. (Key point: Charging VAR formula)

3. Per-phase VAR relates to the three-phase value by:
   A. Per phase = three-phase x 3
   B. Per phase = three-phase / sqrt3
   C. Per phase = three-phase / 3
   D. Same value
   Answer: C. The three-phase value is shared equally among three phases. (Key point: Per phase vs total)

## Related in Power System Analysis I

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- [Charging current of 132 kV line, 6 m equilateral](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aequilateral-132kv-6m-charging) — Power System Analysis I, chapter 4
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