---
title: "Capacitance of line on 5-6-7 m triangle"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:capacitance-triangle-5-6-7m"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/capacitance-triangle-5-6-7m
site: BCT Engineering
---

# Capacitance of line on 5-6-7 m triangle

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2075 Baisakh, 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/capacitance-triangle-5-6-7m
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

Find the capacitance of phase to neutral per kilometre of a three phase line having conductors of 2 cm diameter placed at the corners of a triangle with sides 5 m, 6 m, 7 m respectively. Assume that the line is fully transposed and carries balanced load.

## Short answer

Compute equivalent spacing D_eq = (5×6×7)^(1/3) = 5.944 m, then C_n = 2πε₀ / ln(D_eq/r) with r = 0.01 m. C_n = 8.709 nF/km.

## Answer

For a fully transposed three-phase line with unsymmetrical spacing, the capacitance per phase to neutral is

$$
C_n = \frac{2\pi\varepsilon_0}{\ln(D_{eq}/r)}\ \text{F/m}, \qquad D_{eq} = \sqrt[3]{D_{12}D_{23}D_{31}}
$$

**Data:** diameter 2 cm, so $r = 1$ cm $= 0.01$ m; sides 5 m, 6 m, 7 m; $\varepsilon_0 = 8.854\times10^{-12}$ F/m.

#### Equivalent spacing

$$
D_{eq} = \sqrt[3]{5 \times 6 \times 7} = \sqrt[3]{210} = 5.9439\ \text{m}
$$

#### Capacitance

$$
\begin{aligned}
C_n &= \frac{2\pi \times 8.854\times10^{-12}}{\ln(5.9439/0.01)} = \frac{55.63\times10^{-12}}{\ln 594.39} \\
&= \frac{55.63\times10^{-12}}{6.3875} = 8.709\times10^{-12}\ \text{F/m}
\end{aligned}
$$

Per km: $8.709\times10^{-12} \times 1000 = 8.709\times10^{-9}$ F/km.

**Answer:** $C_n = 8.709$ nF/km $= 0.00871\ \mu$F/km (phase to neutral).

## Check yourself

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

1. For a transposed line with sides 5, 6 and 7 m, the equivalent spacing D_eq is:
   A. 6 m (arithmetic mean)
   B. Cube root of 5 x 6 x 7, about 5.94 m
   C. 7 m
   D. Square root of 5 x 7
   Answer: B. D_eq is the geometric mean (D12 D23 D31)^(1/3) = 210^(1/3) = 5.9439 m. (Key point: Equivalent spacing)

2. The conductor diameter is 2 cm. What value is used in ln(D_eq/r)?
   A. r = 2 cm
   B. r = 1 cm = 0.01 m
   C. r = 0.7788 cm
   D. r = 4 cm
   Answer: B. The radius is half the diameter, and capacitance uses the actual radius, so r = 0.01 m. (Key point: Radius from diameter)

3. Which formula gives capacitance to neutral per metre?
   A. 2*pi*eps0 / ln(D_eq/r)
   B. 2*pi*eps0 x ln(D_eq/r)
   C. pi*eps0 / ln(D_eq/r')
   D. 2e-7 ln(D_eq/r)
   Answer: A. C_n = 2*pi*eps0/ln(D_eq/r) = 55.63e-12/6.3875 = 8.709 pF/m. (Key point: Formula)

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