---
title: "Capacitance with and without earth effect"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:capacitance-with-earth-effect-13m"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/capacitance-with-earth-effect-13m
site: BCT Engineering
---

# Capacitance with and without earth effect

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2075 Bhadra, 4+4 marks
- Video: [PS21 Effect of Earth on Capacitance of Transmission Line](https://www.youtube.com/watch?v=c8F2M1F4FtE) — Lectures in Electrical Engineering
- 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-with-earth-effect-13m
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

Figure below shows the arrangement of a three phase transmission line. The line is transposed and the diameter of each conductor is 2.6 cm. Calculate the capacitance of the transmission line per phase for the following two cases: (i) neglecting the effect of ground (ii) considering the effect of earth. [Figure: conductors R, Y, B in a horizontal row, R–Y = 8 m, Y–B = 8 m, all 13 m above ground.]

## Short answer

Neglecting earth: C_n = 2πε₀ / ln(D_eq/r) = 8.361 nF/km. Including earth via method of images adds a correction term, giving C_n = 8.468 nF/km.

## Answer

For a transposed line, the capacitance to neutral is $C_n = 2\pi\varepsilon_0/\ln(D_{eq}/r)$ when ground is neglected. The method of images adds a correction term when ground is included.

```text
   R o----8 m----o Y----8 m----o B
     |13 m       |13 m         |13 m
 ======================================= ground
     |13 m       |13 m         |13 m
   R'o           o Y'          o B'   (images)
```

**Data:** diameter 2.6 cm, $r = 0.013$ m; $D_{RY} = D_{YB} = 8$ m, $D_{BR} = 16$ m; $h = 13$ m.

$$
D_{eq} = \sqrt[3]{8 \times 8 \times 16} = 10.0794\ \text{m}, \qquad \ln\frac{D_{eq}}{r} = \ln\frac{10.0794}{0.013} = 6.6533
$$

#### (i) Neglecting the effect of ground

$$
C_n = \frac{2\pi\varepsilon_0}{\ln(D_{eq}/r)} = \frac{55.63\times10^{-12}}{6.6533} = 8.361\times10^{-12}\ \text{F/m} = 8.361\ \text{nF/km}
$$

#### (ii) Considering the effect of earth

Distances to images:

| Distance | Value (m) |
|---|---|
| $H_1 = H_2 = H_3$ (own image) | $2h = 26$ |
| $H_{12} = H_{23}$ | $\sqrt{8^2+26^2} = 27.203$ |
| $H_{31}$ | $\sqrt{16^2+26^2} = 30.529$ |

$$
\begin{aligned}
\sqrt[3]{H_{12}H_{23}H_{31}} &= \sqrt[3]{27.203^2 \times 30.529} = 28.269\ \text{m} \\
\sqrt[3]{H_1H_2H_3} &= 26\ \text{m} \\
\text{Correction} &= \ln\frac{28.269}{26} = 0.08368
\end{aligned}
$$

$$
\begin{aligned}
C_n &= \frac{2\pi\varepsilon_0}{\ln\dfrac{D_{eq}}{r} - \ln\dfrac{\sqrt[3]{H_{12}H_{23}H_{31}}}{\sqrt[3]{H_1H_2H_3}}} = \frac{55.63\times10^{-12}}{6.6533 - 0.0837} \\
&= \frac{55.63\times10^{-12}}{6.5696} = 8.468\times10^{-12}\ \text{F/m} = 8.468\ \text{nF/km}
\end{aligned}
$$

**Answer:** (i) $C_n = 8.361$ nF/km; (ii) $C_n = 8.468$ nF/km. The earth raises the capacitance by about 1.3%.

## Check yourself

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

1. Which formula is used for C_n when the ground is neglected?
   A. 2*pi*eps0 / ln(D_eq/r)
   B. 2*pi*eps0 / ln(2h/r)
   C. pi*eps0 / ln(D_eq/r)
   D. 2e-7 ln(D_eq/r)
   Answer: A. For a transposed line, C_n = 2*pi*eps0/ln(D_eq/r) with D_eq = (8 x 8 x 16)^(1/3) = 10.08 m. (Key point: Capacitance neglecting ground)

2. How is the ground effect included in the denominator?
   A. It is subtracted: ln(Deq/r) - ln[(H12 H23 H31)^(1/3)/(H1 H2 H3)^(1/3)]
   B. It is added to ln(Deq/r)
   C. It multiplies the whole expression by 2
   D. It replaces D_eq with h
   Answer: A. The positive correction 0.0837 is subtracted from 6.6533, reducing the denominator. (Key point: Earth correction term)

3. What are H1, H2, H3 in the earth-effect formula?
   A. Distances between phases
   B. Distances from each conductor to its own image, 2h = 26 m
   C. Distances from conductor to ground, 13 m
   D. Distances to the other phases' images
   Answer: B. H1 = H2 = H3 = 2h = 26 m, while H12, H23, H31 are distances to images of other conductors. (Key point: Image distances)

4. The results are 8.361 nF/km (no ground) and 8.468 nF/km (with ground). What does this show?
   A. Ground decreases capacitance by 1.3%
   B. Ground increases capacitance by about 1.3%
   C. Ground has no effect
   D. Ground doubles capacitance
   Answer: B. The reduced denominator raises capacitance by about 1.3%. (Key point: Effect of ground)

## Related in Power System Analysis I

- [Effect of neglecting earth on capacitance](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aearth-effect-horizontal-13m) — Power System Analysis I, chapter 4
- [Inductance and capacitance of 4-conductor bundle line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aquad-bundle-220kv-l-c) — Power System Analysis I, chapter 4
- [Single-phase line capacitance, 6 mm, 1.3 m](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Asingle-phase-capacitance-6mm-1-3m) — Power System Analysis I, chapter 4
- [Effect of earth on line capacitance](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aearth-effect-capacitance) — Power System Analysis I, chapter 4, asked 2×
- [Inductance of composite-conductor single-phase line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Asingle-phase-composite-conductors) — Power System Analysis I, chapter 4
- [Transposition; single-phase capacitance with earth](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atransposition-single-phase-earth-capacitance) — Power System Analysis I, chapter 4

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