---
title: "Capacitance depends on conductor size and spacing"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:capacitance-depends-size-spacing"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/capacitance-depends-size-spacing
site: BCT Engineering
---

# Capacitance depends on conductor size and spacing

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2080 Chaitra, 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-depends-size-spacing
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

Show that the capacitance per phase of a three-phase overhead transmission line is primarily determined by the conductor size and conductor spacings.

## Short answer

For a transposed three-phase line, capacitance per phase to neutral is C = 2*pi*epsilon0 / ln(Deq/r). This expression contains only the conductor radius r and equivalent spacing Deq, with epsilon0 constant. Hence conductor size and spacing are the primary factors; voltage, current and frequency do not appear.

## Answer

The capacitance to neutral of a transposed three-phase line is $C_n = \frac{2\pi\varepsilon_0}{\ln(D_{eq}/r)}$. It contains only the conductor radius r and the spacings (through $D_{eq}$), apart from the constant $\varepsilon_0$. The derivation below shows this.

#### Potential difference between two conductors
For a long conductor with charge q C/m, the potential difference between two points at distances $D_1$ and $D_2$ is $V_{12} = \frac{q}{2\pi\varepsilon_0}\ln\frac{D_2}{D_1}$. For a group of conductors, using superposition:
$$
V_{ab} = \frac{1}{2\pi\varepsilon_0}\left[q_a\ln\frac{D_{ab}}{r} + q_b\ln\frac{r}{D_{ab}} + q_c\ln\frac{D_{bc}}{D_{ca}}\right]
$$

#### Equilateral spacing D
$$
\begin{aligned}
V_{ab} &= \frac{1}{2\pi\varepsilon_0}\left[q_a\ln\frac{D}{r} + q_b\ln\frac{r}{D}\right]\\
V_{ac} &= \frac{1}{2\pi\varepsilon_0}\left[q_a\ln\frac{D}{r} + q_c\ln\frac{r}{D}\right]
\end{aligned}
$$

Adding and using $q_b + q_c = -q_a$ (balanced charges):
$$
V_{ab} + V_{ac} = \frac{3q_a}{2\pi\varepsilon_0}\ln\frac{D}{r}
$$

For a balanced system, $V_{ab} + V_{ac} = 3V_{an}$. So
$$
V_{an} = \frac{q_a}{2\pi\varepsilon_0}\ln\frac{D}{r}, \qquad C_n = \frac{q_a}{V_{an}} = \frac{2\pi\varepsilon_0}{\ln(D/r)}\ \text{F/m}
$$

#### Unsymmetrical spacing (transposed)
Averaging $V_{ab}$ over the three transposition sections gives the same form, with D replaced by the geometric mean spacing:
$$
C_n = \frac{2\pi\varepsilon_0}{\ln\dfrac{D_{eq}}{r}}, \qquad D_{eq} = \sqrt[3]{D_{ab}D_{bc}D_{ca}}
$$

For bundled conductors, r is replaced by the bundle's self-GMD for capacitance, for example $\sqrt{rd}$.

#### Conclusion
- $\varepsilon_0$ is a constant (air). So C depends only on the **ratio $D_{eq}/r$**, that is, on conductor size and spacing.
- Larger r (or bundling) gives higher C. Larger spacing gives lower C. The dependence is logarithmic, so a big change in geometry causes only a modest change in C.
- Current, voltage and frequency do not appear in C. The earth's effect (images) is small for normal conductor heights and is neglected. That is why size and spacing are the **primary** factors.

Example: r = 1 cm and D = 4 m give $C_n = \frac{5.563\times10^{-11}}{\ln 400} = 9.29$ nF/km. Doubling D to 8 m gives 8.32 nF/km.

## Check yourself

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

1. What is the capacitance to neutral of a transposed three-phase line?
   A. 2 pi eps0 / ln(Deq/r)
   B. pi eps0 / ln(Deq/r)
   C. 2 pi eps0 ln(Deq/r)
   D. 2 pi eps0 / ln(r/Deq)
   Answer: A. Only radius r and the spacings (through Deq) appear, besides eps0. (Key point: Capacitance formula)

2. Which relation gives V_an from V_ab and V_ac for balanced charges?
   A. V_ab + V_ac = 3 V_an
   B. V_ab - V_ac = 3 V_an
   C. V_ab + V_ac = V_an
   D. V_ab V_ac = V_an
   Answer: A. For a balanced system the sum equals 3 V_an, using q_b + q_c = -q_a. (Key point: Key step)

3. How does C change with r and D?
   A. Larger r raises C, larger D lowers C, only logarithmically
   B. Larger r lowers C, larger D raises C
   C. C is linear in D
   D. C depends on voltage and frequency
   Answer: A. C depends on 1/ln(Deq/r), so it varies weakly (log) with geometry. (Key point: Trend with geometry)

4. For capacitance of a two-conductor bundle, what replaces r?
   A. sqrt(r d), using the actual radius
   B. sqrt(r' d), using the GMR
   C. d
   D. 2r
   Answer: A. The bundle self-GMD for capacitance uses the actual radius r, e.g. sqrt(r d). (Key point: Bundled conductors)

## Related in Power System Analysis I

- [Inductance of 3-conductor bundled line, 6.1 m spacing](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ainductance-three-subconductor-bundle-6-1m) — 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
- [Inductance of 4-conductor bundle line, 6 m spacing](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aquad-bundle-200km-12mm-6m) — 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
- [Capacitance of double circuit line, 6 mm radius](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Adouble-circuit-capacitance-6mm) — Power System Analysis I, chapter 4

## Related in other subjects

- [Capacitance matrix of 400 kV bundled line](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Acapacitance-matrix-400kv-15m) — High Voltage Engineering, chapter 5, asked 2×
- [Minimum conductor spacing for 230 kV corona limit](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Acorona-spacing-3cm-230kv) — High Voltage Engineering, chapter 5, asked 4×
- [Fault at load end of overhead line](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-load-end-33kv-line) — Power System Analysis II, chapter 3
- [6° phase shift per 100 km of lossless line](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Aphase-shift-6-degree-100km) — High Voltage Engineering, chapter 3