---
title: "Capacitance of double circuit line, 9 mm radius"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:double-circuit-capacitance-9mm"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/double-circuit-capacitance-9mm
site: BCT Engineering
---

# Capacitance of double circuit line, 9 mm radius

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2068 Magh, 6 marks
- Video: [44.  Capacitance of Double Circuit Line in Hexagonal Configuration](https://www.youtube.com/watch?v=NuFnQ1AL3YE) — Bright Engineers
- 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/double-circuit-capacitance-9mm
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

Compute the capacitance per phase per unit length for the 3-phase parallel lines shown in the figure below. Effective radius of each conductor is 9 mm. [Figure: vertical double-circuit arrangement; left column top to bottom a, b, c and right column a', b', c'; vertical spacing 3.05 m between rows; horizontal distances a–a' = 5.25 m, b–b' = 6.40 m, c–c' = 5.25 m.]

## Short answer

For the double-circuit line, mutual GMD Dm = 5.2292 m, self GMD Ds = 0.22467 m. Per-phase capacitance to neutral = (2πε0)/ln(Dm/Ds) = 17.68 pF/m = 17.68 nF/km.

## Answer

The line is transposed, so each phase pair (a, a'), (b, b'), (c, c') is treated as a two-conductor group. $C_n = 2\pi\varepsilon_0/\ln(D_m/D_s)$ with $D_s$ using the actual radius.

```text
   a o------ 5.25 m ------o a'   --
                                  3.05 m
  b o------- 6.40 m -------o b'  --
                                  3.05 m
   c o------ 5.25 m ------o c'   --
```

**Data:** $r = 0.009$ m. Horizontal offset of b from a $= (6.40-5.25)/2 = 0.575$ m.

#### Distances

| Pair | Distance (m) |
|---|---|
| ab, bc, a'b', b'c' | $\sqrt{3.05^2+0.575^2} = 3.1037$ |
| ab', a'b, bc', b'c | $\sqrt{3.05^2+5.825^2} = 6.5752$ |
| ac, a'c' | 6.10 |
| ac', a'c | $\sqrt{6.1^2+5.25^2} = 8.0481$ |
| aa', cc' | 5.25 |
| bb' | 6.40 |

#### Mutual GMD

$$
\begin{aligned}
D_{AB} &= (3.1037^2 \times 6.5752^2)^{1/4} = 4.5175\ \text{m} \\
D_{BC} &= 4.5175\ \text{m} \\
D_{CA} &= (6.10^2 \times 8.0481^2)^{1/4} = 7.0067\ \text{m} \\
D_m &= \sqrt[3]{4.5175^2 \times 7.0067} = 5.2292\ \text{m}
\end{aligned}
$$

#### Self GMD (capacitance)

$$
\begin{aligned}
D_{sA} &= \sqrt{0.009 \times 5.25} = 0.21737\ \text{m} \\
D_{sB} &= \sqrt{0.009 \times 6.40} = 0.24000\ \text{m} \\
D_{sC} &= 0.21737\ \text{m} \\
D_s &= \sqrt[3]{0.21737^2 \times 0.24} = 0.22467\ \text{m}
\end{aligned}
$$

#### Capacitance per phase

$$
\begin{aligned}
C_n &= \frac{2\pi\varepsilon_0}{\ln(D_m/D_s)} = \frac{55.63\times10^{-12}}{\ln(5.2292/0.22467)} = \frac{55.63\times10^{-12}}{3.1474} \\
&= 17.675\times10^{-12}\ \text{F/m}
\end{aligned}
$$

**Answer:** $C_n = 17.68$ pF/m $= 17.68$ nF/km per phase to neutral.

## Check yourself

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

1. In the double-circuit line, each phase is treated as:
   A. A two-conductor group (a with a', etc.)
   B. A single conductor
   C. A three-conductor bundle
   D. Independent of the other circuit
   Answer: A. The two parallel conductors of each phase act as a bundle of two. (Key point: Grouping)

2. The capacitance self-GMD of phase A is:
   A. sqrt(r' x D_aa')
   B. sqrt(r x D_aa') with the actual r
   C. r x D_aa'
   D. sqrt(r/D_aa')
   Answer: B. Capacitance uses the real radius; the two conductors are 5.25 m apart. (Key point: Self GMD)

3. D_AB is found as:
   A. Distance a-b only
   B. Arithmetic mean of ab and a'b'
   C. Geometric mean of the four distances between group A and group B
   D. Product of all 6 distances
   Answer: C. Four distances ab, ab', a'b, a'b' under a fourth root. (Key point: Mutual GMD)

4. The capacitance per phase is:
   A. 2*pi*eps0 / ln(D_s / D_m)
   B. pi*eps0 / ln(D_m / D_s)
   C. 2*pi*eps0 * ln(D_m / D_s)
   D. 2*pi*eps0 / ln(D_m / D_s)
   Answer: D. Larger D_m/D_s lowers C; this is the standard form. (Key point: Formula)

## Related in Power System Analysis I

- [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
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