---
title: "Inductance of transposed double circuit, 0.9 cm radius"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:double-circuit-inductance-0-9cm"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/double-circuit-inductance-0-9cm
site: BCT Engineering
---

# Inductance of transposed double circuit, 0.9 cm radius

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2080 Chaitra, 10 marks
- Video: [Inductance of three phase double circuit lines](https://www.youtube.com/watch?v=zo9HhiMnYeI) — Surumi Shafi
- 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-inductance-0-9cm
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

A part of the transposition cycle of a 3-phase double circuit line is shown in figure. Radius of each conductor is 0.9 cm. The conductors are solid copper. Find the inductance per phase per km of the line. [Figure: vertical double-circuit arrangement; left column top to bottom a, b, c with 3 m vertical spacing between rows; right column top to bottom c', b', a'; horizontal distances a–c' = 6 m, b–b' = 7 m, c–a' = 6 m.]

## Short answer

For the double-circuit line, self GMD per phase Ds = ∛(Dsa Dsb Dsc) = 0.2362 m, with Dsa = √(r' Daa') etc. Mutual GMD between phases Dm = ∛(DAB DBC DCA) = 5.074 m. Inductance L = 2×10⁻⁷ ln(Dm/Ds) = 0.614 mH/km per phase.

## Answer

For a transposed double-circuit line the inductance per phase is

$$
L = 2\times10^{-7}\ln\frac{D_m}{D_s}\ \text{H/m}
$$

Here $D_m$ is the equivalent mutual GMD between phase groups and $D_s$ is the self-GMD of each phase (its two parallel conductors). Both are worked out for the section shown. By transposition, the other sections give the same result.

#### Geometry (section shown)
The tower is symmetrical, so the middle conductors b and b' each stand $(7-6)/2 = 0.5$ m further out than the top and bottom conductors. Coordinates in metres, with c at the origin:

```text
   a (0,6) ------- 6 m ------- c' (6,6)
    |3 m
 b (-0.5,3) ------ 7 m ------- b' (6.5,3)
    |3 m
   c (0,0) ------- 6 m ------- a' (6,0)
```

| Distance | Value (m) |
|---|---|
| ab, bc, a'b', b'c' | $\sqrt{3^2+0.5^2} = \sqrt{9.25} = 3.041$ |
| ab', a'b, bc', b'c | $\sqrt{3^2+6.5^2} = \sqrt{51.25} = 7.159$ |
| ac, a'c', ac', a'c | 6, 6, 6, 6 |
| aa', cc' | $\sqrt{72} = 8.485$ |
| bb' | 7 |

#### Self-GMD
$r' = 0.7788\times0.009 = 0.007009$ m (solid copper).

$$
\begin{aligned}
D_{sa} &= \sqrt{r'\,D_{aa'}} = \sqrt{0.007009\times8.485} = 0.2439\ \text{m}\\
D_{sb} &= \sqrt{r'\,D_{bb'}} = \sqrt{0.007009\times7} = 0.2215\ \text{m}\\
D_{sc} &= \sqrt{r'\,D_{cc'}} = 0.2439\ \text{m}\\
D_s &= \sqrt[3]{0.2439\times0.2215\times0.2439} = 0.2362\ \text{m}
\end{aligned}
$$

#### Mutual GMD
$$
\begin{aligned}
D_{AB} &= \sqrt[4]{D_{ab}D_{ab'}D_{a'b}D_{a'b'}} = \sqrt[4]{3.041\times7.159\times7.159\times3.041} = 4.666\ \text{m}\\
D_{BC} &= \sqrt[4]{D_{bc}D_{bc'}D_{b'c}D_{b'c'}} = \sqrt[4]{3.041\times7.159\times7.159\times3.041} = 4.666\ \text{m}\\
D_{CA} &= \sqrt[4]{6\times6\times6\times6} = 6\ \text{m}\\
D_m &= \sqrt[3]{4.666\times4.666\times6} = 5.074\ \text{m}
\end{aligned}
$$

#### Inductance
$$
\begin{aligned}
L &= 2\times10^{-7}\ln\frac{5.074}{0.2362} = 2\times10^{-7}\times3.0673\\
&= 0.6135\times10^{-6}\ \text{H/m} = 0.6135\ \text{mH/km per phase}
\end{aligned}
$$

At 50 Hz this gives $X_L = 314.16\times0.6135\times10^{-3} = 0.193\ \Omega$/km per phase.

**Answer:** $D_s = 0.236$ m, $D_m = 5.074$ m, and the inductance is **0.614 mH per phase per km**.

## Check yourself

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

1. In L = 2e-7 ln(Dm/Ds) for the double-circuit line, what is Ds?
   A. Self-GMD of each phase (its two parallel conductors)
   B. Distance between the two circuits
   C. Radius of one conductor
   D. Height of the tower
   Answer: A. Each phase has two paralleled conductors, so Ds is the geometric mean of their self and mutual GMRs, averaged over phases. (Key point: Formula roles)

2. How is Dsa computed for phase a with conductors a and a'?
   A. sqrt(r' x Daa')
   B. r' + Daa'
   C. Daa'
   D. sqrt(r x Dab)
   Answer: A. Two conductors give a square root of r' times their separation, e.g. sqrt(0.007009 x 8.485) = 0.2439 m. (Key point: Phase self-GMD)

3. How is the GMD between phase groups A and B found?
   A. Fourth root of the four distances ab, ab', a'b, a'b'
   B. Average of ab and a'b'
   C. Smallest of the four distances
   D. Sum of the four distances
   Answer: A. Each phase has two conductors so there are four conductor-to-conductor distances, giving a fourth root. (Key point: Mutual GMD between phases)

4. What are Dm, Ds and the inductance per phase per km?
   A. Dm = 5.093 m, Ds = 0.236 m, L = 0.614 mH/km
   B. Dm = 0.236 m, Ds = 5.093 m, L = 0.614 mH/km
   C. Dm = 5.093 m, Ds = 0.236 m, L = 6.14 mH/km
   D. Dm = 4.692 m, Ds = 0.007 m, L = 1.1 mH/km
   Answer: A. L = 2e-7 ln(5.093/0.2362) = 0.614e-6 H/m = 0.614 mH/km. (Key point: Final result)

## Related in Power System Analysis I

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