---
title: "Inductance of unsymmetrical transposed line"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:unsymmetrical-inductance-10mm"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/unsymmetrical-inductance-10mm
site: BCT Engineering
---

# Inductance of unsymmetrical transposed line

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2068 Bhadra, 5 marks
- Video: [Inductance calculations for unsymmetrical 3-phase system](https://www.youtube.com/watch?v=T3d27nmIoVI) — Upendra Bhusan
- 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/unsymmetrical-inductance-10mm
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

A 3-phase, 50 Hz, overhead line has three phase conductors with the following configuration. Compute the inductance per phase per km if each of the phase conductors is round solid of 10 mm diameter. Assume completely transposed line. [Figure: conductors a, b, c at the corners of a triangle with Dab = 4.5 m, Dbc = 5.5 m, Dca = 7.5 m.]

## Short answer

Deq = 5.7044 m, r' = 0.7788×5 = 3.894 mm. Per-phase inductance L = 2×10⁻⁷ ln(Deq/r') = 1.458 mH/km.

## Answer

For a completely transposed line with unsymmetrical spacing, the inductance per phase is $L = 2\times10^{-7}\ln(D_{eq}/r')$ H/m, where $D_{eq} = \sqrt[3]{D_{ab}D_{bc}D_{ca}}$.

```text
          a
         / \
   4.5 m/   \7.5 m
       /     \
      b-------c
        5.5 m
```

**Data:** diameter 10 mm, $r = 5$ mm $= 0.005$ m, $r' = 0.7788 \times 0.005 = 0.003894$ m.

#### Equivalent spacing

$$
D_{eq} = \sqrt[3]{4.5 \times 5.5 \times 7.5} = \sqrt[3]{185.625} = 5.7044\ \text{m}
$$

#### Inductance per phase

$$
\begin{aligned}
L &= 2\times10^{-7}\ln\frac{5.7044}{0.003894} = 2\times10^{-7}\times7.2896 \\
&= 1.4579\times10^{-6}\ \text{H/m} = 1.458\ \text{mH/km}
\end{aligned}
$$

**Answer:** $L = 1.458$ mH per phase per km.

## Check yourself

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

1. With D_ab = 4.5, D_bc = 5.5 and D_ca = 7.5 m, D_eq is:
   A. cube root of (4.5 x 5.5 x 7.5) = 5.70 m
   B. Average 5.83 m
   C. 7.5 m
   D. 4.5 m
   Answer: A. Geometric mean for a transposed line. (Key point: D_eq)

2. Inductance per phase per metre is:
   A. 2e-7 ln(D_eq / r)
   B. 2e-7 ln(r' / D_eq)
   C. 2e-7 ln(D_eq / r') with r' = 0.7788r
   D. 2e-7 ln(D_eq) r'
   Answer: C. r' accounts for internal flux linkage. (Key point: Formula)

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