---
title: "Inductance of composite-conductor single-phase line"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:single-phase-composite-conductors"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/single-phase-composite-conductors
site: BCT Engineering
---

# Inductance of composite-conductor single-phase line

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2076 Baisakh, 10 marks
- Video: [Transmission Line Composite Conductors](https://www.youtube.com/watch?v=eeKN_HJATKU) — Wisdom is Power
- 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/single-phase-composite-conductors
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

One circuit of a single-phase transmission line is composed of three solid 0.25-cm-radius wires. The return circuit is composed of two 0.5-cm-radius wires. The arrangement of the conductors is shown below. Find the inductance due to the current in each side of the line and the inductance of the complete line in henrys per metre. [Figure: side X has conductors a, b, c in a vertical line, 6 m apart (a–b = 6 m, b–c = 6 m); side Y has conductors d and e, 9 m to the right of the side X column, d level with a and e level with b.]

## Short answer

Compute self GMD of each side using r' = 0.7788r and mutual GMD from distances. L_X = 6.212×10^-7 H/m, L_Y = 8.505×10^-7 H/m, total loop inductance L = 1.472×10^-6 H/m.

## Answer

Each side of the line is a composite conductor. The inductance of side X is $L_X = 2\times10^{-7}\ln(D_m/D_{sX})$, and similarly for side Y; the loop inductance is $L = L_X + L_Y$.

```text
  side X           side Y
   a o------9 m------o d
     |6 m
   b o------9 m------o e
     |6 m
   c o
```

**Data:** side X: $r = 0.25$ cm, $r' = 0.7788 \times 0.0025 = 0.001947$ m. Side Y: $r = 0.5$ cm, $r' = 0.003894$ m.

#### Distances

| Pair | Distance (m) |
|---|---|
| ad, be | 9 |
| ae, bd, ce | $\sqrt{9^2+6^2} = 10.817$ |
| cd | $\sqrt{9^2+12^2} = 15$ |
| ab, bc (X) | 6 |
| ac (X) | 12 |
| de (Y) | 6 |

#### Mutual GMD (6 distances)

$$
\begin{aligned}
D_m &= (D_{ad}D_{ae}D_{bd}D_{be}D_{cd}D_{ce})^{1/6} \\
&= (9 \times 10.817 \times 10.817 \times 9 \times 15 \times 10.817)^{1/6} = 10.743\ \text{m}
\end{aligned}
$$

#### Self GMD of side X ($3^2 = 9$ terms)

$$
\begin{aligned}
D_{sX} &= \left[(r')^3 \times 6^4 \times 12^2\right]^{1/9} \\
&= \left[0.001947^3 \times 6^4 \times 12^2\right]^{1/9} = 0.4810\ \text{m}
\end{aligned}
$$

#### Self GMD of side Y ($2^2 = 4$ terms)

$$
D_{sY} = \left[(0.003894)^2 \times 6^2\right]^{1/4} = \sqrt{0.003894 \times 6} = 0.15285\ \text{m}
$$

#### Inductances

$$
\begin{aligned}
L_X &= 2\times10^{-7}\ln\frac{10.743}{0.4810} = 6.212\times10^{-7}\ \text{H/m} \\
L_Y &= 2\times10^{-7}\ln\frac{10.743}{0.15285} = 8.505\times10^{-7}\ \text{H/m} \\
L &= L_X + L_Y = 14.718\times10^{-7}\ \text{H/m}
\end{aligned}
$$

**Answer:** $L_X = 6.212\times10^{-7}$ H/m, $L_Y = 8.505\times10^{-7}$ H/m, complete line $L = 1.472\times10^{-6}$ H/m.

## Check yourself

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

1. How is the inductance of the complete line obtained from the two composite sides X and Y?
   A. L = L_X x L_Y
   B. L = L_X + L_Y
   C. L = (L_X + L_Y)/2
   D. L = L_X L_Y/(L_X + L_Y)
   Answer: B. The loop inductance is the sum L = L_X + L_Y, each found from 2e-7 ln(D_m/D_s). (Key point: Loop inductance)

2. Side X has 3 conductors and side Y has 2. How many distances make up the mutual GMD D_m?
   A. 5
   B. 6
   C. 9
   D. 4
   Answer: B. D_m is the (3 x 2) = 6th root of the product of the 6 distances between X and Y conductors. (Key point: Number of terms in mutual GMD)

3. The self-GMD of side X (3 strands) is the root of what order, and which radius is used?
   A. 9th root, with r' = 0.7788r
   B. 3rd root, with r
   C. 9th root, with actual radius r
   D. 6th root, with r'
   Answer: A. D_sX uses n^2 = 9 terms, with each self distance r' because inductance is being found. (Key point: Self GMD of side X)

4. L_X = 6.212e-7 and L_Y = 8.505e-7 H/m. Why is L_Y larger?
   A. Side Y has a smaller D_s (0.153 m versus 0.481 m), so ln(D_m/D_s) is larger
   B. Side Y has a larger radius so it has more inductance
   C. Side Y carries more current
   D. D_m is larger for side Y
   Answer: A. D_m is common, and the smaller self-GMD of side Y gives the larger inductance, despite the larger wire radius. (Key point: Which side has larger L)

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