---
title: "Inductance of 3-conductor bundled line, 6.1 m spacing"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:inductance-three-subconductor-bundle-6-1m"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/inductance-three-subconductor-bundle-6-1m
site: BCT Engineering
---

# Inductance of 3-conductor bundled line, 6.1 m spacing

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2082 Kartik (new course), 8 marks
- Video: [Power Systems: Inductance of Three-Phase Transmission Line (with 3-bundled conductors/phase) TAGALOG](https://www.youtube.com/watch?v=muFGHiPEZ9A) — Engr. Aaron Pineda
- 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/inductance-three-subconductor-bundle-6-1m
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

A 3-phase single circuit bundled conductor line with three sub conductors per phase has a horizontal configuration with spacing of 6.1 m between the center lines of adjacent phases as shown in figure. The distance between the sub conductors of each phase is 30.5 cm and each sub conductor has a diameter of 2.54 cm. Find the inductance per phase per km. [Figure: three phases side by side in a horizontal row, each a triangular bundle of three sub-conductors 30.5 cm apart; phase centres 6.1 m apart, outer phases 12.2 m apart.]

## Short answer

For a three-subconductor bundle per phase, the self GMD of the bundle is Ds^b = ∛(r' d²) = 0.09726 m, with r' = 0.009891 m and d = 0.305 m. The mutual GMD for the horizontal transposed line is Dm = ∛(6.1×6.1×12.2) = 7.686 m. Inductance L = 2×10⁻⁷ ln(Dm/Ds^b) = 0.874 mH/km per phase.

## Answer

For a bundled, transposed line the inductance per phase is

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

Here $D_s^b$ is the self-GMD (GMR) of the bundle and $D_m$ is the mutual GMD between phases. The bundle spacing (0.305 m) is much smaller than the phase spacing, so distances between phases are taken centre to centre. The line is assumed to be fully transposed.

#### Data
- Sub-conductor radius: $r = 2.54/2 = 1.27$ cm $= 0.0127$ m
- GMR of a solid sub-conductor: $r' = 0.7788r = 0.7788\times0.0127 = 0.009891$ m
- Sub-conductor spacing: $d = 0.305$ m (equilateral triangle)

#### Self-GMD of the three-conductor bundle
$$
\begin{aligned}
D_s^b &= \sqrt[9]{(r'\,d\,d)^3} = \sqrt[3]{r'd^2}\\
&= \sqrt[3]{0.009891\times0.305^2}\\
&= 0.09726\ \text{m}
\end{aligned}
$$

#### Mutual GMD
Horizontal arrangement: $D_{ab} = D_{bc} = 6.1$ m and $D_{ca} = 12.2$ m.
$$
D_m = \sqrt[3]{6.1\times6.1\times12.2} = 6.1\times\sqrt[3]{2} = 7.686\ \text{m}
$$

#### Inductance
$$
\begin{aligned}
L &= 2\times10^{-7}\ln\frac{7.686}{0.09726}\ \text{H/m}\\
&= 2\times10^{-7}\times\ln(79.02) = 2\times10^{-7}\times4.3697\\
&= 0.8739\times10^{-6}\ \text{H/m} = 0.8739\ \text{mH/km}
\end{aligned}
$$

If the frequency is 50 Hz, the reactance is $X_L = 2\pi\times50\times0.8739\times10^{-3} = 0.2746\ \Omega$/km per phase.

| Quantity | Value |
|---|---|
| $r'$ | 0.009891 m |
| $D_s^b$ | 0.09726 m |
| $D_m$ | 7.686 m |
| L | 0.8739 mH/km per phase |

**Answer:** $L \approx 0.874$ mH per phase per km ($X_L \approx 0.275\ \Omega$/km at 50 Hz).

## Check yourself

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

1. Which formula gives inductance per phase of the bundled transposed line?
   A. L = 2e-7 ln(Dm/Ds_bundle) H/m
   B. L = 2e-7 ln(Ds_bundle/Dm) H/m
   C. L = 4e-7 ln(Dm/r) H/m
   D. L = 2e-7 ln(Dm/r) H/m using the actual radius of one sub-conductor
   Answer: A. Inductance uses the mutual GMD between phases over the bundle's self-GMD (GMR). (Key point: Formula)

2. For a three-conductor (triangular) bundle with sub-conductor GMR r' and spacing d, the bundle Ds is:
   A. sqrt(r' d)
   B. cube root of (r' d^2)
   C. 1.09 fourth root of (r' d^3)
   D. r' d
   Answer: B. Three sub-conductors in an equilateral triangle give Ds = (r' d^2)^(1/3) = 0.09726 m here. (Key point: Bundle GMR)

3. Phases are 6.1 m, 6.1 m and 12.2 m apart. What is Dm?
   A. 6.1 m
   B. 7.686 m
   C. 8.13 m
   D. 12.2 m
   Answer: B. Dm = cube root of (6.1 x 6.1 x 12.2) = 6.1 x cube root of 2 = 7.686 m. (Key point: Mutual GMD)

4. What is the inductance per phase per km?
   A. About 0.874 mH/km
   B. About 1.25 mH/km
   C. About 0.437 mH/km
   D. About 8.74 mH/km
   Answer: A. L = 2e-7 ln(7.686/0.09726) = 2e-7 x 4.3697 = 0.874e-6 H/m = 0.874 mH/km. (Key point: Final value)

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