---
title: "Inductance and capacitance of 4-conductor bundle line"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:quad-bundle-220kv-l-c"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/quad-bundle-220kv-l-c
site: BCT Engineering
---

# Inductance and capacitance of 4-conductor bundle line

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2078 Chaitra, 10 marks
- Video: [UNIT-1 LECTURE 20.Problem on bundled conductor](https://www.youtube.com/watch?v=iOtuS6JCNoI) — M. V. Sudarshan PS-2
- 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/quad-bundle-220kv-l-c
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

A 220 kV, 50 Hz, 200 km long 3-phase line has conductors spaced as shown in the figure below. Compute the inductance and capacitance per phase. Each sub-conductor is of 12 mm radius. Assume transposed configuration of the line. [Figure: horizontal single-circuit line; each phase A, B, C is a square bundle of four sub-conductors (1,2,3,4) with 45 cm sides; centres of adjacent phases 6 m apart.]

## Short answer

Mutual GMD = 7.5595 m. Inductance: bundle GMR using r'=0.009346 m is 0.18629 m, L = 2e-7 ln(7.5595/0.18629)=0.7407 mH/km. Capacitance: bundle GMR using r=0.012 m is0.19830 m, C = 2*pi*epsilon0/ln(7.5595/0.19830)=15.28 nF/km. For 200 km, L=148.1 mH, C=3.056 µF.

## Answer

Each phase is a 4-conductor square bundle, so we find the bundle GMR (self-GMD) and the mutual GMD between phase centres, then use the standard formulas for a transposed line.

```text
   o o        o o        o o
   o o        o o        o o
    A    6 m   B    6 m   C
   |<------- 12 m ------->|
 square side d = 0.45 m, r = 12 mm
```

**Data:** $r = 0.012$ m, $d = 0.45$ m, $D_{AB} = D_{BC} = 6$ m, $D_{CA} = 12$ m, length 200 km.

#### Mutual GMD

$$
D_m = \sqrt[3]{6 \times 6 \times 12} = 7.5595\ \text{m}
$$

#### Self GMD of a 4-conductor bundle

For a square bundle, $D_s = (r_x \cdot d \cdot d \cdot \sqrt{2}d)^{1/4} = 1.09\,(r_x d^3)^{1/4}$.

- Inductance: $r' = 0.7788 \times 0.012 = 0.009346$ m
$$
D_{sL} = 1.09\,(0.009346 \times 0.45^3)^{1/4} = 0.18629\ \text{m}
$$
- Capacitance (uses actual radius $r$):
$$
D_{sC} = 1.09\,(0.012 \times 0.45^3)^{1/4} = 0.19830\ \text{m}
$$

#### Inductance per phase

$$
\begin{aligned}
L &= 2\times10^{-7}\ln\frac{D_m}{D_{sL}} = 2\times10^{-7}\ln\frac{7.5595}{0.18629} \\
&= 2\times10^{-7}\times 3.7033 = 7.4065\times10^{-7}\ \text{H/m} \\
&= 0.7407\ \text{mH/km}
\end{aligned}
$$

For 200 km: $L = 0.7407 \times 200 = 148.13$ mH.

#### Capacitance per phase (to neutral)

$$
\begin{aligned}
C &= \frac{2\pi\varepsilon_0}{\ln(D_m/D_{sC})} = \frac{2\pi \times 8.854\times10^{-12}}{\ln(7.5595/0.19830)} \\
&= \frac{55.63\times10^{-12}}{3.6408} = 15.28\times10^{-12}\ \text{F/m} \\
&= 15.28\ \text{nF/km}
\end{aligned}
$$

For 200 km: $C = 15.28 \times 200 = 3056$ nF $= 3.056\ \mu$F.

**Answer:** $L = 0.7407$ mH/km per phase (148.13 mH for 200 km); $C = 15.28$ nF/km per phase to neutral (3.056 µF for 200 km).

## Check yourself

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

1. What is the self-GMD of a four-conductor square bundle with side d?
   A. 1.09 (r d^3)^(1/4)
   B. sqrt(r d)
   C. (r d^2)^(1/3)
   D. 1.09 (r d)^(1/2)
   Answer: A. The product r x d x d x sqrt(2) d, fourth-rooted, equals 1.09 (r d^3)^(1/4). (Key point: Quad bundle GMD)

2. Which radius goes into the bundle GMD for inductance and for capacitance?
   A. Inductance uses r' = 0.7788 r; capacitance uses actual r
   B. Both use r'
   C. Both use actual r
   D. Inductance uses actual r; capacitance uses r'
   Answer: A. Internal flux adds to inductance (GMR), whereas capacitance depends only on the physical surface radius. (Key point: r' vs r)

3. With phase spacing 6 m, 6 m and 12 m, Dm is:
   A. 7.56 m
   B. 8 m
   C. 6 m
   D. 12 m
   Answer: A. Cube root of (6 x 6 x 12) = 7.5595 m. (Key point: Mutual GMD)

4. Which pair gives the per-phase L and C per km?
   A. 0.7407 mH/km and 15.28 nF/km
   B. 7.407 mH/km and 1.528 nF/km
   C. 0.7407 mH/km and 1.528 nF/km
   D. 0.1863 mH/km and 15.28 nF/km
   Answer: A. L = 2e-7 ln(7.5595/0.18629) = 0.7407 mH/km and C = 55.63e-12/3.6408 = 15.28 nF/km. (Key point: Results)

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