---
title: "L and C of triple-bundle line, D=8 m"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:triple-bundle-18mm-45cm-8m"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/triple-bundle-18mm-45cm-8m
site: BCT Engineering
---

# L and C of triple-bundle line, D=8 m

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2071 Bhadra, 5 marks
- 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/triple-bundle-18mm-45cm-8m
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

Compute the inductance and capacitance per phase for a 200 km line using phase conductors of 18 mm radius and configured as given below. Assume complete transposition of the line. [Figure: horizontal single-circuit line; each phase is a triangular bundle of three sub-conductors with spacing d = 45 cm; centres of adjacent phases D = 8 m apart.]

## Short answer

Using Deq = 10.08 m; bundle self-GMDs for inductance 0.1416 m and capacitance 0.1539 m. Inductance = 0.853 mH/km (170.6 mH for 200 km), capacitance = 13.30 nF/km (2.66 µF for 200 km).

## Answer

Each phase is a 3-conductor bundle (equilateral triangle, side $d$). The bundle self-GMD is $\sqrt[3]{r_x d^2}$, with $r_x = r'$ for inductance and $r$ for capacitance.

```text
     o          o          o
    o o        o o        o o
     a   8 m    b   8 m    c
   d = 45 cm, r = 18 mm
```

**Data:** $r = 0.018$ m, $r' = 0.7788 \times 0.018 = 0.014018$ m, $d = 0.45$ m, $D = 8, 8, 16$ m, 200 km.

#### GMDs

$$
\begin{aligned}
D_m &= \sqrt[3]{8 \times 8 \times 16} = 10.0794\ \text{m} \\
D_{sL} &= \sqrt[3]{0.014018 \times 0.45^2} = 0.14159\ \text{m} \\
D_{sC} &= \sqrt[3]{0.018 \times 0.45^2} = 0.15390\ \text{m}
\end{aligned}
$$

#### Inductance per phase

$$
\begin{aligned}
L &= 2\times10^{-7}\ln\frac{10.0794}{0.14159} = 2\times10^{-7}\times4.2653 = 0.8531\ \text{mH/km} \\
L_{200} &= 0.8531 \times 200 = 170.61\ \text{mH}
\end{aligned}
$$

#### Capacitance per phase

$$
\begin{aligned}
C &= \frac{55.63\times10^{-12}}{\ln(10.0794/0.15390)} = \frac{55.63\times10^{-12}}{4.1820} = 13.30\ \text{nF/km} \\
C_{200} &= 13.30 \times 200 = 2660.5\ \text{nF} = 2.661\ \mu\text{F}
\end{aligned}
$$

**Answer:** $L = 170.6$ mH per phase (0.853 mH/km); $C = 2.661\ \mu$F per phase (13.30 nF/km).

## Check yourself

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

1. For a 3-conductor triangular bundle of side d, the self-GMD is:
   A. cube root of (r_x * d^3)
   B. cube root of (r_x * d^2)
   C. square root of (r_x * d)
   D. 1.09 x fourth root of (r_x d^3)
   Answer: B. Three sub-conductors give a cube root of r_x times d*d (two distances to the other sub-conductors). (Key point: Bundle self-GMD)

2. Which radius is used in the bundle GMD for inductance and for capacitance?
   A. Actual r for both
   B. r' for both
   C. r' = 0.7788r for inductance, actual r for capacitance
   D. r' for capacitance, r for inductance
   Answer: C. Internal flux affects inductance only, so r' appears in L and the true radius r in C. (Key point: r' vs r)

3. For horizontal spacing of 8 m between phases, D_m is:
   A. cube root of (8 x 8 x 16) = 10.08 m
   B. 8 m
   C. 12 m, the mean of 8 and 16
   D. cube root of (8 x 8 x 8) = 8 m
   Answer: A. The extreme phases are 16 m apart, so the transposed geometric mean is 10.08 m. (Key point: Mutual GMD)

4. How is capacitance per phase computed for the bundle?
   A. C = 2*pi*eps0 / ln(D_m / D_sL)
   B. C = 2*pi*eps0 / ln(D_m / D_sC)
   C. C = 2*pi*eps0 * ln(D_m / D_sC)
   D. C = pi*eps0 / ln(D_m/ D_sC)
   Answer: B. Capacitance uses the capacitance GMD D_sC (built from r, not r') in the denominator of the logarithm. (Key point: Capacitance formula)

## Close variants set in other exams

- [Inductance of twin-bundle line, 16 mm radius](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwin-bundle-150km-16mm) — Power System Analysis I, chapter 4, asked 3×

## Related in Power System Analysis I

- [Inductance and capacitance of 4-conductor bundle line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aquad-bundle-220kv-l-c) — Power System Analysis I, chapter 4
- [Capacitance of double circuit line, 6 mm radius](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Adouble-circuit-capacitance-6mm) — Power System Analysis I, chapter 4
- [Inductance of 4-conductor bundle line, 6 m spacing](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aquad-bundle-200km-12mm-6m) — Power System Analysis I, chapter 4, asked 2×
- [Capacitance of double circuit line, 9 mm radius](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Adouble-circuit-capacitance-9mm) — Power System Analysis I, chapter 4
- [Single-phase line capacitance, 6 mm, 1.3 m](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Asingle-phase-capacitance-6mm-1-3m) — Power System Analysis I, chapter 4
- [Inductance of composite-conductor single-phase line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Asingle-phase-composite-conductors) — Power System Analysis I, chapter 4

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- [Initial line design for 200 MW over 120 km](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aline-design-200mw-120km) — Transmission and Distribution System Design, chapter 2
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