---
title: "400 kV line with 3-conductor bundles: reactances and VAR"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:triple-bundle-400kv-reactances"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/triple-bundle-400kv-reactances
site: BCT Engineering
---

# 400 kV line with 3-conductor bundles: reactances and VAR

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2074 Bhadra, 10 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-400kv-reactances
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

Determine the inductive and capacitive reactance, charging current and charging VAR per phase per km of the 400 kV single circuit, 3-φ 50 Hz fully transposed transmission line having three bundle conductors per phase with spacing of sub conductors as d = 40 cm and radius of each sub conductor is 20 mm as shown in figure below. [Figure: horizontal single-circuit line; each phase is a triangular bundle of three sub-conductors with 40 cm sides; centres of adjacent phases 10 m apart.]

## Short answer

Compute bundle self GMDs: D_sL = (r' d²)^(1/3) = 0.1356 m, D_sC = (r d²)^(1/3) = 0.1474 m, D_m = 12.599 m. Then X_L = 0.2847 Ω/km, X_C = 2.545×10^5 Ω·km, I_c = 0.907 A/km, Q_c = 209.5 kVAR/km per phase.

## Answer

Each phase is a 3-conductor bundle at the corners of an equilateral triangle of side $d$. We find the bundle self-GMD for inductance and capacitance, the mutual GMD between phases, and then the reactances.

```text
     o          o          o
    o o        o o        o o
     A   10 m   B   10 m   C
   d = 40 cm, r = 20 mm
```

**Data:** $r = 0.02$ m, $r' = 0.7788 \times 0.02 = 0.015576$ m, $d = 0.4$ m, $D = 10, 10, 20$ m, 400 kV, 50 Hz.

#### GMDs

$$
\begin{aligned}
D_m &= \sqrt[3]{10 \times 10 \times 20} = 12.599\ \text{m} \\
D_{sL} &= \sqrt[3]{r'd^2} = \sqrt[3]{0.015576 \times 0.4^2} = 0.13558\ \text{m} \\
D_{sC} &= \sqrt[3]{rd^2} = \sqrt[3]{0.02 \times 0.4^2} = 0.14736\ \text{m}
\end{aligned}
$$

#### Inductive reactance per km

$$
\begin{aligned}
L &= 2\times10^{-7}\ln\frac{12.599}{0.13558} = 2\times10^{-7}\times4.5318 = 0.9064\ \text{mH/km} \\
X_L &= 2\pi f L = 2\pi \times 50 \times 0.9064\times10^{-3} = 0.2847\ \Omega/\text{km}
\end{aligned}
$$

#### Capacitive reactance per km

$$
\begin{aligned}
C &= \frac{55.63\times10^{-12}}{\ln(12.599/0.14736)} = \frac{55.63\times10^{-12}}{4.4485} = 12.506\ \text{nF/km} \\
X_C &= \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 12.506\times10^{-9}} = 2.545\times10^5\ \Omega\text{-km}
\end{aligned}
$$

#### Charging current and VAR per phase per km

$$
\begin{aligned}
V_p &= 400/\sqrt{3} = 230.94\ \text{kV} \\
I_c &= \frac{V_p}{X_C} = \frac{230\,940}{254\,533} = 0.9073\ \text{A/km} \\
Q_c &= V_p I_c = 230.94 \times 0.9073 = 209.53\ \text{kVAR/km per phase}
\end{aligned}
$$

(Three phases together: 628.6 kVAR/km.)

**Answer:** $X_L = 0.2847\ \Omega$/km, $X_C = 2.545\times10^5\ \Omega$-km, $I_c = 0.907$ A/km, $Q_c = 209.5$ kVAR/km per phase.

## Check yourself

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

1. Which expressions give D_s for a three-conductor bundle for L and for C?
   A. cube root of (r' d^2) and cube root of (r d^2)
   B. sqrt(r' d) and sqrt(r d)
   C. 1.09 x 4th root(r' d^3) for both
   D. cube root of (r d^2) and cube root of (r' d^2)
   Answer: A. For an equilateral 3-bundle, D_sL = (r' d^2)^(1/3) and D_sC = (r d^2)^(1/3). (Key point: 3-bundle self-GMD)

2. With L = 0.9064 mH/km, X_L at 50 Hz is:
   A. 0.2847 ohm/km
   B. 2.847 ohm/km
   C. 0.0285 ohm/km
   D. 0.9064 ohm/km
   Answer: A. X_L = 2*pi x 50 x 0.9064e-3 = 0.2847 ohm/km. (Key point: Inductive reactance)

3. How is the charging current per phase found?
   A. I_c = V_line / X_C
   B. I_c = V_phase / X_C with V_phase = 400/sqrt(3) kV
   C. I_c = V_phase x X_C
   D. I_c = 3 V_phase / X_C
   Answer: B. I_c = V_p/X_C = 230.94 kV/254,533 = 0.907 A/km. (Key point: Charging current)

4. The per-phase charging VAR per km is:
   A. Q = V_p I_c = about 209.5 kVAR/km
   B. Q = V_L I_c = about 363 kVAR/km
   C. Q = I_c^2 / X_C
   D. Q = about 628.6 kVAR/km
   Answer: A. Q per phase = 230.94 kV x 0.9073 A = 209.5 kVAR/km; 628.6 kVAR/km is for all three phases. (Key point: Charging VAR per phase)

## Close variants set in other exams

- [400 kV twin-bundle line: reactances and charging VAR](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atwin-bundle-400kv-reactances) — Power System Analysis I, chapter 4, asked 2×
- [Capacitive reactance and charging of 2-conductor bundle line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Abundle-capacitance-9mm-40cm-10m) — Power System Analysis I, chapter 4, asked 2×

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