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

# 400 kV twin-bundle line: reactances and charging VAR

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2073 Magh, 8 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/twin-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 and charging VAR per phase per km of a 400 kV single circuit, 3-φ, 50 Hz transposed transmission line having two bundle conductors per phase with spacing of sub conductors as d = 40 cm and radius of each sub-conductor is 20 mm. [Figure: horizontal single-circuit line, phases A, B, C each a two-conductor bundle (a–a', b–b', c–c') with 40 cm between sub-conductors; 10 m between centres of adjacent phases.]

## Short answer

With D_sL = √(r′d) = 0.07893 m, D_sC = √(rd) = 0.08944 m, mutual GMD = 12.599 m, the per km per phase inductive reactance is 0.3187 Ω/km, capacitive reactance 2.831×10⁵ Ω·km, and charging reactive power 188.4 kVAR/km.

## Answer

Each phase is a two-conductor bundle. The self-GMD of a 2-bundle is $\sqrt{r'd}$ for inductance and $\sqrt{rd}$ for capacitance.

```text
   o--o        o--o        o--o
   a  a'       b  b'       c  c'
    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

$$
\begin{aligned}
D_m &= \sqrt[3]{10 \times 10 \times 20} = 12.599\ \text{m} \\
D_{sL} &= \sqrt{0.015576 \times 0.4} = 0.07893\ \text{m} \\
D_{sC} &= \sqrt{0.02 \times 0.4} = 0.08944\ \text{m}
\end{aligned}
$$

#### Inductive reactance per phase per km

$$
\begin{aligned}
L &= 2\times10^{-7}\ln\frac{12.599}{0.07893} = 2\times10^{-7}\times5.0728 = 1.0146\ \text{mH/km} \\
X_L &= 2\pi \times 50 \times 1.0146\times10^{-3} = 0.3187\ \Omega/\text{km}
\end{aligned}
$$

#### Capacitive reactance per phase per km

$$
\begin{aligned}
C &= \frac{55.63\times10^{-12}}{\ln(12.599/0.08944)} = \frac{55.63\times10^{-12}}{4.9478} = 11.244\ \text{nF/km} \\
X_C &= \frac{1}{2\pi \times 50 \times 11.244\times10^{-9}} = 2.831\times10^5\ \Omega\text{-km}
\end{aligned}
$$

#### Charging VAR per phase per km

$$
\begin{aligned}
V_p &= 400/\sqrt{3} = 230.94\ \text{kV} \\
I_c &= V_p/X_C = 230\,940/283\,101 = 0.8158\ \text{A/km} \\
Q_c &= \frac{V_p^2}{X_C} = \frac{(230\,940)^2}{283\,101} = 188.39\ \text{kVAR/km}
\end{aligned}
$$

**Answer:** $X_L = 0.3187\ \Omega$/km, $X_C = 2.831\times10^5\ \Omega$-km, charging $Q_c = 188.4$ kVAR per phase per km (565.2 kVAR/km for all three phases).

## Check yourself

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

1. What are D_sL and D_sC for the twin bundle (d = 0.4 m, r = 0.02 m)?
   A. sqrt(r' d) = 0.0789 m and sqrt(r d) = 0.0894 m
   B. r' = 0.0156 m and r = 0.02 m
   C. sqrt(r' d) = 0.0894 m and sqrt(r d) = 0.0789 m
   D. 0.4 m for both
   Answer: A. D_sL = sqrt(0.015576 x 0.4) = 0.07893 m and D_sC = sqrt(0.02 x 0.4) = 0.08944 m. (Key point: 2-bundle self-GMD)

2. With L = 1.0146 mH/km, X_L is about:
   A. 0.3187 ohm/km
   B. 3.187 ohm/km
   C. 0.0319 ohm/km
   D. 1.0146 ohm/km
   Answer: A. X_L = 2*pi x 50 x 1.0146e-3 = 0.3187 ohm/km. (Key point: Inductive reactance)

3. Which formula gives the charging VAR per phase?
   A. Q = V_p^2 / X_C = 188.4 kVAR/km
   B. Q = V_L^2 / X_C per phase
   C. Q = V_p x X_C
   D. Q = I_c / V_p
   Answer: A. Q per phase = V_p^2/X_C with V_p = 230.94 kV, giving 188.4 kVAR/km (565.2 for three phases). (Key point: Charging VAR formula)

4. Compared with the twin bundle, a single conductor of the same radius would have:
   A. Smaller X_L and larger charging VAR
   B. Larger X_L and smaller charging VAR
   C. Same X_L and VAR
   D. Smaller X_L and smaller VAR
   Answer: B. Without bundling D_s is smaller, so L is larger and C (hence charging VAR) is smaller. (Key point: Compare with single conductor)

## Close variants set in other exams

- [400 kV line with 3-conductor bundles: reactances and VAR](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atriple-bundle-400kv-reactances) — Power System Analysis I, chapter 4, asked 3×

## Related in Power System Analysis I

- [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×
- [400 kV quadruple bundle line, 20 m spacing](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aquad-bundle-400kv-20m) — Power System Analysis I, chapter 4
- [330 kV quad-bundle line: inductance and charging](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aquad-bundle-330kv-6m) — Power System Analysis I, chapter 4
- [Charging VAR of 132 kV equilateral line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Acharging-var-132kv-6m-1-1cm) — Power System Analysis I, chapter 4
- [Inductive reactance of vertical 132 kV line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Avertical-132kv-12mm-5m) — Power System Analysis I, chapter 4
- [Charging current of 60 Hz horizontal 132 kV line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ahorizontal-60hz-132kv-charging) — Power System Analysis I, chapter 4

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