---
title: "220 kV line on 6-6-12 m triangle"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:triangle-6-6-12m-220kv"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/triangle-6-6-12m-220kv
site: BCT Engineering
---

# 220 kV line on 6-6-12 m triangle

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2070 Magh, 6 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/triangle-6-6-12m-220kv
- 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 its conductors on the corners of a triangle with sides 6 m, 6 m, and 12 m. The conductor radius is 1.81 cm. Find the inductance and capacitance per phase per km, capacitive reactance per phase, charging current and total charging MVAR.

## Short answer

Inductance 1.257 mH/km, capacitance 9.219 nF/km, capacitive reactance 1726.4 Ω, charging current 73.6 A, total charging MVAR 28.0 MVAR.

## Answer

Sides of 6, 6 and 12 m mean the conductors are actually in a straight line (flat configuration with 6 m between adjacent conductors). For a transposed line use $D_{eq} = \sqrt[3]{D_{ab}D_{bc}D_{ca}}$.

**Data:** $r = 1.81$ cm $= 0.0181$ m, $r' = 0.7788 \times 0.0181 = 0.014096$ m, 220 kV, 50 Hz, 200 km.

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

#### Inductance per phase per km

$$
\begin{aligned}
L &= 2\times10^{-7}\ln\frac{7.5595}{0.014096} = 2\times10^{-7}\times6.2847 \\
&= 1.2569\times10^{-6}\ \text{H/m} = 1.257\ \text{mH/km}
\end{aligned}
$$

#### Capacitance per phase per km

$$
\begin{aligned}
C &= \frac{55.63\times10^{-12}}{\ln(7.5595/0.0181)} = \frac{55.63\times10^{-12}}{6.0347} \\
&= 9.219\times10^{-12}\ \text{F/m} = 9.219\ \text{nF/km}
\end{aligned}
$$

#### Capacitive reactance per phase (200 km)

$$
\begin{aligned}
C_{200} &= 9.219 \times 200 = 1843.7\ \text{nF} = 1.8437\ \mu\text{F} \\
X_C &= \frac{1}{2\pi \times 50 \times 1.8437\times10^{-6}} = 1726.4\ \Omega
\end{aligned}
$$

#### Charging current

$$
I_c = \frac{V_p}{X_C} = \frac{220\,000/\sqrt3}{1726.4} = \frac{127\,017}{1726.4} = 73.57\ \text{A}
$$

#### Total charging MVAR

$$
Q = 3V_pI_c = \frac{V_L^2}{X_C} = \frac{(220\times10^3)^2}{1726.4} = 28.03\ \text{MVAR}
$$

**Answer:** $L = 1.257$ mH/km, $C = 9.219$ nF/km, $X_C = 1726.4\ \Omega$, $I_c = 73.6$ A, $Q = 28.0$ MVAR.

## Check yourself

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

1. Sides 6 m, 6 m, 12 m mean the conductors are:
   A. In a straight line (flat) since 6 + 6 = 12
   B. In an equilateral triangle
   C. In a right triangle
   D. In an isosceles triangle with apex up
   Answer: A. Two sides sum to the third, so the triangle degenerates into a straight line. (Key point: Geometry recognition)

2. Which radius is used in L = 2e-7 ln(D_eq / ...)?
   A. r
   B. r' = 0.7788r
   C. 2r
   D. r'/0.7788
   Answer: B. Inductance includes internal flux through r'; capacitance uses r. (Key point: Inductance radius)

3. Charging current for the 220 kV, 200 km line uses:
   A. 220 kV divided by X_C
   B. 220/sqrt3 kV divided by X_C per km
   C. V_p = 220/sqrt3 kV divided by total X_C of 200 km
   D. 220 kV x sqrt3 divided by X_C
   Answer: C. I_c = V_p/X_C with X_C for the full 200 km (1726 ohm) gives 73.6 A. (Key point: Charging current)

4. Total three-phase charging MVAR equals:
   A. V_p^2 / X_C
   B. V_L I_c
   C. V_L^2 x X_C
   D. V_L^2 / X_C (= 3 V_p I_c)
   Answer: D. Three phases each supply V_p I_c, so Q = 3 V_p I_c = V_L^2/X_C = 28 MVAR. (Key point: Total charging MVAR)

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