---
title: "Inductance of a single-phase two-wire line"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:single-phase-inductance-derivation"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/single-phase-inductance-derivation
site: BCT Engineering
---

# Inductance of a single-phase two-wire line

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2075 Baisakh, 5 marks
- Video: [Inductance of Single Phase Two Wire Line - Transmission Line Parameters - Power System Engineering 1](https://www.youtube.com/watch?v=XoLZru1dQU0) — Ekeeda
- 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/single-phase-inductance-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

Derive the expression for the inductance per metre of a single phase two wire transmission line.

## Short answer

Inductance per conductor is L = 2×10^-7 ln(D/r') H/m, where r' = 0.7788r accounts for internal flux. The loop inductance is twice that, L_loop = 4×10^-7 ln(D/r') H/m.

## Answer

The inductance of a single-phase two-wire line is found by adding the internal and external flux linkages of each conductor due to its own current and the return current.

```text
     I -->                    <-- I
   (  a  )<------- D ------->(  b  )
   radius r1                 radius r2
```

#### Step 1: Internal flux linkage

Inside a conductor of radius $r$ at radius $x$, the enclosed current is $I_x = I x^2/r^2$. From Ampere's law, $H_x = \dfrac{Ix}{2\pi r^2}$. The flux in a tube of thickness $dx$ links only the fraction $x^2/r^2$ of the current:

$$
\lambda_{int} = \int_0^r \frac{x^2}{r^2}\cdot\frac{\mu_0 I x}{2\pi r^2}\,dx = \frac{\mu_0 I}{8\pi} = \frac{I}{2}\times10^{-7}\ \text{Wb-T/m}
$$

#### Step 2: External flux linkage

Outside the conductor, $H_x = I/2\pi x$, and the flux between radii $D_1$ and $D_2$ links the full current:

$$
\lambda_{12} = \frac{\mu_0 I}{2\pi}\ln\frac{D_2}{D_1} = 2\times10^{-7} I\ln\frac{D_2}{D_1}
$$

#### Step 3: Flux linkage of conductor a

Flux due to $I_a = I$ out to a distant point P (distance $D_{1P}$), plus flux due to $I_b = -I$:

$$
\begin{aligned}
\lambda_a &= 2\times10^{-7}\left[I\left(\frac{1}{4} + \ln\frac{D_{1P}}{r_1}\right) - I\ln\frac{D_{2P}}{D}\right]
\end{aligned}
$$

As P moves to infinity, $D_{1P}/D_{2P} \to 1$, and using $\tfrac14 = \ln e^{1/4}$:

$$
\lambda_a = 2\times10^{-7} I\ln\frac{D}{r_1 e^{-1/4}} = 2\times10^{-7} I\ln\frac{D}{r_1'}
$$

where $r_1' = 0.7788 r_1$ is the GMR.

#### Step 4: Inductances

$$
L_a = \frac{\lambda_a}{I} = 2\times10^{-7}\ln\frac{D}{r_1'}\ \text{H/m}, \qquad L_b = 2\times10^{-7}\ln\frac{D}{r_2'}\ \text{H/m}
$$

Loop inductance:

$$
L = L_a + L_b = 4\times10^{-7}\ln\frac{D}{\sqrt{r_1' r_2'}}\ \text{H/m}
$$

For identical conductors ($r_1 = r_2 = r$):

$$
L = 4\times10^{-7}\ln\frac{D}{r'}\ \text{H/m}, \quad r' = 0.7788r
$$

$L_a$ is the inductance per conductor; the loop inductance is twice that value.

## Check yourself

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

1. What is the internal flux linkage per metre of a solid conductor carrying current I?
   A. mu0 I/(2 pi)
   B. mu0 I/(8 pi), independent of radius
   C. mu0 I/(4 pi r)
   D. mu0 I r/(8 pi)
   Answer: B. Integrating the partial-linkage flux gives mu0 I/8pi = (I/2) x 1e-7 Wb-T/m. (Key point: Internal flux linkage)

2. Inside a conductor at radius x, why does the flux tube link only the fraction x^2/r^2 of the current?
   A. Only the current enclosed within radius x produces that flux
   B. The current is zero inside
   C. The flux is shielded by the surface
   D. Because the current flows only at the surface
   Answer: A. The tube at x links only the current enclosed within x, which is I x^2/r^2 for uniform current density. (Key point: Fraction of current linked)

3. How does the 1/4 term from internal flux become r' = 0.7788 r?
   A. 1/4 = ln e^(1/4), so ln(D/(r e^(-1/4))) with e^(-1/4) = 0.7788
   B. 1/4 is added to r
   C. r' = r/4
   D. It comes from the external flux
   Answer: A. Writing 1/4 as ln e^(1/4) combines it with ln(D/r) to give ln(D/(r e^(-1/4))) = ln(D/r'). (Key point: GMR origin)

4. For two identical conductors, how are L_a and the loop inductance related?
   A. L = L_a/2
   B. L = L_a
   C. L = 2 L_a = 4e-7 ln(D/r')
   D. L = 8e-7 ln(D/r')
   Answer: C. L_a = 2e-7 ln(D/r'), and the loop L = L_a + L_b = 4e-7 ln(D/r'). (Key point: Loop vs per-conductor)

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