---
title: "Internal inductance independent of current"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:internal-inductance-independent-current"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/internal-inductance-independent-current
site: BCT Engineering
---

# Internal inductance independent of current

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2070 Bhadra, 6 marks; 2069 Poush, 4 marks; 2068 Magh, 5 marks (asked 3 times)
- Video: [Internal Flux Linkage with in conductor, full derivation | electrical power system](https://www.youtube.com/watch?v=yGergiJFLFk) — ELECTRICAL HUB
- 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/internal-inductance-independent-current
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

Prove that the inductance of a conductor due to its internal flux linkage is independent of the current flowing through it.

## Short answer

Internal flux linkage λ_int = μ0 I/(8π) gives L_int = λ_int/I = μ0/(8π) = 0.5×10⁻⁷ H/m. Since λ_int is directly proportional to current I, the inductance is constant and independent of current magnitude.

## Answer

**Statement:** the internal inductance of a straight round conductor is $L_{int} = \frac{\mu_0}{8\pi} = \frac{1}{2}\times10^{-7}$ H/m. It contains no current term, so it is independent of the current.

#### Derivation
Take a long round conductor of radius r carrying current I, with uniform current density (skin effect neglected) and $\mu_r = 1$.

```text
        .---------.
      /   .-----.   \     r = conductor radius
     |   /   x   \   |    x = radius of the circular path
     |  |  <--dx  |  |    dx = thin tubular element
     |   \       /   |
      \   '-----'   /
        '---------'
```

1. **Current enclosed** by a circle of radius x < r:
$$
I_x = \frac{\pi x^2}{\pi r^2}I = \frac{x^2}{r^2}I
$$

2. **Ampere's law** around that circle:
$$
H_x(2\pi x) = I_x \;\Rightarrow\; H_x = \frac{x}{2\pi r^2}I, \qquad B_x = \mu_0H_x = \frac{\mu_0 x I}{2\pi r^2}
$$

3. **Flux in the tube** of thickness dx and length 1 m:
$$
d\phi = B_x\,dx = \frac{\mu_0 x I}{2\pi r^2}dx
$$

4. **Flux linkage.** This flux links only the fraction $x^2/r^2$ of the total current:
$$
d\lambda = \frac{x^2}{r^2}d\phi = \frac{\mu_0 I x^3}{2\pi r^4}dx
$$

5. **Total internal flux linkage:**
$$
\lambda_{int} = \int_0^r \frac{\mu_0 I x^3}{2\pi r^4}dx = \frac{\mu_0 I}{2\pi r^4}\cdot\frac{r^4}{4} = \frac{\mu_0 I}{8\pi}\ \text{Wb-turn/m}
$$

6. **Internal inductance:**
$$
L_{int} = \frac{\lambda_{int}}{I} = \frac{\mu_0}{8\pi} = \frac{4\pi\times10^{-7}}{8\pi} = \frac{1}{2}\times10^{-7}\ \text{H/m}
$$

#### Conclusion
The flux linkage $\lambda_{int}$ is directly proportional to I, so the ratio $\lambda/I$ is a constant. The current cancels and $L_{int} = 0.5\times10^{-7}$ H/m for any value of current. This is true as long as μ is constant, which holds for non-magnetic conductors such as copper and aluminium. (The same result also shows that $L_{int}$ does not depend on the radius r.)

Physical reason: doubling I doubles both the field H at every point and the flux linkage, so their ratio stays the same.

## Check yourself

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

1. For a round conductor of radius r with uniform current density, what current is enclosed within radius x < r?
   A. I x / r
   B. I x^2 / r^2
   C. I x^3 / r^3
   D. I
   Answer: B. Enclosed current is proportional to area: pi x^2 / pi r^2 times I. (Key point: Enclosed current)

2. Why is the flux in the tube at radius x multiplied by x^2/r^2 to get flux linkage?
   A. Because the flux links only the fraction of the current enclosed by that circle
   B. Because skin effect reduces the flux
   C. Because the tube has area x^2/r^2
   D. Because the permeability changes with radius
   Answer: A. Flux inside radius x is linked only with the current within x, which is the fraction x^2/r^2 of the total. (Key point: Partial linkage)

3. What is the total internal flux linkage per metre?
   A. mu0 I / (2 pi)
   B. mu0 I / (4 pi)
   C. mu0 I / (8 pi)
   D. mu0 I / (8 pi r)
   Answer: C. Integrating mu0 I x^3 / (2 pi r^4) from 0 to r gives mu0 I / (8 pi). (Key point: Result of integration)

4. Why is L_int independent of current?
   A. Flux linkage is proportional to I, so lambda/I is a constant 0.5e-7 H/m
   B. Current is assumed zero
   C. Flux linkage is constant whatever the current
   D. The conductor is superconducting
   Answer: A. lambda_int is directly proportional to I, so the current cancels in L = lambda/I (valid when mu is constant). (Key point: Independence of current)

## Related in Power System Analysis I

- [Internal inductance independent of geometry and current](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ainternal-inductance-geometry-current) — Power System Analysis I, chapter 4
- [Internal inductance independent of geometry](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ainternal-inductance-independent-geometry) — Power System Analysis I, chapter 4, asked 2×
- [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
- [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×
- [Inductance of 3-conductor bundled line, 6.1 m spacing](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ainductance-three-subconductor-bundle-6-1m) — Power System Analysis I, chapter 4
- [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

## Related in other subjects

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- [Inductance from flux linkages](https://www.bctengineer.com/question.md?key=semester_1%2Fbasic_electrical_engineering%3Ainductance-flux-linkages-derivation) — Basic Electrical Engineering, chapter 4
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