---
title: "Optimum conductor radius for minimum cable stress"
subject: "Power System Analysis I"
chapter: 2
key: "semester_4/power_system_analysis_i:cable-optimum-conductor-radius"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/cable-optimum-conductor-radius
site: BCT Engineering
---

# Optimum conductor radius for minimum cable stress

Power System Analysis I · Chapter 2: Overhead and Underground Transmission

- Asked in: 2076 Baisakh, 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/cable-optimum-conductor-radius
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-2

## Question

Derive a relation between the conductor radius and inside sheath radius of a single core cable so that the electric stress at the conductor surface may be minimum.

## Short answer

For a single-core cable, maximum stress is g_max = V/(r ln(R/r)). Setting derivative zero gives ln(R/r)=1, so R/r = e ≈ 2.718. The conductor radius should be r = R/e, yielding minimum stress g_max = V/r.

## Answer

In a single-core cable, the electric stress is greatest at the conductor surface. For a fixed sheath radius, this maximum stress is lowest when $R/r = e$.

#### Stress in a single-core cable
Let the conductor radius be $r$, the inside radius of the sheath $R$, and the voltage between core and sheath $V$. The charge per metre is $q$. By Gauss's law, the field at radius $x$ is:

$$
g_x = \frac{q}{2\pi\varepsilon_0\varepsilon_r x}
$$

Integrating from $r$ to $R$:

$$
V = \frac{q}{2\pi\varepsilon_0\varepsilon_r}\ln\frac{R}{r}
\;\Rightarrow\; g_x = \frac{V}{x\ln(R/r)}
$$

The stress is largest at $x = r$:

$$
g_{max} = \frac{V}{r\ln(R/r)}
$$

#### Condition for minimum $g_{max}$
With $V$ and $R$ fixed, $g_{max}$ is least when the denominator $f(r) = r\ln(R/r)$ is greatest:

$$
\begin{aligned}
\frac{df}{dr} &= \ln\frac{R}{r} + r\cdot\left(-\frac{1}{r}\right) = \ln\frac{R}{r} - 1 = 0\\
\ln\frac{R}{r} &= 1 \;\Rightarrow\; \frac{R}{r} = e = 2.718
\end{aligned}
$$

The second derivative, $d^2f/dr^2 = -1/r$, is negative, so this point is a maximum of $f$ and therefore a minimum of $g_{max}$.

**Result:**

$$
r = \frac{R}{e}, \qquad g_{max}\big|_{min} = \frac{V}{r} = \frac{eV}{R}
$$

The insulation thickness is then $R - r = r(e-1) = 1.718r$.

#### Practical note
The conductor radius from $r = R/e$ is often larger than the radius needed to carry the current. In that case a hollow conductor or a larger-diameter (for example aluminium) conductor is used to obtain the optimum radius.

#### Example
Take $V = 66$ kV (core to sheath) and $R = 2$ cm. Then $r = 2/2.718 = 0.736$ cm and $g_{max} = 66/0.736 = 89.7$ kV/cm.

## Check yourself

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

1. For fixed V and R, which quantity is minimised by choosing r?
   A. The capacitance of the cable
   B. The stress at the sheath
   C. The maximum stress at the conductor surface, g_max = V/(r ln(R/r))
   D. The charge per metre
   Answer: C. The design aim is to make the highest stress as small as possible. (Key point: Quantity to minimise)

2. How is the optimum r found?
   A. Set d/dr [ln(R/r)] = 0
   B. Set d/dr [r ln(R/r)] = 0, which gives ln(R/r) = 1
   C. Set r = R
   D. Set V = 0
   Answer: B. g_max is least when f(r) = r ln(R/r) is greatest: ln(R/r) - 1 = 0. (Key point: Optimisation step)

3. What is the optimum ratio R/r?
   A. 2
   B. e = 2.718
   C. 1
   D. 10
   Answer: B. ln(R/r) = 1 gives R/r = e. (Key point: Optimum ratio)

4. What is the minimum value of g_max?
   A. V/R
   B. V/(e R)
   C. V ln(R/r)
   D. e V / R, i.e. V/r with r = R/e
   Answer: D. With r = R/e, g_max = V/(r x 1) = V/r = eV/R. (Key point: Minimum stress value)

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