---
title: "Kelvin's law and its modified form"
subject: "Transmission and Distribution System Design"
chapter: 4
key: "semester_8/transmission_and_distribution_system_design:kelvins-law-modified-above-33kv"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/kelvins-law-modified-above-33kv
site: BCT Engineering
---

# Kelvin's law and its modified form

Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection

- Asked in: 2067 Mangsir (old course), 4+2 marks
- Video: [L49: Kelvin's Law of Economic Conductor Size in Hindi | Economic Transmission | Power System](https://www.youtube.com/watch?v=up9SsirOpsY) — TechInsight
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/kelvins-law-modified-above-33kv
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4

## Question

State and prove the Kelvin's law for most economical size of the conductor? Also discuss why this method is used in modified form for a line above 33 kV?

## Short answer

Kelvin's law: the most economical conductor size makes the annual variable cost of conductor (interest & depreciation) equal the annual cost of energy losses. Above 33 kV a modified form is used because tower and insulator costs depend on conductor size, corona sets a minimum size, loads vary, and future costs are uncertain.

## Answer

**Kelvin's law:** the most economical conductor size is the one for which the **annual cost of energy lost** in the conductor equals the **annual interest and depreciation on the part of the capital cost that varies with conductor size**.

#### Proof
Let $a$ be the conductor cross-section.

1. The capital cost of the line is $P_1 + P_2a$. Here $P_1$ is the part that does not depend on size (towers, insulators, erection, approximately) and $P_2a$ is the conductor cost. The annual charge at a rate $r$ (interest + depreciation) is
$$
C_1 = r(P_1 + P_2a) = k_0 + k_1a
$$
2. The resistance is $R = \rho l/a$, so the annual energy loss is $3I^2R\,t$, where $t$ is the equivalent full-load hours, and its cost is
$$
C_2 = \frac{k_2}{a}
$$
3. The total annual cost is
$$
C = k_0 + k_1a + \frac{k_2}{a}
$$
4. For minimum cost:
$$
\frac{dC}{da} = k_1 - \frac{k_2}{a^2} = 0 \;\Rightarrow\; k_1a = \frac{k_2}{a}
$$
Also $\dfrac{d^2C}{da^2} = \dfrac{2k_2}{a^3} > 0$, so this is a minimum.

So the variable annual capital charge equals the annual cost of losses, which proves the law. The economical size is $a = \sqrt{k_2/k_1}$.

#### Why a modified form is used above 33 kV
Kelvin's simple law has several limits:
- It assumes the **tower, insulator and erection costs** do not depend on conductor size. At high voltage these costs dominate, and they do depend on conductor weight, diameter and sag.
- It ignores **technical limits**. Above 33 kV the minimum size is often set by **corona** (minimum diameter), by **voltage regulation and stability**, or by the **thermal rating**, not by economics.
- It assumes a constant load and constant interest and energy rates. Real loads vary, so the **loss load factor** must be used.
- It is hard to know the exact interest rate, depreciation and future energy price.

Hence for lines above 33 kV a **modified Kelvin's law** is used. The total annual cost of the whole line (towers + conductors + insulators + losses) is computed for several standard conductors that already meet the corona, thermal and regulation requirements. The conductor with the least total annual cost is then chosen.

## Check yourself

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

1. What does Kelvin's law state for the most economical conductor size?
   A. Cost of losses equals the total capital cost
   B. Conductor area equals the load current
   C. Annual energy-loss cost equals annual interest and depreciation on the size-dependent capital cost
   D. Annual cost is zero at minimum size
   Answer: C. Kelvin's law equates the variable annual capital charge with the annual cost of losses. (Key point: Statement of the law)

2. In the proof, C = k0 + k1*a + k2/a. Why is the loss term k2/a?
   A. Loss is proportional to 1/R, and R is proportional to a
   B. Loss cost is constant
   C. R = rho*l/a so I^2R loss varies as 1/a
   D. Loss varies with a^2
   Answer: C. Resistance varies inversely with area, so energy loss cost is k2/a. (Key point: Proof structure)

3. What does dC/da = 0 give?
   A. a = k1/k2
   B. a = sqrt(k2/k1), where k1*a = k2/a
   C. a = k1*k2
   D. a = 0
   Answer: B. k1 - k2/a^2 = 0 gives a = sqrt(k2/k1); the second derivative is positive, so it is a minimum. (Key point: Economical size)

4. Why is Kelvin's law used in a modified form above 33 kV?
   A. Tower and insulator costs depend on conductor size, and corona, regulation and thermal limits set the minimum size
   B. Losses become zero
   C. Interest rates are constant
   D. Conductor cost is negligible
   Answer: A. At high voltage, tower costs and technical limits matter, so the lowest total annual cost among conductors that satisfy those limits is chosen. (Key point: Why modified above 33 kV)

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