---
title: "Power loss change on doubling line length"
subject: "Transmission and Distribution System Design"
chapter: 7
key: "semester_8/transmission_and_distribution_system_design:power-loss-doubling-line-length"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/power-loss-doubling-line-length
site: BCT Engineering
---

# Power loss change on doubling line length

Transmission and Distribution System Design · Chapter 7: Distribution System Design

- Asked in: 2079 Chaitra, 4 marks
- 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/power-loss-doubling-line-length
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7

## Question

For same load density of the load centre and same conductor what will be the change in power loss by doubling the length of the line?

## Short answer

Doubling length with same load density doubles total load and resistance. Power loss becomes (2I)² × (2rl) = 8 times the original, so loss increases 8-fold (∝l³). Supplied load only doubles, so percentage loss quadruples.

## Answer

With the **same load density** (kW per km) and the **same conductor**, doubling the length of a uniformly loaded line doubles both its total load and its resistance.

**Original line** (length $l$, uniform load density $\lambda$ kW/km, sending-end current $I \propto \lambda l$, resistance $R = rl$):

$$
P_{loss,1} = 3 \times \frac{I^2 R}{3} = I^2 r l
$$

(for a uniformly distributed load the loss is one-third of the end-loaded value).

**Doubled line** (length $2l$): total load $= 2\lambda l$, so current $I_2 = 2I$; resistance $R_2 = 2rl$:

$$
\begin{aligned}
P_{loss,2} &= (2I)^2 \times r(2l) = 8 I^2 r l \\
\frac{P_{loss,2}}{P_{loss,1}} &= 8
\end{aligned}
$$

**Result:**

- **Power loss increases 8 times** ($\propto l^3$).
- The load supplied only doubles, so the **percentage power loss increases 4 times** ($\propto l^2$).
- Similarly, voltage drop ($\propto I \times l$) increases 4 times.

This is why LT feeders must be kept short; in a large area it is better to use more transformers placed near load centres than to extend long LT lines.

## Check yourself

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

1. Doubling the length of a uniformly loaded line at the same load density and conductor makes power loss:
   A. Double
   B. 4 times
   C. 8 times
   D. Unchanged
   Answer: C. Current doubles and resistance doubles: (2)^2 x 2 = 8. (Key point: Result)

2. The percentage power loss becomes:
   A. 2 times
   B. 4 times
   C. 8 times
   D. Same
   Answer: B. Loss grows 8x while load grows 2x, so percentage loss grows 4x. (Key point: Percentage loss)

3. The voltage drop becomes:
   A. 2 times
   B. 8 times
   C. 4 times
   D. Half
   Answer: C. Drop proportional to I x l gives 2 x 2 = 4. (Key point: Voltage drop)

4. What practical lesson follows?
   A. Use longer LT lines
   B. Use more transformers near load centres rather than long LT lines
   C. Use smaller conductors
   D. Lower the voltage
   Answer: B. Loss grows as l^3, so keep LT feeders short. (Key point: Practical)

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