---
title: "T/F: % power loss equals % energy loss"
subject: "Transmission and Distribution System Design"
chapter: 6
key: "semester_8/transmission_and_distribution_system_design:tf-power-loss-energy-loss-same"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/tf-power-loss-energy-loss-same
site: BCT Engineering
---

# T/F: % power loss equals % energy loss

Transmission and Distribution System Design · Chapter 6: Electrical Load Characteristics and Load Forecast

- Asked in: 2069 Bhadra (old course), 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/tf-power-loss-energy-loss-same
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-6

## Question

For a given amount of power to be transmitted over a given distance, state whether the following statement is TRUE or FALSE. Justify your answer with a brief explanation: In practical case, percentage power loss and percentage energy loss in a transmission line is same.

## Short answer

FALSE. Percentage energy loss equals % peak power loss × (LLF/LF), and LLF ≤ LF, so % energy loss is lower. For 0.4 LF, 10% peak power loss yields only 5.8% energy loss.

## Answer

**FALSE.**

The percentage **power** loss is measured at peak load, while the percentage **energy** loss is averaged over time. Because losses vary as the square of the load, they are not equal.

**Justification:**

$$
\begin{aligned}
\%P_{loss} &= \frac{P_{loss,peak}}{P_{max}} \times 100 \\
\%E_{loss} &= \frac{P_{loss,peak} \times LLF \times T}{P_{max} \times LF \times T} \times 100 = \%P_{loss} \times \frac{LLF}{LF}
\end{aligned}
$$

Since $LF^2 \le LLF \le LF$, we get $LLF/LF \le 1$, so % energy loss is less than % peak power loss.

**Example:** % peak power loss = 10%, LF = 0.4. Using $LLF = 0.3LF + 0.7LF^2$:

$$
\begin{aligned}
LLF &= 0.3(0.4) + 0.7(0.16) = 0.232 \\
\%E_{loss} &= 10 \times \frac{0.232}{0.4} = 5.8\%
\end{aligned}
$$

So 10% power loss corresponds to only 5.8% energy loss.

The two are equal only when $LF = LLF = 1$, i.e. a perfectly constant load, which never occurs in practical lines. (If constant losses such as corona or transformer core loss are included, those parts have LLF = 1, but the overall figures still differ.)

## Check yourself

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

1. Is % power loss equal to % energy loss in a practical line?
   A. Yes, always
   B. No, they differ because losses vary as the square of load
   C. Yes, if voltage is constant
   D. No, because % energy loss is always larger
   Answer: B. The statement is false: peak power loss is at peak load while energy loss is averaged over varying load. (Key point: Verdict)

2. Which relation connects them?
   A. %E = %P x LF/LLF
   B. %E = %P x LLF/LF
   C. %E = %P x LF x LLF
   D. %E = %P + LLF
   Answer: B. %E_loss = %P_loss x LLF/LF, and since LLF <= LF the energy loss percentage is smaller. (Key point: Relation between them)

3. With 10% peak power loss and LF = 0.4, LLF = 0.3LF + 0.7LF^2 gives % energy loss of:
   A. 10%
   B. 2.32%
   C. 5.8%
   D. 23.2%
   Answer: C. LLF = 0.232, so %E = 10 x 0.232/0.4 = 5.8%. (Key point: Example numbers)

## Related in Transmission and Distribution System Design

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