---
title: "T/F: LLF unity for constant power loss"
subject: "Transmission and Distribution System Design"
chapter: 6
key: "semester_8/transmission_and_distribution_system_design:tf-llf-unity-constant-loss"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/tf-llf-unity-constant-loss
site: BCT Engineering
---

# T/F: LLF unity for constant power loss

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

- Asked in: 2072 Asoj, 3 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-llf-unity-constant-loss
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-6

## Question

State whether the following statement is TRUE or FALSE. Justify your answer with a brief explanation: Loss of load factor (LLF) is always unity while evaluating energy loss from constant power loss.

## Short answer

TRUE. For constant power loss (e.g., transformer core loss, dielectric loss), average loss equals peak loss, so LLF = 1. These losses depend on nearly constant voltage, not load current.

## Answer

**TRUE.**

Loss of load factor is defined as

$$
LLF = \frac{\text{Average power loss over period}}{\text{Peak power loss}}
$$

If the power loss is constant throughout the period, the average loss equals the peak loss, so $LLF = 1$.

**Justification:**

- **Constant (no-load) losses** such as transformer core (iron) loss, dielectric loss in cables, corona loss in fair weather and losses in meter potential coils depend on voltage, not on load current. Since voltage is nearly constant, these losses are the same every hour.
- Their energy loss is therefore $E = P_{const} \times 8760$ kWh/year, i.e. the loss acts at its full value for all hours (LLF = 1).
- Example: a 100 kVA transformer with 200 W core loss loses $0.2 \times 8760 = 1752$ kWh/year whether it is lightly or fully loaded.
- Only **variable (copper) losses**, which vary as $I^2$, need $LLF < 1$, e.g. $LLF = 0.3LF + 0.7LF^2$.

The statement also follows from the formula: a constant load gives $LF = 1$, and $0.3(1) + 0.7(1)^2 = 1$.

## Check yourself

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

1. Loss of load factor is defined as:
   A. Peak loss / average loss
   B. Average power loss / peak power loss
   C. Average load / peak load
   D. Energy loss / energy supplied
   Answer: B. LLF = average power loss over the period divided by the peak power loss. (Key point: LLF definition)

2. Why is LLF = 1 for constant losses such as transformer core loss?
   A. They are zero at peak
   B. They vary as I^2
   C. They are the same every hour, so average equals peak
   D. They occur only at peak
   Answer: C. Core loss depends on voltage, which is nearly constant, so average loss equals peak loss. (Key point: Constant loss gives LLF = 1)

3. Which losses require LLF < 1 when computing energy loss?
   A. Core loss
   B. Corona loss in fair weather
   C. Dielectric loss in cables
   D. Variable copper loss varying as I^2
   Answer: D. Only I^2 losses vary with load, so LLF = 0.3LF + 0.7LF^2 applies to them. (Key point: Which losses need LLF < 1)

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