---
title: "T/F: peak demand as sum of class peaks"
subject: "Transmission and Distribution System Design"
chapter: 6
key: "semester_8/transmission_and_distribution_system_design:tf-peak-demand-sum-classes"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/tf-peak-demand-sum-classes
site: BCT Engineering
---

# T/F: peak demand as sum of class peaks

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

- Asked in: 2075 Bhadra, 1+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-peak-demand-sum-classes
- 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: The peak demand of a load centre is the sum of the peak demand of different consumer classes.

## Short answer

FALSE. Different classes peak at different times; the coincident peak is the sum of class peaks multiplied by their contribution factors (≤1), which is less than the arithmetic sum.

## Answer

**FALSE.**

The peak demand of a load centre is the **coincident** (simultaneous) maximum of the combined load, not the arithmetic sum of the class peaks.

**Justification:**

- Different consumer classes reach their maximum demand at different hours. Domestic load peaks in the evening (lighting, cooking), commercial load around midday/afternoon, and offices, schools and institutions during working hours.
- At the hour of the load centre peak, each class draws only a fraction of its own peak. This fraction is the **contribution (responsibility) factor** $c_i$.
- Therefore

$$
P_{LC} = \sum c_i \, P_{class,i} \le \sum P_{class,i}
$$

**Example:** domestic peak 20 kW (evening, $c = 1$), commercial peak 6 kW ($c = 0.5$ in the evening), institutional peak 4 kW ($c = 0.1$):

$$
P_{LC} = 20 + 3 + 0.4 = 23.4 \text{ kW, not } 30 \text{ kW}
$$

Adding the class peaks directly would oversize the distribution transformer and feeders, raising capital cost and no-load losses. The plain sum is correct only if all classes peak at the same time (all contribution factors equal to 1), which rarely happens.

## Check yourself

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

1. 'Peak demand of a load centre is the sum of peak demands of the classes' is:
   A. True, always
   B. False, because the classes peak at different times
   C. True, if power factor is 1
   D. False, because the sum is smaller than the true peak
   Answer: B. The load-centre peak is the coincident maximum, not the plain sum of class peaks. (Key point: Verdict)

2. The correct expression for load-centre peak is:
   A. Sum of P_class
   B. Sum of c_i x P_class,i, with c_i the contribution factor
   C. Max of P_class divided by c_i
   D. Average of P_class
   Answer: B. Each class draws only a fraction c_i of its peak at the system peak. (Key point: Formula)

3. Domestic 20 kW (c = 1), commercial 6 kW (c = 0.5), institutional 4 kW (c = 0.1). The load-centre peak is:
   A. 30 kW
   B. 23.4 kW
   C. 26.5 kW
   D. 20 kW
   Answer: B. 20 + 3 + 0.4 = 23.4 kW. (Key point: Example)

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