---
title: "Load sharing between two stations in parallel"
subject: "Electric Machines II"
chapter: 1
key: "semester_5/electric_machines_ii:load-sharing-two-stations"
source: https://www.bctengineer.com/q/semester_5/electric_machines_ii/load-sharing-two-stations
site: BCT Engineering
---

# Load sharing between two stations in parallel

Electric Machines II · Chapter 1: Three Phase Synchronous Generator

- Asked in: 2082 Bhadra (new course), 3 marks
- Video: [Load Sharing Between Parallel Alternators | Electrical Engineering](https://www.youtube.com/watch?v=IBkGj28xugg) — Endless Edutainment
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/electric_machines_ii/load-sharing-two-stations
- Whole chapter: https://www.bctengineer.com/md/semester_5/electric_machines_ii/chapter-1

## Question

Two station generators A and B operate in parallel. Station capacity of A is 50 MW and that of B is 25 MW. Full load speed regulation of station A is 3% and B is 3.5%. Calculate the load sharing if the connected load is 50 MW, no load frequency is 50 Hz. Comment on the answer.

## Short answer

Droop: A: 0.03 Hz/MW, B: 0.07 Hz/MW. Solving f=50-0.03PA=50-0.07PB and PA+PB=50 gives PA=35 MW, PB=15 MW, f=48.95 Hz. Load not proportional to ratings; A with smaller regulation takes larger share. Frequency drops; secondary control needed to restore 50 Hz.

## Answer

Each station follows a linear speed (frequency) droop: frequency falls from the no-load value (50 Hz) by its regulation percentage at its full load.

#### Droop of each station
$$
\begin{aligned}
\text{A: drop at full load} &= 0.03\times50 = 1.5\ \text{Hz for } 50\ \text{MW} \Rightarrow 0.03\ \text{Hz/MW} \\
\text{B: drop at full load} &= 0.035\times50 = 1.75\ \text{Hz for } 25\ \text{MW} \Rightarrow 0.07\ \text{Hz/MW}
\end{aligned}
$$

#### Load sharing
Both run at the same frequency $f$:
$$
\begin{aligned}
f &= 50 - 0.03P_A = 50 - 0.07P_B \\
P_A + P_B &= 50 \\
0.03P_A &= 0.07(50 - P_A) \Rightarrow P_A = \frac{3.5}{0.10} = 35\ \text{MW} \\
P_B &= 15\ \text{MW} \\
f &= 50 - 0.03\times35 = 48.95\ \text{Hz}
\end{aligned}
$$

#### Comment
- A supplies 35 MW (70% of its capacity) and B supplies 15 MW (60% of its capacity).
- The load is not shared in proportion to the ratings (which would be 33.3 : 16.7 MW). B has the steeper droop (larger Hz/MW), and A, with the smaller speed regulation, takes a larger share relative to its rating.
- For load sharing in proportion to ratings, the per-unit droops must be equal.
- The frequency falls to 48.95 Hz; the governor speed settings must be raised (secondary control) to restore 50 Hz.

**Answer: P_A = 35 MW, P_B = 15 MW, common frequency = 48.95 Hz.**

## Check yourself

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

1. What are the droops of stations A and B?
   A. A 0.07 Hz/MW, B 0.03 Hz/MW
   B. A 0.03 Hz/MW, B 0.07 Hz/MW
   C. Both 0.05 Hz/MW
   D. A 1.5 Hz/MW, B 1.75 Hz/MW
   Answer: B. A drops 1.5 Hz over 50 MW = 0.03 Hz/MW; B drops 1.75 Hz over 25 MW = 0.07 Hz/MW. (Key point: Droop slopes)

2. How do A and B share the 50 MW load?
   A. A 25 MW, B 25 MW
   B. A 33.3 MW, B 16.7 MW
   C. A 15 MW, B 35 MW
   D. A 35 MW, B 15 MW
   Answer: D. At common frequency 0.03 PA = 0.07 PB with PA + PB = 50, giving PA = 35 MW and PB = 15 MW. (Key point: Load sharing)

3. What is the key comment on the result?
   A. Load is shared in proportion to ratings
   B. Frequency stays 50 Hz
   C. Sharing is not in proportion to ratings because the per-unit droops differ
   D. B is overloaded
   Answer: C. Proportional sharing needs equal per-unit droops; here the frequency also falls to 48.95 Hz and needs secondary control. (Key point: Comment on sharing)

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