---
title: "Load sharing: 250 MW and 500 MW, 4%/3% droop"
subject: "Power Plant Equipment"
chapter: 2
key: "semester_7/power_plant_equipment:load-sharing-250-500-4-3"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/load-sharing-250-500-4-3
site: BCT Engineering
---

# Load sharing: 250 MW and 500 MW, 4%/3% droop

Power Plant Equipment · Chapter 2: Power/Frequency Control in Hydro Generator System

- Asked in: 2073 Chaitra, 8 marks
- Video: [What is Droop setting in Governor of Generators? How Load of Generators in parallel is controlled?](https://www.youtube.com/watch?v=YXeVxP5-utk) — Electro Globe
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/load-sharing-250-500-4-3
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-2

## Question

Two generator of 250 MW and 500 MW capacities respectively are operating in parallel and connected to common bus. When each generator is half loaded, they operate at a common frequency of 50 Hz. Their drop characteristics are 4% and 3% respectively based on respective ratings. (i) What will be the common frequency if the common load is increased by 100 MW and calculate the new generation of each units.

## Short answer

Common frequency is approximately 49.78 Hz. G1 generates 152.27 MW and G2 generates 322.73 MW.

## Answer

Both units move down their droop lines to a common lower frequency, sharing the extra 100 MW in proportion to $P_r/R$.

**Method:** a unit with droop $R$ (pu on its own rating $P_r$) and rated frequency $f_0$ changes its output by
$$
\Delta P = -\frac{P_r}{R\,f_0}\,\Delta f = -K\,\Delta f
$$
where $K = P_r/(R f_0)$ is its stiffness in MW/Hz. All units run at the same frequency, so $\Delta P_L = -\Delta f \sum K$.

**Data:** G1: 250 MW, 4 %, initial 125 MW; G2: 500 MW, 3 %, initial 250 MW; $f_0 = 50$ Hz; $\Delta P_L = 100$ MW.

$$
\begin{aligned}
K_1 &= \frac{250}{0.04 \times 50} = 125\ \text{MW/Hz} \\
K_2 &= \frac{500}{0.03 \times 50} = 333.33\ \text{MW/Hz} \\
\Delta f &= -\frac{100}{458.33} = -0.2182\ \text{Hz} \\
f_{new} &= 50 - 0.2182 = 49.78\ \text{Hz} \\
\Delta P_1 &= 125 \times 0.2182 = 27.27\ \text{MW} \\
\Delta P_2 &= 333.33 \times 0.2182 = 72.73\ \text{MW}
\end{aligned}
$$

| Generator | Initial (MW) | Increase (MW) | New (MW) |
|---|---|---|---|
| G1 (250 MW, 4 %) | 125 | 27.27 | 152.27 |
| G2 (500 MW, 3 %) | 250 | 72.73 | 322.73 |
| Total | 375 | 100 | 475 |

**Answer:** Common frequency ≈ **49.78 Hz**; G1 generates **152.27 MW** and G2 **322.73 MW**.

## Check yourself

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

1. At 50 Hz, what are K1 (250 MW, 4%) and K2 (500 MW, 3%)?
   A. 125 and 333.33 MW/Hz
   B. 62.5 and 166.7 MW/Hz
   C. 250 and 500 MW/Hz
   D. 125 and 250 MW/Hz
   Answer: A. K1 = 250/(0.04 x 50) = 125 and K2 = 500/(0.03 x 50) = 333.33. (Key point: Stiffness values)

2. What is the common frequency after a 100 MW rise?
   A. 49.78 Hz
   B. 49.50 Hz
   C. 49.90 Hz
   D. 49.22 Hz
   Answer: A. df = -100/458.33 = -0.2182 Hz, so f = 49.78 Hz. (Key point: Frequency drop)

3. How is the 100 MW split between G1 and G2?
   A. 50 and 50
   B. 27.27 and 72.73
   C. 72.73 and 27.27
   D. 33.33 and 66.67
   Answer: B. Shares are proportional to K: 125/458.33 and 333.33/458.33 of 100 MW. (Key point: Increase on each unit)

4. What are the new outputs of G1 and G2 given they started at 125 MW and 250 MW?
   A. 152.27 and 322.73 MW
   B. 225 and 350 MW
   C. 152.27 and 277.27 MW
   D. 125 and 350 MW
   Answer: A. 125 + 27.27 = 152.27 MW and 250 + 72.73 = 322.73 MW. (Key point: New generation)

## Related in Power Plant Equipment

- [Load sharing: 250/500 MW, frequency falls to 49.5 Hz](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-250-500-4-2) — Power Plant Equipment, chapter 2, asked 2×
- [Load sharing: 500 MW and 250 MW, 5%/6% droop](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-500-250-5-6) — Power Plant Equipment, chapter 2, asked 2×
- [Droop for capacity sharing: 500 and 250 MW](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Adroop-capacity-sharing-500-250) — Power Plant Equipment, chapter 2
- [Load sharing: 250 MW and 500 MW, 20%/10% droop](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-sharing-250-500-20-10) — Power Plant Equipment, chapter 2, asked 2×
- [Load decrease: 250 and 500 MW at 49 Hz](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Aload-decrease-250-500-49hz) — Power Plant Equipment, chapter 2
- [Droop for capacity sharing: 250 and 500 MW](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Adroop-capacity-sharing-250-500) — Power Plant Equipment, chapter 2

## Related in other subjects

- [Load sharing between two stations in parallel](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Aload-sharing-two-stations) — Electric Machines II, chapter 1
- [Load sharing of 50 kVA and 25 kVA transformers in parallel](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aparallel-50kva-25kva-60kva-load) — Electrical Machines I, chapter 2