---
title: "Half-load efficiency of 200 kVA transformer"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:efficiency-half-load-200kva-98"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/efficiency-half-load-200kva-98
site: BCT Engineering
---

# Half-load efficiency of 200 kVA transformer

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2080 Chaitra, 4 marks
- Video: [Transformer efficiency: at Full and Half Load problem.](https://www.youtube.com/watch?v=ek2AQfUmAKo) — Dr Bharat Singh Choudhary
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/efficiency-half-load-200kva-98
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

A 200 kVA transformer has an efficiency of 98 % at full load if the maximum efficiency occurs at three quarters of full load, calculate efficiency at half load. Assume negligible magnetizing current and 0.8 pf at all loads.

## Short answer

Using full-load efficiency, total losses are 3.2653 kW; maximum efficiency at 3/4 load gives copper loss 2.0898 kW and iron loss 1.1755 kW. At half load copper loss is 0.5224 kW and efficiency = 80/(80 + 1.1755 + 0.5224) = 97.92%.

## Answer

Maximum efficiency occurs when copper loss equals iron loss. Use this with the full-load efficiency to find both losses.

#### Losses at full load

$$
\begin{aligned}
P_{out,fl} &= 200\times0.8 = 160\ \text{kW} \\
\text{Total losses} &= \frac{160}{0.98} - 160 = 3.2653\ \text{kW} \\
P_i + P_{cu,fl} &= 3.2653
\end{aligned}
$$

#### Condition for maximum efficiency at 3/4 load

$$
\begin{aligned}
P_i &= \left(\tfrac34\right)^2 P_{cu,fl} = 0.5625\,P_{cu,fl} \\
1.5625\,P_{cu,fl} &= 3.2653 \Rightarrow P_{cu,fl} = 2.0898\ \text{kW} \\
P_i &= 3.2653 - 2.0898 = 1.1755\ \text{kW}
\end{aligned}
$$

#### Efficiency at half load

$$
\begin{aligned}
P_{out} &= 0.5\times160 = 80\ \text{kW} \\
P_{cu} &= (0.5)^2\times2.0898 = 0.5224\ \text{kW} \\
\eta &= \frac{80}{80 + 1.1755 + 0.5224} = \frac{80}{81.698} = 97.92\%
\end{aligned}
$$

**Answer: efficiency at half load ≈ 97.92%.**

## Check yourself

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

1. The transformer has maximum efficiency at 3/4 of full load. What does this tell us?
   A. Iron loss is zero at 3/4 load
   B. Iron loss equals full-load copper loss
   C. Iron loss equals copper loss at 3/4 load
   D. Copper loss at full load is zero
   Answer: C. Maximum efficiency occurs when the variable copper loss equals the constant iron loss, here at 3/4 load. (Key point: Max efficiency condition)

2. Which relation between iron loss Pi and full-load copper loss follows from maximum efficiency at 3/4 load?
   A. Pi = 0.5625 Pcu,fl
   B. Pi = 1.25 Pcu,fl
   C. Pi = Pcu,fl
   D. Pi = 0.75 Pcu,fl
   Answer: A. Copper loss at 3/4 load is (3/4)^2 Pcu,fl, and this equals Pi, so Pi = 0.5625 Pcu,fl. (Key point: Relating Pi and Pcu,fl)

3. How is the total full-load loss found from the 98% efficiency?
   A. Output/0.98 minus output, with output = 160 kW
   B. 160 x 0.02 = 3.2 kW using input as 160 kW
   C. Output x 0.98 = 156.8 kW
   D. 200 x 0.02 = 4 kW using kVA as power
   Answer: A. Output is 200 x 0.8 = 160 kW; input = 160/0.98, so loss = 160/0.98 - 160 = 3.2653 kW. (Key point: Total loss at full load)

4. At half load, how do the losses change when finding efficiency?
   A. Copper loss halves and iron loss is neglected
   B. Iron loss becomes 0.25 of its value; copper loss stays
   C. Both iron and copper loss halve
   D. Iron loss stays 1.1755 kW; copper loss becomes 0.25 x 2.0898 kW
   Answer: D. Iron loss is constant while copper loss varies with the square of load, giving 0.5224 kW and efficiency of about 97.92%. (Key point: Half-load efficiency)

## Related in Electrical Machines I

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