--- title: "Efficiency variation with load and maximum efficiency condition" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:efficiency-variation-max-condition" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/efficiency-variation-max-condition site: BCT Engineering --- # Efficiency variation with load and maximum efficiency condition Electrical Machines I · Chapter 2: Transformer - Asked in: 2076 Bhadra, 8 marks - Video: [13 Losses and Efficiency of Transformer](https://www.youtube.com/watch?v=jShXjmkMyB8) — SP Engg. ACADEMY - 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-variation-max-condition - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question Explain how the efficiency of a transformer varies with load. Derive the condition for maximum efficiency. ## Short answer Efficiency is output divided by output plus losses; iron loss is constant while copper loss varies as the square of load. Differentiating efficiency gives maximum when copper loss equals iron loss, at load fraction x=sqrt(Pi/Pcu,FL). ## Answer **Efficiency** of a transformer is the ratio of output power to input power: $\eta = \frac{\text{output}}{\text{output} + \text{losses}}$. Because one loss is constant and the other depends on load, efficiency changes with load. #### Losses in a transformer - **Iron (core) loss $P_i$** = hysteresis + eddy-current loss. It depends on voltage and frequency, which are constant, so $P_i$ is **constant** at all loads. - **Copper loss $P_{cu}$** = $I_1^2R_1 + I_2^2R_2 = I_2^2R_{02}$. It varies with the **square of the load current**. At a fraction $x$ of full load, $P_{cu} = x^2 P_{cu,FL}$. #### Variation of efficiency with load At fraction $x$ of full load: $$ \eta = \frac{x\,S\cos\phi}{x\,S\cos\phi + P_i + x^2 P_{cu,FL}} $$ - At **no load**, output is zero, so $\eta = 0$. - At **light load**, the fixed iron loss is large compared with the output, so $\eta$ is low but rises quickly. - As load increases, $\eta$ rises to a **maximum** where copper loss equals iron loss. - Beyond that, copper loss (growing as $x^2$) dominates and $\eta$ falls slowly. - At lower power factor, output is smaller for the same current, so the whole curve is lower. ```text eta ^ max (Pcu = Pi) | ____ | .-' '--..___ upf | .' ___...--''--.._ 0.8 pf | / .-' | /.' | // +-------------------------> load (x) 0 x_max 1.0 ``` #### Condition for maximum efficiency Write $\eta$ in terms of secondary current $I_2$ (with $V_2$ and $\cos\phi$ constant): $$ \eta = \frac{V_2 I_2 \cos\phi}{V_2 I_2\cos\phi + P_i + I_2^2 R_{02}} $$ Divide numerator and denominator by $I_2$: $$ \eta = \frac{V_2\cos\phi}{V_2\cos\phi + \dfrac{P_i}{I_2} + I_2R_{02}} $$ $\eta$ is maximum when the denominator is minimum: $$ \begin{aligned} \frac{d}{dI_2}\left(V_2\cos\phi + \frac{P_i}{I_2} + I_2R_{02}\right) &= 0 \\ -\frac{P_i}{I_2^2} + R_{02} &= 0 \\ I_2^2 R_{02} &= P_i \end{aligned} $$ So **efficiency is maximum when copper loss equals iron loss** (variable loss = constant loss). The current and load at maximum efficiency are: $$ I_2 = \sqrt{\frac{P_i}{R_{02}}}, \qquad x = \sqrt{\frac{P_i}{P_{cu,FL}}} $$ $$ \eta_{max} = \frac{x\,S\cos\phi}{x\,S\cos\phi + 2P_i} $$ **Example:** if $P_i = 400$ W and $P_{cu,FL} = 1600$ W, maximum efficiency occurs at $x = \sqrt{400/1600} = 0.5$, i.e. at half load. Distribution transformers are designed with low iron loss so that the maximum efficiency occurs at the light loads they carry most of the day. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which statement about transformer losses is correct? A. Iron loss is constant; copper loss varies as the square of load current B. Both vary with the square of load C. Both are constant D. Iron loss varies with load; copper loss is constant Answer: A. Core loss depends on constant voltage and frequency; copper loss is I^2R. (Key point: Constant and variable losses) 2. Why is efficiency low at light loads? A. Voltage drops to zero B. Constant iron loss is large compared with small output C. Leakage flux increases D. Copper loss is very high Answer: B. At light load output is small while iron loss remains the same. (Key point: Efficiency at light load) 3. After dividing by I2, efficiency is maximum when what is minimum? A. P_i/I2 + I2 R02 (the denominator) B. The numerator V2 cos phi C. P_i only D. I2^2 R02 only Answer: A. Setting derivative of the denominator to zero gives -Pi/I2^2 + R02 = 0. (Key point: Derivation step) 4. What is the condition for maximum efficiency? A. Copper loss is half of iron loss B. Copper loss equals iron loss C. Iron loss is zero D. Copper loss is twice iron loss Answer: B. I2^2 R02 = Pi, variable loss equals constant loss. 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