--- title: "T/F: under-loaded distribution transformer losses" subject: "Transmission and Distribution System Design" chapter: 7 key: "semester_8/transmission_and_distribution_system_design:tf-underloaded-distribution-transformer" source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/tf-underloaded-distribution-transformer site: BCT Engineering --- # T/F: under-loaded distribution transformer losses Transmission and Distribution System Design · Chapter 7: Distribution System Design - Asked in: 2080 Chaitra, 4 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-underloaded-distribution-transformer - Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-7 ## Question State whether the following statement is TRUE or FALSE and justify your answer with brief explanation: A distribution transformer operating at significantly under-loading conditions is always beneficial from energy loss prospective. ## Short answer FALSE. Transformers have constant core loss whenever energized. At very light loads, copper loss is small but core loss continues, causing percentage energy loss to rise. Maximum efficiency occurs where copper loss equals core loss (typically 40–60% load). Gross underloading wastes energy. ## Answer **FALSE.** **Justification:** A transformer has two kinds of loss: - **No-load (core) loss** $P_i$: constant whenever the transformer is energised, regardless of load. - **Load (copper) loss**: $P_{cu} = x^2 P_{cu,FL}$, where $x$ is the fraction of rated load. Efficiency is maximum when copper loss equals core loss, i.e. at load fraction $$ x_{\eta,max} = \sqrt{\frac{P_i}{P_{cu,FL}}} $$ which is typically 40–60% of rating for distribution transformers. When the transformer is significantly under-loaded: - Copper loss becomes very small, but the core loss continues for all 8760 hours. - The energy delivered is small, so the **percentage energy loss rises**. **Example:** 100 kVA unit, $P_i = 200$ W, $P_{cu,FL} = 1100$ W, LLF taken as 0.3 for illustration. | Peak load | Core loss (kWh/yr) | Cu loss (kWh/yr) | Total | |---|---|---|---| | 20 kVA | 1752 | 1100 × 0.04 × 0.3 × 8.76 = 116 | 1868 | | 60 kVA | 1752 | 1100 × 0.36 × 0.3 × 8.76 = 1041 | 2793 | The 20 kVA case carries one-third of the load but has two-thirds of the losses, so loss per kWh delivered is about twice as high. Therefore gross under-loading (oversized transformers) wastes energy through core loss, as well as capital. Moderate under-loading near the maximum-efficiency point is beneficial; very light loading is not. This is why transformers should be sized for the realistic design-year demand and why low-loss cores are used where load factor is low. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The statement that significant under-loading is always beneficial for energy loss is: A. True, copper loss falls B. False, core loss continues so percentage loss rises C. True, efficiency is maximum at no load D. False, because copper loss rises Answer: B. Core loss persists for 8760 h while delivered energy is small. (Key point: Truth value) 2. Efficiency is maximum when: A. Core loss equals full-load copper loss B. Copper loss equals core loss C. Copper loss is zero D. Load is 100% Answer: B. At max efficiency copper loss equals iron loss, at x = sqrt(Pi/Pcu,FL). (Key point: Max efficiency) 3. Core (no-load) loss is: A. Proportional to load squared B. Constant whenever energised C. Zero at light load D. Proportional to load Answer: B. It is independent of load while the transformer is energised. (Key point: Core loss nature) 4. In the example, the 20 kVA case compared with 60 kVA had: A. One-third load and two-thirds of the losses B. Equal losses C. Double the load and half the losses D. One-third load and one-third losses Answer: A. Losses 1868 vs 2793 kWh, about two-thirds, so loss per kWh doubles. 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