--- title: "Copper saving in auto transformer" subject: "Electrical Machines I" chapter: 2 key: "semester_4/electrical_machines_i:autotransformer-copper-saving" source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/autotransformer-copper-saving site: BCT Engineering --- # Copper saving in auto transformer Electrical Machines I · Chapter 2: Transformer - Asked in: 2071 Magh, 8 marks - Video: [15 Copper Saving in Aurotransformer Compare with Two Winding Transformer - Derivation and Equation](https://www.youtube.com/watch?v=ZlHx1DKC5ME) — 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/autotransformer-copper-saving - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2 ## Question Prove with suitable assumption that copper saving in auto transformer is significant when the transformation ratio is nearly equal to unity. ## Short answer For same rating and current density, copper weight is proportional to ampere-turns. For an autotransformer W_auto = (1-K)W_two, so copper saving is K times the two-winding copper. Thus saving is largest when K is near unity, where only a small fraction of power is transformed inductively. ## Answer An **autotransformer** has a single winding, part of which is common to both primary and secondary. Power is transferred partly by conduction and partly by induction, so it needs less copper than a two-winding transformer of the same rating. #### Assumptions 1. Same rating (same $V_1$, $V_2$, $I_1$, $I_2$) for both transformers. 2. Same current density in all sections, so copper weight ∝ (current × number of turns) of each section (length ∝ turns, cross-section ∝ current). 3. No-load current and losses are neglected, so $N_1I_1 = N_2I_2$. 4. Step-down case with $K = \frac{N_2}{N_1} = \frac{V_2}{V_1} = \frac{I_1}{I_2} < 1$. ```text I1 o------>----+ A S section AB: (N1-N2) turns, V1 S current I1 +---------+---> I2 o S B | S section BC: N2 turns, V2 S current (I2 - I1) S | o-----------+ C ------+---------o ``` #### Two-winding transformer Primary: $N_1$ turns carrying $I_1$; secondary: $N_2$ turns carrying $I_2$. $$ W_{tw} \propto N_1I_1 + N_2I_2 $$ #### Autotransformer - Section AB has $(N_1 - N_2)$ turns carrying $I_1$. - Section BC has $N_2$ turns carrying $(I_2 - I_1)$. $$ W_{auto} \propto (N_1 - N_2)I_1 + N_2(I_2 - I_1) $$ #### Ratio of copper weights $$ \begin{aligned} \frac{W_{auto}}{W_{tw}} &= \frac{N_1I_1 - N_2I_1 + N_2I_2 - N_2I_1}{N_1I_1 + N_2I_2} \end{aligned} $$ Using $N_2I_2 = N_1I_1$ and $N_2 = KN_1$: $$ \begin{aligned} \frac{W_{auto}}{W_{tw}} &= \frac{2N_1I_1 - 2KN_1I_1}{2N_1I_1} = 1 - K \\ W_{auto} &= (1 - K)\,W_{tw} \end{aligned} $$ Therefore $$ \text{Saving of copper} = W_{tw} - W_{auto} = K\,W_{tw} $$ #### Significance when K is near unity The saving is $K$ times the copper of the two-winding transformer, so it grows as $K \to 1$: | K = V2/V1 | Copper in auto (× W_tw) | Saving | |---|---|---| | 0.1 | 0.9 | 10% | | 0.5 | 0.5 | 50% | | 0.8 | 0.2 | 80% | | 0.95 | 0.05 | 95% | Physically, when $K \approx 1$ the common section BC carries only the small difference $(I_2 - I_1)$, and section AB has very few turns, so very little copper is needed. Only the fraction $(1-K)$ of the power is transformed inductively; the rest is conducted directly. Hence autotransformers are economical when the voltage ratio is close to 1 (e.g. 400/440 V, 220/132 kV interconnection, variacs, starting of induction motors), but not for large ratios, where saving is small and the lack of electrical isolation is a safety risk. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Under equal current density, the copper weight of a winding section is proportional to: A. Turns x current B. Voltage x turns C. Current only D. Turns only Answer: A. Length goes with turns and cross-section with current, so weight is proportional to N x I. (Key point: Copper weight proportional) 2. In a step-down autotransformer, the common section BC carries: A. I1 B. I2 C. I1 + I2 D. I2 - I1 Answer: D. The common winding carries the difference of the output and input currents. (Key point: Current in common section) 3. The ratio W_auto / W_two-winding is: A. K B. 1 - K C. 1 + K D. 1/K Answer: B. Substituting N2 I2 = N1 I1 gives (2N1I1 - 2K N1I1)/(2N1I1) = 1 - K. (Key point: Weight ratio) 4. The saving of copper is K times the two-winding copper, so it is large when: A. K is close to 0 B. K equals 0.1 C. K is close to 1 D. The load is light Answer: C. Saving = K W_tw increases as K approaches 1, since little power is transformed inductively. 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