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Autotransformer working and Cu saving

Short answer

A single winding with a common section carries only the difference current I₂−I₁. Copper weight: Wₐ ∝ (N₁−N₂)I₁ + N₂(I₂−I₁) = 2N₁I₁(1−K). Two-winding transformer requires 2N₁I₁. Saving = K·Wₜ, a fraction K of the copper is saved.

Explain working principle of an auto-transformer. Derive an expression for Cu saving in an auto-transformer.

An autotransformer has only one winding. A part of it is common to both the input and the output circuits. Power passes from input to output partly by conduction (direct electrical connection) and partly by transformer action.

Working principle

 I1 -->  A o---------+---------o  (input)
                     |
             N1-N2   |
             turns   | I1
                 B o-+-----+---o  (output tap)
             N2      |     | I2
             turns   |     v
                 C o-+---------o  (common)
          V1 across AC, V2 across BC
  • The supply V1 is applied across the full winding AC (N1 turns). The alternating flux in the core induces the same emf per turn all along the winding.
  • The output is taken between a tap B and the common terminal C (N2 turns), so V2/V1 = N2/N1 = K.
  • By ampere-turn balance, I1/I2 = N2/N1 = K.
  • For a step-down unit, the section AB carries I1; the common section BC carries only (I2 − I1), the difference, which is why it can use thinner copper.

Copper saving

Weight of copper ∝ turns × current.

Auto:      Wa ∝ (N1 − N2) I1 + N2 (I2 − I1)
                = N1 I1 + N2 I2 − 2 N2 I1
Two-wdg:   Wt ∝ N1 I1 + N2 I2
Using N1 I1 = N2 I2:
           Wa ∝ 2 N1 I1 (1 − K),   Wt ∝ 2 N1 I1
           Wa / Wt = 1 − K

Saving of copper = Wt − Wa = K · Wt, i.e. a fraction K of the copper of an equivalent two-winding transformer. For K = 0.5 the saving is 50 %; for K = 0.9 it is 90 %.

Advantages: less copper, smaller size, lower losses, better efficiency and regulation. Disadvantage: no electrical isolation, so a fault on the high-voltage side can reach the low-voltage side. It is used for starting induction motors, voltage regulators (variacs), and joining systems of nearly equal voltage.

✓ Answer checked by Claude Opus 5.5 (high effort)

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