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Auto transformer merits and Cu saving

Short answer

An auto transformer has a single winding, part common to primary and secondary. Merits: less copper (saving fraction K), smaller size, higher efficiency and better regulation. Demerits: no isolation, high short-circuit current, uneconomical for large ratios. Copper saving = K × W_two-winding, derived from ampere‑turn analysis of winding sections.

What is an auto transformer? State its merits and demerits over a two winding transformer. Derive an expression of cu-saving in auto transformer.

An auto transformer is a transformer with only one winding, part of which is common to both the primary and the secondary circuits. Power is transferred partly by conduction (the electrical connection) and partly by induction.

   I1                I2
 o--->---+          +---<--- o
         |          |
         |   N1-N2  |  load
         +----------+---o
         |   N2     |
 o-------+----------+---o

Merits over a two-winding transformer

  1. Less copper is needed (the saving is a fraction KK, derived below), so it is cheaper and lighter.
  2. Smaller size and lower weight for the same rating, since the core is also smaller.
  3. Lower copper and core losses, so higher efficiency.
  4. Better voltage regulation, as the leakage reactance and resistance are lower.
  5. Continuously variable output is possible with a sliding tap.

Demerits

  1. There is no electrical isolation between primary and secondary. A fault (for example an open circuit in the common winding) can put the full HV-side voltage on the LV side, which is dangerous to the user and the equipment.
  2. The short-circuit current is high because of the low internal impedance.
  3. It is not economical when the ratio is far from 1 (for example 10:1).
  4. For three-phase units the neutral must be solidly earthed, and special protection is needed.

Copper saving

Let K=N2N1=V2V1=I1I2<1K = \dfrac{N_2}{N_1} = \dfrac{V_2}{V_1} = \dfrac{I_1}{I_2} < 1. Section AB (of N1−N2N_1 - N_2 turns) carries I1I_1 and section BC (of N2N_2 turns) carries (I2−I1)(I_2 - I_1). Copper weight is proportional to the ampere-turns of each section.

For the two-winding transformer: W2w∝N1I1+N2I2=2N1I1W_{2w} \propto N_1I_1 + N_2I_2 = 2N_1I_1.

For the autotransformer (using N2I2=N1I1N_2I_2 = N_1I_1 and N2I1=KN1I1N_2I_1 = KN_1I_1):

Wauto∝(N1−N2)I1+N2(I2−I1)=N1I1−N2I1+N2I2−N2I1=2N1I1−2N2I1=2N1I1(1−K)\begin{aligned} W_{auto} &\propto (N_1 - N_2)I_1 + N_2(I_2 - I_1) \\ &= N_1I_1 - N_2I_1 + N_2I_2 - N_2I_1 \\ &= 2N_1I_1 - 2N_2I_1 = 2N_1I_1(1 - K) \end{aligned} WautoW2w=1−K,Cu saving=K W2w\frac{W_{auto}}{W_{2w}} = 1 - K,\qquad \boxed{\text{Cu saving} = K\,W_{2w}}

For example, with K=0.8K = 0.8 the autotransformer needs only 20% of the copper of an equivalent two-winding transformer, a saving of 80%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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