Copper saving in auto-transformer

Consider a single-phase autotransformer with $N_1$ turns across the primary and $N_2$ turns across the secondary ($N_2 < N_1$), with

$$K = \frac{N_2}{N_1} = \frac{V_2}{V_1} = \frac{I_1}{I_2}.$$

 I1 o-->---+        section AB: (N1-N2) turns, current I1
           |        section BC: N2 turns, current (I2 - I1)
           +---o V2, I2 to load
           |
 o---------+---o

Copper weight is proportional to the length of conductor times its cross-section, which is proportional to turns × current. Hence:

Autotransformer

$$\begin{aligned} W_{auto} &\propto (N_1 - N_2)I_1 + N_2(I_2 - I_1) \\ &= N_1I_1 - 2N_2I_1 + N_2I_2 \end{aligned}$$

With $N_2I_2 = N_1I_1$ and $N_2I_1 = KN_1I_1$:

$$W_{auto} \propto 2N_1I_1 - 2KN_1I_1 = 2N_1I_1(1 - K)$$

Equivalent two-winding transformer

$$W_{2w} \propto N_1I_1 + N_2I_2 = 2N_1I_1$$

Ratio and saving

$$\frac{W_{auto}}{W_{2w}} = 1 - K$$ $$\boxed{\text{Saving in copper} = W_{2w} - W_{auto} = K\,W_{2w}}$$

The saving is a fraction $K$ of the copper of the two-winding transformer. It is large when $K$ is near 1 and small when $K$ is small.