Autotransformer, a basic transformer connected as an autotransformer, "EXO 4 -Video 1, 3 "

A load R operates at 480 [Veff] at 50 [Hz]. We have a distribution line with a voltage amplitude of V=848.5 [Vmax] and a basic transformer with a ratio of n=600/120. The apparent power is S=15000 [VA] at 50 [Hz]. This transformer is connected as an autotransformer to obtain 480 [Vrms] to supply R. Show the connections and evaluate the apparent power Smax available when R is supplied by the autotransformer. What is the value of R corresponding to this maximum apparent power?
Note:
1) This application exercise allows us to see the reason why the diameters of the windings at the primary and secondary of an autotransformer are different.
2) The maximum apparent power is calculated using the effective values Smax=Veff.Ieff; the direction of the currents is not required.
3) The direction of the current is then evaluated when the circuit is not closed, at no load. Thus, to determine the direction of current i2 in the secondary of the autotransformer, we assume, when the circuit is connected to the source, that the voltage alternation is "n" and that the directions of flow of i0, i1, and i2 are those assumed at the outset, without phase shift, i.e., the phase angles of i0, i1, i2 are all zero. We then apply the rules i1/i2 = -n and V2/V1 = +n via the modules and phases of i1, i2, and V1, V2 to deduce the modulus and phase of i2. There must always be 2 incoming currents and one outgoing current during one alternation, then 2 outgoing currents and one incoming current during the next alternation. This explains the reason why the diameters of the windings at the primary and secondary of an autotransformer are different.