Autotransformer, "EXO 4 -Vidéo 2/3".

“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 estimate the apparent power Smax available when R is supplied by the autotransformer. What is the value of R corresponding to this maximum apparent power?
Remark:
1) Maximum apparent power is calculated using RMS values Smax=Veff.Ieff; current direction is not required.
2) Current direction is then evaluated when the circuit is not closed, at no load. So, 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, with no 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 moduli 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 half-wave, then 2 outgoing currents and one incoming current during the next half-wave. The 2 alternations considered can be “n” and “n+1” or “n+13” and “n+14”. it doesn't matter.”