PHASE CONVERSION 223

it is, then, choosing the diagrammatic representation, Fig. 69D:

/o - -EoFo = h + Erf, = A, (9)

Eo = Et + 2 Z (I, + $2y0), (10)

h = EtYf, (11)

substituting (11) into "(10) and transposing, gives:

if the diagram, Fig. 692?, is used, it is :

which differs very little from (12).
And, substituting (11) and (12) into (9):

h = #2 (7o + Fa) + #oY"o,
• « P 72+27o+2Zro(y0+72)

(14)

Equations (11), (12) and (13) give for any value of load, 72,
on the quadrature phase, the values of voltage, #2, and current,
/2, of this phase, and the supply current, I0, at supply voltage, $0.
It must be understood, however, that the actual quadrature
voltage is not #2, but is ^2, carried a quarter phase forward by
the rotation, as discussed before.
132. As instance, consider a phase converter operating at con-
stant supply voltage :
$0 = e0 = 100 volts;
of the constants :
YQ = 0.01 - O.f j,
Z0 = Zi = 0.05 + 0.15 j;
thus:
Z = 0.1 + 0.3 j;
and let:
Yz = a (p - jq)
= a (0.8 - 0.6 j),
that is, a load of 80 per cent, power-factor, which corresponds
about to the average power-factor of an induction motor,