224

ELECTRICAL APPARATUS

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It is, then, substituted into (11) to (13):

100

(1.062 + 0.52-a) + j (0.36 a - 0.028)

___________(80 - 60 j) a__________

(1.062 + 0.52 a) + j (0.36 a - 0.028);

for:

a = 0, or no-load, this gives:

62 = 94.1,
<* = 0,
io = 19.5;

for:

a = oo 9 or short-circuit, this gives:

02 = 0,
12 = 159,
ii = 169.

The voltage diagram is shown in Fig. 70, and the load char-
acteristics or regulation curves in Fig. 71.

As seen: the voltage, 62, is already at no-load lower than the
supply voltage, e0, due to the drop of voltage of the exciting cur-
rent in the self-inductive impedance of the phase converter.

In Fig. 70 are marked by circles the values of voltage, 62, for
every 20 per cent, of the short-circuit current.

Fig. 71 gives the quadrature component of the voltage, 62, as
e"i, and the apparent efficiency, or ratio of volt-ampere output
to volt-ampere input:

62^*2
r =

and the primary supply current, i0.
It is interesting to compare the voltage diagram and especially
the load and regulation curves of the induction phase converter,
Figs. 70 and 71, with those of the monocyclic square, Figs. 67
and 68.
As seen, in the phase converter, the supply current at no-load
is small, is a mere induction-machine exciting current, and in-
creases with the load and approximately proportional thereto.
The no-load input of both devices is practically the same, but
the voltage regulation of the phase converter is very much better:
the voltage drops to zero at 159 amp. output, while that of the