124

ELECTRICAL APPARATUS

tive impedance, Z\ = 0.1 + 0.3 j; supply voltage, -eQ = 110 volts,
and rated output, 5000 watts per phase.
Assuming this motor to be operated: •

1. By transformers of about 2 per cent, resistance and 4 per
cent, reactance voltage, that is, transformers of good regulation,
with constant voltage at the transformer terminals.

2. By transformers of about 2 per cent, resistance and 15 per
cent, reactance voltage, that is, very poorly regulating trans-
formers, at constant supply voltage at the transformer primaries.

3. With constant voltage at the generator terminals, and
about 8 per cent, resistance, 40 per cent, reactance voltage in
line and transformers between generator and motor.

This gives, in complex quantities, the impedance between the
motor terminals and the constant voltage supply:

1. Z = 0.04 + 0.08j,

2. Z = 0.04 + 0.3 j,

3. Z = 0.16 + 0.8 j.

It is assumed that the constant supply voltage is such as to
give 110 volts at the motor terminals at full-load.

The load and speed curves of the motor, when operating under
these conditions, that is, with the impedance, Z, in series between
the motor terminals and the constant voltage supply, e\9 then can
be calculated from the motor characteristics at constant termi-
nal voltage, CQ, as follows:

At slip, s, and constant terminal voltage, e0, the current in the
motor is f 0, its power-factor p = cos 6. The effective or equiva-

lent impedance of the motor at this slip then is 2° = -A, and, in

^o

complex quantities, Z° = -r- (cos 6 + j sin 0), and the total im-

IQ

pedance, including that of transformers and line, thus is:

or, in absolute values :

^ •+«)'•
and, at the supply voltage, e1; the current thus is: