SINGLE-PHASE INDUCTION MOTOR

119

It is then, counting the voltages and currents in the direction
indicated by the arrows of Fig. 42:

/O — / 1 — f 2 = f'2 —

substituting:

(81)

(82)

gives:
thus:

F2

+

•• m (cos <t> + j sin

(83)

This gives the phase angle, <j>, between the voltages, $1 and $2,
of the monocyclic triangle. Since:

it is, by (83):

= e

Yl

2 Y

(84)
(85)

anid the quadrature voltage:

s

I — F2

_

_ _ _________

and the total current input into the motor, inclusive starting
device:

/»

= e

= e-

F2 + 2 F

I

F2)+3F2

/1+y2 + 2F (87)
As with the balanced three-phase motor, the quadrature com-
P
ponent of voltage numerically is ~ A/3, it is, when denoting by: