SINGLE-PHASE INDUCTION MOTOR 107

only due to its simplicity and cheapness, where moderate start-
ing torques are sufficient, and thus the starting-torque efficiency
less important. For producing high starting torque with high
starting-torque efficiency, thus, only capacity and inductance
would come into consideration.

Assume, then, that the one impedance is a capacity:

x2 = — k, or: Z* = — jk, (45)

while the other, $1, may be an inductance or also a capacity, what-
ever may be desired:

Zl = +jxlt (46)

where xi is negative for a capacity.
It is, then:

(35): m (cos <j> + j sin 0) =

r + j (xi + a?) [r2 - (xl + x)(k ~- x)] + jrxik . .
r-j(k-x) ~~ r2 + (k - x)2 * l j

True quadrature relation of the voltages, e\ and e2, or angle,
$ = ff requires:

JU

cos 0 = 0,

thus:

(xx + a) (fc - a?) = r2 (48)

and the two voltages, ei and e2, are equal, that is, a true quarter-
phase system of voltages is produced, if in

(34): [Z + Zi] - [Z + Z*],

where the [ ] denote the absolute values.
This gives:

r2 + (xl + xY = r2 + (* - x)\
or:

Xl + x = k - x, (49)

hence, by (48):

Xi + x = k — x = r,
k = r + x,

(50)
XL = r - x. '
Thus, if x > r, or in a low-resistance motor, the second reactance,
\j also must be a capacity.