r,

I !

410

ELECTRICAL APPARATUS

Substituting these trigonometric values into the expression
(121) of the voltage ratio for minimum commutation current,
it is:

__ j^ sin (o- — 3)

u •—-• „

(127)

Substituting (117) into (106) and expanding gives a relatively
simple value, since most terms eliminate:

J0 = e {[cos2 (a — d) + b\ (sin a sin (a — 5) — cos 5)]

y [ sin (0- — 8) cos (cr — 6) — &X4 (sin cr cos (cr — g) —• sin 3)]}

(1 — c0X4, (GO

and the absolute value:

* e ^ CQS ^ "" ^ ""

cos

(128)
(129)

"0 ri A i

CQ& [ 1 — CoA4J

or, resubstituting for a and 8:

, = ^X^-co^-X/)} f .

^0 n *y fl /> \ 1 {/* 2 -.1-- C2^ ' ^lOUy

CQ^/ [JL CoA4J ^Co "T" ^ /

From (129) and (130) follows, that iffQ = 0, or the commutation
current vanishes, if:

cos (a — <5) — 6X4 cos (7 = 0, (131)

or:

SX"4 - co (X426 - V4) = 0.
This gives, substituting, X;/4 = A/X42 — XX42, and expanding:

{&X4c02ąJ

Co2 +

Co2 (62X42"::.....1)7

COS (cr — 5) =

Vc02 + S2
From (131) follows:

cos (cr — d) = 6X4 cos a.
Since cos (a — 5) must be less than one, this means:

6X4 COS cr < 1,

or:

1

(132)

X4<

b cos

or:

X4<

or, inversely:

Cob

S > Co V62X42 - 1.

(133)