of the Synchronous Motor. 59 



(6) Also from (4) we see that 



e — E according as I = 2R, 



that is according as (27mL) 2 + R 2 = 4R 



3R 2 



" » L2 <4ttV 



< 2™ * 



Running Light. 

 5. We have 



p + c 2 R=cE COS'v/r ^ 



and p = ce cos </> J • 



If we neglect the friction of the bearings &c, we may, in 

 this case, put p = ; we then have 



and 



cR=E cost/t. 



Hence the maximum value of c is (putting -^=0) 



E 



C= R' 



the same as would he produced by a constant E.M.F. E in a 

 non-inductive circuit of resistance R. 



7T 



Also putting ijr= ± J-? the minimum current is zero. 



Now, from fig. 1, we get (of course E and c are not now in 

 phase) 



E 2 = e 2 -rIV + 2L*cos (0-0); 



when <f> = + ~~ , this becomes 



E 2 =£ 2 + IV±2L*?sin0 



=« 2 + lV±2Sw, (5) 



• a S 

 since smc7= j. 



The upper sign in (5) corresponds to the machine running 

 as a generator, and the lower sign as a motor. 



We also see from (5) that corresponding to c= vr we have 



_ ES 



