530 BEPOET — 1881. 



oiVby(3),(l),and(2), 



r^(^^')^ . . . .(13) 



KK'^H + R + R') 

 Hence either c=0 (14) 



, KK'{E+R + R') f,f.. 



^= V{Rlt')v (^^> 



The case of c = ia that in which 



v<.— — ^^ — i ..... (lb) 



■where I^ denotes the value of J for c = O. To understand it, remember we are 

 supposing no residual magnetism. For any speed subject to (^16), the dynamo 

 produces no current. When this limit is exceeded the electric equilibrium in 

 the circuit becomes unstable ; an infinitesimal current started in either direction 

 rises rapidly in strength, till it is limited by equation (15), through the diminu- 

 tion of I, which it produces. Thus, regarding 7 as a function of c, we have 

 in (15) the equation mathematically expressing the strength of the current main- 

 tained by the dynamo when its regular action is reached. Using (15) in (10) 

 we find — 



»- = ^ (17) 



which we all knew forty years ago from Joule. 



In the shunt-dynamo the whole current, t-', of the working coil branches into 

 two streams, c through the electro-magnet, and c' — c through the external circuity 

 whose strengths are inversely as the resistances of their channels. Still calling 



the resistance of the external circuit E, we therefore have — 



■p 

 c 2t=(c'-c) E, which gives c = „ „ c' . . . (18) 



Hence, by Joule's original law, the expenditures of work per unit of time in the 

 three channels are respectively 



R' c'"^ . , . . working coil 1 



^(^P" • -l^^tro-magnet j_ _ ^ ^^^^ 



■^(.ff — f)^'^ ' external circuit 



Hence, denoting as above by r the ratio of the whole work to the work developed 

 in the external circuit, we have — 



. = U.^j Kr.e) ^ ^ ^(20) 



^{r^J 



whence R'' r= R'^-^^-^ + R {R + E) 



E 



^'^' (R + R')E+R(2R' + R) 



(21) 



E 



Suppose now R and R' given, and E to be found ; to make r a minimum. The 



solution is- 



aud this makes 



Put now 



^=^li5' (^^> 



. g , R'(R + R ') 2R' + R ^qq\ 



f- ^''^ 



