454 Induction of Currents in Linear Circuits [CH. xiv 



As in 514, it is simplest to assume an electromotive force Ee ipt : the 

 solution actually required will be obtained by ultimately rejecting the 

 imaginary terms in the solution obtained. 



The equations to be solved are now 



~(Li l -}-Mi. 2 ) = Ri l (450), 



(Mi 1 + Ni,) = Si a (451). 



As before both ^ and i 2) as given by these equations, will involve the 



d 

 dt 



time only through a factor e ipt , so that we may replace -j- by ip, and the 



equations become 



Rii + Lip\ + Mipi z = 



Si. 2 + Mipi 1 4- Nipi 2 = 0, 

 from which we obtain 



8 + Nip -Mip (R 

 The current ^ in the primary is given, from these equations, by 



Lip 



T . 



S + Nip 



M *p* (S - Nip) 



R' + L'ip' 



R 



The case of no secondary circuit being present is obtained at once by 

 putting 8 = oo , and the solution for ^ is seen to be the same as if no 

 secondary circuit were present, except that R', L' are replaced by R and L. 

 Thus the current in the primary circuit is affected by the presence of the 

 secondary in just the same way as if its resistance were increased from 

 R to R', and its coefficient of self-induction decreased from L' to L. 



The amplitudes of the two currents are j i\ and i 2 \ , so that the ratio of 

 the amplitude of the current in the secondary to that in the primary is 



M ip 



S + Nip 

 _ M P_ (452) 



" 



