Potential of an Induction- Coil at "make." 325 



Since each particular solution satisfies equations! (4) and (5) 

 Ave have, on substituting and adding, 



LiSAs+LjASB* 2 + R,2A=0, 

 L 2 C 2 SBc 2 + L 21 2 As + JR 2 C,2B.r + SB = 0, 



which by (11) give 



2Ar=- V Cs 2B: 2 , 



and 



SBc 2 =- 



EL 21 



__ EL 21 



OJ^L^l-F) 



= G, say. 

 Here k is the coefficient of coupling of the primary and 

 secondary circuits, that is, k= \ / L l2 L 2 ilL 1 L 2 ~ 

 equations for the B's are 



Bi +B 2 +B 3 =0 1 



lV~i-fB 2 r 2 + B 3 ~ 3 = .... y m 



Thus the 



B^+B 2 4+B 3 4=G 

 Similarly, for the A's the equation 



J 



:»2) 



s are 



Ai + A 2 + A 3 = ELj 

 z l z 2 z s H* 



Ai + A, + A 3 = - 

 A 1 2 1 + A 2 s 2 +,A 3 ^ 3 = 



E 



Ri •* 



E 



1^(1 -/:*)• 



(13) 



Along with (6) equations (13) enable us to find the primary 

 current and equations (12) the secondary potential at any 

 time after make ; it is the latter quantity that we are con- 

 cerned with here. 



Solving (12) for B„ we find, after reduction, 



B 1= ( L___ 



which by (7) and (8) becomes 



G j 



±7rn ' 'lira + {k x — $)i 



_G_ 2Trn-(Jc 1 -8)i 

 i-rrn 47rV } 4-(£ 1 --SV' 



B, = - 



