Self-induction of Wires. 



205 



beginning : investigated the differential equations concerned, 

 verified my suspicions, and gave the results in a Postscript. 

 I have since further found that, when using the only practi- 

 cable method of equal ratio, there are no other ways than those 

 described in the paper referred to of getting a true balance of 

 induction by variation of a single L or M, after the resistance- 

 balance has been secured. This will appear in the following 

 investigation, which, though it may look complex, is quite 

 mechanical in its simplicity. 



Write down the equations of electromotive force in the 

 three circuits 6 + 1 + 3, 1 + 5—2, and 3—4—5, when there 

 is impressed force in branch 6 only. They are (p standing 

 for d/dt), 



e 6 =(R 6 + L 6 p)C 6 +(E 1 + L 1 p)0 1 + (R3 + L 3 ^)C 3 

 +p (MeA + M 62 C 2 + M 63 3 + M 6 A + M 65 C 5 ) 

 + P (M 12 C 2 + M 13 3 + M 14 C 4 + M 15 C 5 + M 16 C 6 ) 

 + i 9(M 31 C 1 + M 32 2 + MsA + M 35 C 5 + M 36 C 6 ). 



= (E, + L lP ) d + (E 6 + L 5i >)C 5 - (R 2 + L 2i ?)C 2 

 + i 9(M 12 2 + M 13 3 + M 14 d + M 15 C 5 + M 16 C 6 ) 

 + j p(M 51 C 1 + M 52 C 2 + M 53 C 3 + M 54 C 4 + M 56 C 6 ) 

 -i>(M 2 A + M 23 C 3 + M 24 C 4 + M 25 C 5 + M 26 C 6 ) 



= (R 3 + L 3 p)C 3 -(R 4 + L 4i 9)C 4 -(R 5 + L 5i 9)C 5 

 +p(M 31 C 1 + M 32 C 2 + M^d + M 35 C 5 + M 36 C 6 ) 

 -p(M*fii + M^C. + M 43 C 3 + M 45 5 + M 46 C 6 ) 

 -p(M 51 d + M 52 d + M 53 C 3 + M 54 d + M 56 C 6 ) 



Now, eliminate C x , d> d by the continuity conditions 

 C^Ca+C,,, C 2 =C 4 -C 5 , C 6 =C 3 +C 4 , . . (66c) 

 giving us 



e 6 = X U C 3 + X 12 C 4 + X 13 d, ^ 



0=X 21 d + X 2 A + X 23 d, > . . . . (676-) 



— X 31 C 3 + X 32 d + X 33 dj ) 



where the X's are functions of p and constants. Solve for 

 C 5 . Then we see that 



> (65c) 



.X91 Xqo — X>oX 



-21^32 



22^31 



.... (68c) 



is the complex condition of conjugacy of branches 5 and 6. 

 This could be more simply deduced by assuming C 5 =0 at 



