248 Dr. J. A. Fleming on the Distribution of 



or 



L -t- L 4(*-^= s ^-°*- 



Now y and x—y represent the strengths of the currents flowing 

 through the galvanometer and the shunt at any instant. 



If we integrate both sides of the equation from to oo, we 

 have 



[Liy-L 2 (#-y)]" = Sl x-ydt-G^ ydt. 



Now the left-hand side of the equation is zero because quan- 

 tities of the form of Ly represent the number of lines of force 

 which are added into the circuit of the galvanometer, and the 

 discharge may be divided into two parts, during one of which 

 lines of force are being added to, and in the other of which sub- 

 tracted from, the circuits of the galvanometer and shunt; and 



/» oo /% ao 



the sum of these is zero. Again, I (x-y)dt and L ydt re- 

 Jo ^Jo 

 present the whole quantities </ 2 an d q x of electricity which 

 have flowed respectively through the galvanometer and the 

 shunt. Hence we arrive at the conclusion that 



S^ — Gr2i = 0, 



or 



that is, the total quantity of the discharge is divided between 

 the two circuits inversely as their resistances. We see there- 

 fore that self-induction does not affect the ratio of division of 

 a discharge in a divided circuit, provided that no external 

 work, such as the moving of magnets or circuits conveying 

 currents, absorbs current energy. Hence, if a ballistic gal- 

 vanometer is shunted and a discharge- sent through it, if the 

 needle has sufficient moment of inertia and the discharge is 

 sufficiently short, so that the needle has not perceptibly moved 

 from its position before the discharge is over, then the whole 

 quantity of electricity is divided between the galvanometer 

 and the shunt in the inverse ratio of their resistances. 



§ 18. To complete the solution we have to calculate the 

 current flowing through the galvanometer and shunt at any 

 instant. 



Taking the two cycle equations 



L .f +0 H « 



