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



The solution of these for as and y is 



E B + S 



-s 



y = 



B 



4 +G 



and 



B 



4 +a 



B + S 



-S 



B 







= current through 

 [galvanometer, 



x = 



B 



B+S 



l£+g -s 



at 



= current througl 

 [the shunt. 



Writing out this differential equation for y 1 we have, 



B + S 



y+ 



B 

 -G 



B + S 



s 



y= 



E 







-B + S 



s 



or 



dy 



or 



B + SLg + (BS + RG + SG)y=ES, 



<fy BS+BG+SG _ ES 

 dt + (B + S)L y ~(B + S)I/ 

 The solution of this differential equation is 



Eg / BG+GS+BS 



y- 



l— e ~ (B+8)L" 



BG + GS + BS^ 



This gives the value of the current through the galvanometer 

 at any time, t, after starting the flow by making the connec- 

 tion with the battery. 



When £ = 0, then y=Q, and as t increases y increases, and 



finally, when t= oo, y= „p ,n<s ■ pg ? or j as ^ mav De 



S E 



written, 3/=^-^ 



B + 



SG ' 

 G+S 



This last is the ordinary formula given for the current 

 through a shunted galvanometer ; but we see that when self- 

 induction is taken into account, it is not until after an infinite 

 time that the current rises to this value, 



