of Measuring the Self- Inductance of a Coil. 851 



Differentiating (9), and eliminating a and — between the 



resulting equation together with (8) and (9), we finally 

 obtain 



f ~ M /' /T r>i 9\1 ^ m 



+ ^ + ,„C +7+ £(L-Gr-)J^ + -* 



= *T< C, *- I| )'5— T—^" • • (10) 



The solution of this differential equation will consist of 

 two parts, a complementary function and a particular 

 integral. The former is the solution obtained by putting the 

 right-hand side of (10) equal to zero. This part will, 

 consequently, be the complete solution in the case for which 

 E is constant. 



In this case the value of x is, therefore, 



x = Ax*?"* 1 * + A 2 e~ °* + A 3 <?-°- 3 ', 

 where « 1? a 2 , and « 3 are the roots of the cubic equation 



LC W '-{M0+L0£+L{Qy 



+ r /)+mC+ M + Z(L-(V)J y -»'=o, 



and Aj, A 2 , and A 3 are constants determined by the initial 

 conditions. 



From the above value of x — the current through the 

 galvanometer — the value of a, the current through the 

 shunted resistance, can be obtained. 



For, from (1) and (8) above, we have 



da, , _ / . £BE . dx 



dt 

 dx 



Cr dt+*= a =)r v+ -ir+i> 



Putting in the values of x and -^~. and solving for a. we 



find dt 



. _L &BE , . If \ e-«« 



a =A 5 e c, + -- + A^- -j^j—g- 



. ff \ e~ a * , A // "\ e-^ 



A 5 being another constant determined by the initial 

 conditions. 



