138 Mr. 0. Heaviside on the Extra Current* 



In the intermediate case, when ??? = m / = 0, 





cos I 

 and 



dQ, r* -- VlTT COS WTCC _ — t 



eft A-Z sm / 



a 



(12) 



The current Ce * does not influence the potential in any 

 way. The above solutions suppose that the initial current is 



Sill VTTX 



C, and the initial potential u=V" -y 1 -, and give the poten- 



VTTX 



tial and current at any time after. When sin -j— is taken, the 



potential at the ends of the wire is always zero ; and when 



cos —j— is taken, the current is always zero at the ends. 



After this preliminary we can pass to more practical cases. 



In the first place, let a constant current -j-j be flowing through 



the wire, caused by a battery of negligible resistance and elec- 

 tromotive force V ; and let the potential of the wire be 



V(l- ?Y so that it is V at the end P and at the end Q. 



By Fourier's theorem, 



2V«> 1 . forx 



„ / xs. 2V21 . 

 V (1— 7 )= — Z^si 



'-T ; 



therefore, if the end P is put to earth at the time t = 0. the po- 

 tential at the time t is 



2V„ 



v= — Z 



IT 



1 . form --L (1+nii ^ 1— m, { J™\ 



-. Sill —f— . e 2« . < — - , 6 2a . e 2a \ 



% I i. zm i ziiii ) 



4 Z-sm-j— . € 2a ( cos + — T sm ) -~ . (13) 



it % I \ m' { J 2cc v ' 



by (7) and (8), where the first series includes all integral 



values of i which make 4i 2 ir 2 -^ — 1 negative, and the second 



series all the rest up to i = cc ■ , And by (9) and (10), 



dt ~ hi ' € 



2V _ form e * / *™i _*»£\ 



2V ^ iirx 2e 2a . tm' ' 4 

 —j- . sin -^ 



-w 2cos ~r-^7- sin: 2T • ( 14 ) 



