130 Mr. 0. Heaviside on the Extra Current. 



be adopted to solve the equation under consideration, viz. 



l v , dv d 2 v 



=ck di +cs d? < 3 > 



Let the potential of the wire at any moment be 



sin iirx 

 cos "7 



''=v--F-/(0, (*) 



where f(t) is a function of t only, and V is constant, From 

 (4) by differentiation, 



d 2 v 

 therefore by (3), 



dx 2 ~ I 2 V; 



d 2 v 7 dv ^ 2 7^ 2 . 



cs d? +ck d> + ir v=0 ' 



the solution of which is 



if 4z' 2 7r 2 ^ < 1, and 



? ? = 6~2S(A / cos + B'sin) ^A/^'V 2 ^ —1 



if 42 2 7r 2 ^ > 1. Here A, B, A / and B / are constants, and a = t 



and f3 = ckl 2 , both time-constants. Therefore if the potential 

 when t = is 



TT sin rrrx 



t,=v co S -r 



the potential of the wire at the time t is 



Vi-*«v« ," AX -l\/i- 



tf = V C0S -v-. 6 2«.|A.€2« V / 3+(l_A> 2« V i8} ? W 



or 



according as 4zV 2 ~ < or > 1. The remaining constants A and 



B ; must be determined from the value of the current at some 



dv 

 fixed time. By solving equation (2), where — — is to be 



found from (5) and (6), we shall find 



