166 Mr. 0. Heaviside on the 



(24). It is necessary in this instance, to first find the odd a's 

 from the second equation (21) and Tables. 



18. Now for working with condensers at both ends. Let 

 orti = 0, m a = 0, Wj = oo ? n 2 = oo , 

 and let ^ and r 2 be both very small. At time t after the in- 

 troduction of E at P, the potential of the line is 



ax 

 C0S -T aH 



v = 2Er£ r— e-^r (28) 



1 , sin a v 7 



a 

 from ,v = to x — = , and 



ax 1 

 cos vjt cos -y- a2; 



« = 2Er 1 S : -e"^ . . . (29) 



sin a v 7 



a 

 from a/=0 to ss'= ^-, where x' — l—x. 

 The a's are the positive roots of 



sin a cos 2 ^=0; (30) 



za 2 



or 



a ' . a 1 

 cos 2=0, tan g = g-, 



£#/ being the resistance of the fault in the centre. 



The current Y arriving at x—l is T = r 2 cl~j~; that is, 



-T, 2E ^— a 2 COS27T -^ /qix 



r= ir^ 2 —^r £ T - • • • (31) 



1 +• 



a 



When there is no fault, c = co , a,=?7r, and equations (28) and 



(29) both become 



CO • i2 ff 2< 



v = Er 1 + 2Er 1 2cos^pe "r (32) 



i ( 



Here Er x is placed outside the %, because « =0, and the value 



f . i s i for a and 1 for the rest. The current leav- 



1 sm a. 



mg x=0 is — ripl^r-'f oi 



T, =a =-fry^e-^;. .... (33) 



