438 Mr. 0. Heaviside on the 



steady, and put on everywhere at the time t = 0, reaching the 

 final value C , 



C=C -2(-^) -, j^fe.^ ; . . . (191) 



$> x — § finding the jo's. We can find Y at distance z by inte- 

 grating the second of (189) with respect to z from to z ; 

 thus 



V=f «fe+(Z -f V(fe)0, . . . (192) 



Jo Jo 



wherein C is to be the right member of (191). This finds V 

 by differentiations with respect to t performed on C. In the 

 final state put R 7/ for R" and — R for Z , steady-flow resist- 

 ances. Y will usually vary with the time until the steady 

 state is reached ; but if the line is homogeneous, with only 

 the two constants R and L, and if also Z and Zj are zero, 

 V will be independent of f, and instantly assume its final 

 distribution. 



Thus, on these assumptions, we shall have 



c=(r^/RQ(i-e-^ L ), 



V= p«fe-(*/Z)f «fer. 



Jo Jo 





(193) 



showing the current to rise independently of the distribution 

 of £, and V to have its final distribution from the first moment, 

 which, when the impressed force is wholly at z = 0, of amount 

 e oy is e (l—z/l). This infinitely rapid propagation of V is 

 common sense according to the prescribed conditions, but 

 absolute nonsense physically considered, especially in view of 

 the transfer of energy. The question then arises, How does V 

 really set itself up, when the line is so short that the current 

 rises sensibly according to the electromagnetic theory ? 



To examine this, let the line constants be R, 8, L (inde- 

 pendent of d/dt), and Z^Zo^O. Put on e at z=0 at time 

 2 = 0. Y and C will rise thus (a special case of (183) and (187)), 



C = |(l-e-^) + |e-B^ S | i eos^si„^ 



where m has the values tt/Z, 27t/Z, &c, and 

 m'=(4m 2 L/R 2 S-l)*. 



