353-355] Generation of Heat 309 



The problem in this form can also be attacked by the methods of the 

 infinitesimal calculus. Let V be the potential at a distance x along the 

 cable, V now being regarded as a continuous function of x. Let the 

 resistance of the cable be supposed to be R per unit length, then the re- 

 sistance from x to x + dx will be Rdx. The resistance of the insulation from 



Si 



x to x 4- dx, being inversely proportional to dx, may be supposed to be -7- . 



Let C be the current in the cable at the point x, so that the leak from 



dC 



the cable between the points x and x 4- dx is =- dx. This leak is a current 



dx 



a 



which flows through a resistance -?- with a fall of potential V. Hence by 

 Ohm's Law, 



j 



= - -T- dx -j- , 



dx \dx 



dC 



Also, the fall of potential along the cable from x to x 4- dx is -7- cfo, the 



current is C, and the resistance is Rdx. Hence by Ohm's Law, 



< 282 >- 



Eliminating G from equations (280) and (281), we find as the differential 

 equation satisfied by F, 



d_ ( I dV\ _ V 

 dx\Rdx)~ S' 



If R and 8 have the same values at all points of the cable, the solution 

 of this equation is 



VA cosh A/ -s x + B sinh ^J -~ x, 

 which is easily seen to be the limiting form assumed by equation (280). 



GENERATION OF HEAT IN CONDUCTORS. 



The Joule Effect. 



355. Let P, Q be any two points in a linear conductor, let V P , V Q be 

 the potentials at these points, R the resistance between them, and x the 

 current flowing from P to Q. Then, by Ohm's Law, 



V P -V Q = Rx ....... . ...................... (283). 



In moving a single unit of electricity from Q to P an amount oi work is 

 done against the electric field equal to Vp Vq. Hence when a unit of 

 electricity passes from P to Q, there is work done on it by the electric field 



