70 TRANSMISSION LINE FORMULAS 



II CoVp 8 IO~ 45 



and 



* 

 8640 r 4 



Let the resistance of the wire per centimeter be R, where 



R = - ohms per cm., 



Z 



and let m 



IO~ 9 



r 



co IP" 9 



_ 



R 

 Then the total drop in the wire is 



= IR + /to/ I0~ 9 ( 2 Iog 6 - H --- / 

 \ p 2 12 



volts per centimeter. 



The total drop in phase with the current is 



IR(I + m 2 --^-m 4 + ' } (6) 



\ 12 180 / 



The total copper loss due to all the currents in the wire is 

 therefore equal to 



12 180 



This can be checked by integrating the losses due to the 

 total in-phase and quadrature currents in all parts of the 

 section of the wire, the above result being obtained by 

 this method also. Thus, in every respect, both as to volt- 

 age drop and watts loss, the resistance of the wire to the 

 alternating current is 



+ -w 2 --|-w 4 ' -V 



12 I 80 / 



