62 



Mr. 0. 11 



eavisiue on iiic 



th 



per unit of length. For a cable 2000 miles long, with c = J 

 microfarad, and £ = 6 ohms, we have ekP= 8 seconds, and 

 a =*1866 sec. The current, though calculable from the first 

 instant, is relatively insensible for some little time. Thus, 

 when t= l*5a, it has only reached *0047 of its final strength, 

 but, thereafter increasing much more rapidly, reaches half its 



IT 

 final strength in (>a, '98 in 20a, and its final strength — after 



the lapse of a theoretically infinite time. 



M 



Fio-. 1. 



Now compare curve 1 with curve 3. In the latter all the 

 circumstances are the same, with the exception that there is a 

 fault of infinitely small resistance situated at the middle of 

 the cable. Of course such a fault could not be worked through, 

 since no current would arrive at the receiving end. Never- 

 theless this is not by any means a case of an unpractical 

 nature; for it is quite possible to work, and very well too, a 

 cable containing a fault of next to no resistance. It will be 

 seen that with the fault of no resistance the current becomes 

 sensible sooner, and increases much more rapidly. It reaches 

 •0017 of its final strength in la, -044 in 1±, -1318 in 2a, *4274 

 in 3a, *68 in 4«, *8357 in 5a, and '9826 in 8a. Half the final 

 strength is reached in 3'3a, as against 6« with no fault. 



When the fault has a finite resistance the arrival curve of 

 the current is intermediate between curve 1 and curve 3. 

 The one shown by curve 2 corresponds to the case of a fault 

 having a resistance equal to one fourth of the cables. This 



makes the final strength of current = --- one half its value 



when there is no fault. In 2a the current reaches '0429 of 



