i I! It 'KM M"II"N 



to a lir>' "f con-taut ti-iisioii. the only variable being tin* 

 load, which may change in sign and amount NV-- must be 

 careful n- .'"ii-- the results obtained by alt- 



f tin- loail with thos- olitaiip-d by changing 

 the value of the induction factor or tin- ten-in of the 

 lii'..-; these cases will be < 



>presents the current taken by a motor when 

 running on a line of t. 'he aiva ' >IIKI> r'|tn8ents 



the product /.'. and thus expresses graphically the energy 

 per second supplied T,, th.- motor \\><\\\ the line. Sim-.- 

 7/OF=tan '/,'. tho area Ollt'H is the product r 1 /.'. or the 

 energy per second \p"ml--l in hrat. Furtln-r. .inn- lll\ 

 in th.- t.'iision of thi- lin-. ami ///-'i-ijual to the ppnlti 

 or the hent drop, it follows that /'A is tin- induced tension. 

 The area i>l'l\l> i> t!i--n t ; \ per srcoml cx|>ende<] 



. and this in shown graphically to be equal 

 difference 1- in* m.-rgy per second supplied 



to the motor from the lin- and the energy per second 

 s|*M>t in heating the resistance. 



Thin diagram nhowu how we may find ^niphically th- 

 tension of the line required to do work at a given rate. 

 When the tonju and the induction factor are given, we 

 can obtain the curr.-nt. l,.-t this In- hl\ If the speed is 

 given, knowing M, we can obtain tin- induced tension. 

 Ix-t this I).- A/' Tli.- iin-a './'A// is thus equal to th. 

 given rale of work: - The energy expended per second 

 in heat depends on the resistance, since e is now ' 

 Draw fil-'i) equal to tan '//; OD will thn !> tin* re juin^l 



-ion of the line. 

 We have suppose! th- K>il to vary from thr 



/ \! 



maximum 1-M ' t.. nothing, and have shown that if the 

 It 



