116 TREATISE ON ALTERNATING CURRENTS. 



where t M is the magnetizing wattless component, and i h is the 

 hysteretic energy component. 



The component i M is determined as follows. We have 



B = fiH 



where JJL is the permeability of the iron core, and H is the mag- 

 netizing force. Also 



H_ _ 4ir N^ 



\/2 = 10 ' L 



where t M is the R.M.S. value of the wattless component, and H the 

 maximum value of the magnetizing force. Therefore 



BL 



Substituting the values B = 2500, L = 65, N = 230, and 

 H = 2390, we get 



2500 x 65 

 V "" 1-76 x 2390 x 400 



= 0167 ampere (17) 



Also, i h is calculated from the fact that the primary impressed P.D. 

 multiplied by i k is equal to the total iron loss, so that 



1000i = 150 



in = 015 (18) 



Now, i' M and i h are in quadrature ; therefore 



i = x/015 2 4^01672 

 = 0-225 ampere ...... (19) 



The power factor on open secondary circuit is given by 



cos = ^ 

 t 



015 

 ~ 0-225 

 = 0-67 (20) 



To find the watts lost by radiation and conduction of heat per 

 square inch of cooling surface at no-load and full-load respectively. 



