TRANSFORMERS 303 



Primary impedance is now z\ 0.6 + 0.8 j. 



Secondary impedance is z 2 = 50 + 80 j. 



Primary exciting admittance is y ' = 10 2 y = (2 6j)10~ 4 . 



Let the secondary terminal e.m.f. be E and take it as the real 

 axis, the other e.m.fs. and currents may then be expressed in 

 rectangular coordinates. 



Secondary terminal e.m.f. 



E = E + Oj. 

 Secondary current 

 J 2 = + 5j. 

 Secondary impedance e.m.f. 



JEV" = 7 2?2 - 5 j (50 + 80 j) - - 400 + 250 j. 

 Secondary induced e.m.f. 



E 2 = E + E 2 " f = E - 400 + 250 j. 



Primary induced eoin.f. 



E' = T V# 2 = 0.1 E - 40 + 25 j. 

 Primary load current 



/' = 10/2 = + 50.;. 



Primary exciting current 



J ' = E'y ' = (0.1 E - 40 + 25 j) (2 - 6 j)W~* 



= 1(0.2 E + 70) - (0.6 E - 290) JJ10- 4 . 

 Total primary current 



/! =/'+/= [(0.2 # + 70) - |(0.6 E- 290) HH-SOjj]. 

 Primary impedance e.m.f,, 



E"' =/ 1?1 = [(0.2#+70)-K0.6#-290)10- 4 -50b](0.6+0.8j) 

 = K(0.6 E - 190) 10~ 4 - 40j + K- 0.2 E + 230) 10~ 4 + 30}]. 

 Primary impressed e.m.f. 



E 1 = E f + EJ"' = (0.1 E - 80) - 55 j. 



The absolute value of primary impressed e.m.f. is 



E l = V(0.1 #-80) 2 +(55) 2 = 2000 

 and solving this gives the secondary terminal e.m.f. 

 E = 23,160. 



