TRANSFORMERS 293 



is leading) because a larger component of impressed e.m.f. is 

 consumed in driving the current through the primary impedance, 

 thus, E' = - E b = E, - 7 1?1 (278) 



A smaller flux, is, therefore, required and a smaller exciting 

 current. The decrease in flux from no-load to full-load non-in- 

 ductive is 1 or 2 per cent and for an inductive load of 50 per cent 

 power factor is only 5 or 6 per cent. With anti-inductive load the 

 flux increases. 



The secondary induced e.m.f., which is proportional to the 

 primary, decreases with it. 



The secondary terminal e.m.f. E is less than the secondary 

 induced e.m.f. by the e.m.f. consumed by the secondary impedance ; 

 thus, E = E 2 - I 2 z 2 . (279) 



178. Vector Diagrams for the Transformer. Fig. 267 is the 

 vector diagram of e.m.f. 's and currents in a transformer with a 

 load impedance Z = R -f- jX and a load power factor 



R 



cos 6 = 



FIG. 267. Vector diagram of a transformer with inductive load. 



OE = E = secondary terminal e.m.f. 



0/2 =1.2 = secondary load current lagging behind E by 

 angle 6. 



OE 2 = E 2 = 7 2 r 2 = e.m.f. consumed by secondary resistance, 

 in phase with 7 2 . 



OEj r = E 2 " = jI 2 X2 = e.m.f. consumed by secondary react- 

 ance, in quadrature ahead of 7 2 . 



OE 2 " f = E 2 " f = 7 2 z 2 = e.m.f. consumed by secondary impedance. 



OE 2 = E 2 = E + EJ" = e.m.f. induced in the secondary wind- 

 ing by the alternating flux $. 



OE b = E b = e.m.f. induced in the primary winding by the 



flux $# fe = -# 2 . 



