TRANSFORMER ON OPEN SECONDARY CIRCUIT. 95 



W the number of ergs per cycle lost in hysteresis, n the frequency 

 of the supply, and E Q the primary applied P.D. Then (neglecting 

 eddy currents) 



nWV 



or 



. nWV 



..... (8) 



We will show later (see 61) how the loss due to eddy 

 currents may be approximately calculated. The current necessary 

 to supply energy for this loss must be included in i h . 



TRANSFORMER WORKING UNDER LOAD. 



6O. We now pass on to consider the working of a transformer 

 under load. If a transformer is used to supply power for incan- 

 descent lamps only, the external circuit may be treated as non- 

 inductive; if, however, it supplies arc lamps requiring choking 

 coils, or induction motors, the load is inductive, and the external 

 reactance must be taken into account. 



We shall assume that the maximum induction in the iron core 

 is constant at all loads. 



The method of vector algebra supplies a concise means of 

 performing the necessary calculations. 



We shall for the present assume that both the reactances of 

 the two circuits and their mutual induction are constant. 



Let the resistance of the primary circuit be r\, its reactance Si, 

 and the E.M S. primary current i lt and let the corresponding 

 quantities for the secondary circuit (internal and external) be r. 2 , 

 82, and 1*2, the coefficient of mutual induction between the two 

 circuits M, and the E.M.S. potential difference between the primary 

 terminals e. 



Then, following the argument of 47, Chap. VIII., the vector 

 equation of E.M.F.s in the primary circuit is 



Tiii -}- ksiii + "kpMi% = e (9) 



The E.M.F. in the secondary circuit due to mutual induction 

 is - kpMii. This has to furnish a component, r 2 H in phase with 

 the secondary current to drive the current against the ohmic 

 resistance of the circuit, and also a component, ks&z, to balance the 



