154 



THE TRANSFORMER. 



curve is seen to be part of a parabola. This is due to the fact 

 that the voltage drop due to resistance is proportional to the 

 load current, and the watts spent in the losses are the product 

 of current and voltage drop. Thus the watts are proportional 

 to the square of the load current. 



DETERMINATION OF COPPER LOSSES IN A TRANSFORMER. 



Transformer No .Type 



Output kw cycles per second. 



Transformation . . volts to . . . . volts. 



A slight modification of the method just given avoids 

 the measurement of the loss in instruments. By putting an 

 ammeter in the primary circuit between the supply 

 terminals and the wattmeter, the observations of the power 

 supplied at a given primary current may be observed. The 

 secondary winding is then short-circuited by a stout conductor 

 of negligible resistance. The ratio of the primary to secondary 

 currents may be determined by calculation or by a separate 

 measurement. 



Effect of Magnetic Leakage. The alternating field pro- 

 duced by the current in the primary winding will only produce 

 the voltage given by the formulae on page 126, if the whole of 

 the magnetic lines follow the' magnetic path formed by the 

 core, so as to pass through both primary and secondary 

 windings. Any lines which are formed in the primary which 

 do not pass through the secondary will produce a back 

 electromotive force in the primary in the same way as the 

 remainder of the field, but will not produce any effect on the 

 secondary voltage. Similarly, the current in the secondary 

 may produce some magnetic lines which do not pass through 

 the primary. These lines will act in opposition to the main 

 primary field in the secondary winding, lessening the flux 



