THE TRANSFOEMER. 185 



primary back voltage, i.e., it passes through its maximum 

 and minimum values at the same time, but the direction of 

 the electromotive force is always opposite in the two coils. 



This bears out the fact, which was stated earlier, that the 

 secondary current acts as a demagnetising current on the 

 transformer core. 



The no-load current of the transformer will consist partly 

 of the idle magnetising current and partly of the energy 

 current, which is spent in overcoming the iron losses. The 

 iron losses are equivalent to a demagnetising action on the 

 magnetic circuit, in fact they are partly due to eddy currents, 

 which are demagnetising currents in the same way as the 

 load currents in the secondary winding. It is consequently 

 convenient to consider the iron losses as equivalent to a 

 current acting on the transformer as a load current, i.e., as 

 a demagnetising current. The value of this imaginary 

 current is equal to 



watts spent in iron losses, 



5 TT~ - an d is represented in Fig. 85 by 



primary voltage 



the line C K . The actual primary current overcoming this 

 demagnetising action is equal and opposite in phase to this. 

 Thus the total no-load primary current is the resultant of 

 the magnetising current C m and of a current which is 

 equal and opposite to (7,.. This resultant is shown as 

 C u . 



The iron losses in the transformer under con- 

 sideration were 21 watts, which corresponds to an 



21 



energy current of =^ = = -3 amperes, since the second- 

 ary voltage is taken as 70. The loss current must be 

 in phase with the secondary voltage and is accordingly repre- 

 sented by the line OC e . The resultant of this current 

 reversed and the current C m is C n , which is the total no- 

 load current of the transformer. To produce the current 

 C o requires a voltage G x R, which is shown on the diagram 

 by E r , R being 5-8 ohms, the resistance of the primary 

 winding. The volts given to the transformer are the resul- 

 tant of the back voltage E l and the resistance electromotive 

 force E r . Thus the total impressed voltage is shown in 

 magnitude and phase by the line E. 



The angle of lag of the primary circuit is the angle E E r , 

 since it is the angle between current and voltage. 



