THE TRANSFORMER 181 



ance drop i^A'o, due to $2, in quadrature with I* and loading. 

 As both the primary and secondary induced voltages are induced 

 by the same llux, and both windings have the same number of 

 turns, since the ratio i< 1 to 1. the primary and secondary in- 

 duced voltages will be equal in magnitude and will be in phase 

 with each other. Therefore, EI = E 2 . It has already been 

 demonstrated that an emf. induced by a flux varying sinusoid- 

 ally with time is a sine wave and lags the flux 90 (page 27, Par. 

 1 I . Then-fore, in Fig. 174 (a) the mutual flux <t> kads the 

 induced emfs. by 90, as shown. 



The line must first supply a voltage at least equal to the 

 primary induced voltage, and in opposition thereto, before 

 current can flow into the primary. This is analogous to the 

 direct -current motor, where the line must first supply a voltage 

 equal to the back electromotive force, and in opposition thereto, 

 before any current can flow into the armature. Therefore, a 

 voltage Ei opposite and equal to EI must first be supplied 

 by the line. The primary must furnish at least a sufficient 

 number of ampere-turns to balance the ampere-turns of the 

 la ry. These primary ampere-turns and the secondary 

 ampere-turns are equal and opposite. Therefore, if there are 

 urns in t he secondary, there must be an equal num- 

 ber of ampere-turns in the primary to balance these. These 

 primary ampere-turns NI/'I, Fig. 174 (a), are 180 from N 2/2- 

 not customary to show the ampere-turns on the diagram, 

 however, but only the currents, as in Fig. 174. The ampere- 

 turns may then be obtained by multiplying each current by its 

 pro|>er number of turns. 



In addition to /',, the no-load current / must exist to produce 

 the mutual flux, 0, and to supply the no-load losses. This current 

 would IK- in phase with the ilux <}> were it not for (he core losses. 

 The-.- loftSefl require that /,, have an energy component shown by 

 /. in \'\x. 171 '//i. That 18, / i> resolved into two components, 

 a mairneti/irm: component / , in phase with <j>, and an energy 

 component / in phase with 1 he primary emf. I-'., and leading 

 H). 



The total primary current j< /,, the vector >um of / , and /',. 



The primary leakage tlux <;> } i- in pha-e with /,. and induce^ t|,,. 

 which i d by /,.\ . 



