THK TRANSFORM I:' If 



I'M 



To obtain tho true value of the exciting current, the current 7 

 measured by the ammeter, in Fig. 177, should be resolved into 

 two components, one of which lies along the voltage E\, or V, 

 and is shown as 7, in Fig. 179 ( EI and V are practically equal 

 at no load). This current /, = 7 cos 6 is the energy component 

 of the current and supplies the core losses. The quadrature 

 component 7 m = 7 sin 6 is the true magnetizing current, shown 

 plotted in Fig. 178 (6). In most commercial transformers 7 = 

 Im, very nearly. 



84. Short-circuit Test. Figure 180 shows the transformer of 

 Fiir. 177 reversed and the low side short-circuited. The reversal 

 is made in order that the line current may not be excessive, and 



VW\A I 



High 



Low 



IK. 179. Magnetizing and core 

 loss currents in transformer. 



Fio. 180. Connections for short-circuit test 

 of a transformer. 



in order that a reasonable voltage drop may be obtained. 

 In a transformer, the impedance drop seldom exceeds 5 per cent. 

 of the rated voltage. If the 2 .200- volt side of a transformer, Fig. 

 180, be used as the primary, the voltage necessary to send rated 

 current through the windings on short -circuit is about ~) per cent. 

 of 2,200, or 110 volts, which is a standard voltage for instrument 

 OOfl& If the secondary of the transformer were rated :it L'-JO 

 the voltage .-it short-circuit would be only 11 volts and the 

 current would nlso In- high. At this low voltage, high precision 

 could not be obtained with ordinary instruments. 



When a primary current / . ISO. t! dary 



rum- i ; ,l t,/ ( ^ } . Then- is. therefore, no need <>\ u>ing 



an ammel / , The power delivered to the trans- 



former, Fiir. IMI. I 'iPI'ly t)l '"' pnniary copper 



OOpper I- . / /.'-. and tlie core loss at 



