THE TRANSFORMER. 165 



The actual figures taken in Fig. 76 correspond to readings 

 taken on the transformer, whose drop of voltage is shown in 

 Fig. 61. D represents 51-4 volts, the no-load voltage. 

 The volts at the primary terminals were 35, consequently 



35 x 128 128 , 



D E is taken as ~JQ = 18-1 volts, . Q being the ratio or 



the windings. The copper drop is taken as 7-3 volts, as 

 already calculated on page 1 32. This is represented by F E. 

 The terminal voltage E is thus seen to be 41 J volts, which 

 agrees with the value shown from direct measurement in 

 Fig. 61. 



When working on inductive load, the diagram becomes 

 that shown in Fig. 77, where the current lags behind the 

 terminal volts by an angle <. 



In this case the terminal voltage is again obtained from 

 the no-load voltage by constructing the triangle D F E, 

 the side F E of which must be parallel to the current line, since 

 the ohmic drop must be in phase with the current. 



Fig. 77 has been drawn for the same impedance voltage, 

 and the full-load current of 20 amps., as in the previous 

 figure. The angle of lag is taken at about 34, so that the 

 value of the power-factor = cos 34 = -83. At this power- 

 factor, it will be seen that the terminal voltage is only 35 

 volts. 



By drawing the diagram, Fig. 77, for a number of different 

 power-factors, the regulation at any value of cos ^> can be 

 obtained. 



Fig. 78 shows a diagram constructed in this way for a 

 constant primary voltage (i.e., a constant induced secondary 

 voltage of 51-4 volts), and constant secondary load of 20 

 amps., but for four different values of the power-factor. 

 It may be noticed that with a leading current the value of 

 the terminal voltage E ni is greater than the voltage O E 

 on non-inductive load. 



This diagram illustrates a useful approximate method 

 of rapidly determining the drop in volts at any power- 

 factor. By constructing the triangle D F E in the position 

 shown dotted as f e in Fig. 78, we obtain a point e such 

 that a circle drawn from this point with the same radius 

 as D will pass through all the points E. This circle is 

 shown as a dotted line. Thus any line E 1 drawn to cut 

 the dotted circle and the full line circle drawn from with 

 radius equal to the no-load voltage, will show the terminal 



