'9 2 TRANSFORMERS. [Exp. 



For a working range it can be readily constructed, as in Fig. 

 12, which is more nearly to scale, as follows : 



Lay off AD = I (2) for T fo, ^, i, J, f, i and ij load. (It is 

 to be noted that, in Fig. 12, the angle DAP is small ; hence AD 

 is taken as practically equal to AP or / (2 )0 Thus, for a 2,000- 

 volt, 2 K.W., transformer, AD is laid off, successively, equal to 

 .01, o.i, .25, .50, .75, i.o and 1.25 amperes. 



For each value of AD, the point P is located by laying off 



which can be derived from the figure and is the equation of a 

 circle referred to A as an origin. The line DP represents the 

 quadrature component of primary current due to leakage re- 

 actance. This is always small and would be zero when X = b, 

 for the diameter of the semicircle (see Fig. n) is then infinite. 

 The power component AD is, therefore, practically equal to AP. 

 It is to be noted that 



and OC = 

 From these values, compute* for different loads 



Primary current = OP = VOC*+ CP*. 

 Power factor =OC~ OP. 



The curves in Fig. 4, Exp. 5~A were thus computed. Note 

 4 ia, Exp. 5-B. 



(26a). It will be seen, also, that 



Watts input, Wi = O 



Watts output, W 2 = Wi losses ; 



This gives a possible method for determining the total voltage drop. 



