ANALYSIS OF ARMATURE REACTANCE 



161 



ST = f EP. Hence the p.d. is equal to SB - ST a TB, so that the 



1 



point T obviously lies on a curve CTD, obtained by subtracting from 

 each of the ordinates of the open-circuit curve the constant value 



'I 1 1'. = V. Again, since FT = y FP, it follows that T also lies on a 



l 



curve ETF, similar to PBA, the centre of similarity being at P', and 



the radii drawn from P' to ETF being in the constant ratio =r- to those 



* 



drawn to PBA. It is now obvious that the position of T may be 

 obtained by the following construction. Draw the curve CTD, by 

 lowering the open-circuit curve through a vertical distance = V. Next, 

 draw ETF similar to the open-circuit curve, the centre of similarity 

 being at P', and the ratio of the radii vectores drawn from P' to the 



two curves being r . The intersection of CTD and ETF gives T, and 

 A 



we then find E, = -y- , A r = -y. 



The method given by Fischer-Hinnen is as follows. In Fig. 119, 

 let P'V = V be the p.d. corresponding to a wattless current I at 



P' 



FIG. 119. Fischer-Hinnen's Method of determining the Two Components of 

 Armature Reactance. 



excitation OP'. Lay off OB to represent the excitation (obtained from 

 the short-circuit curve) required to produce a short-circuit current I. 

 Join BV.and displace the open-circuit curve (by using a tracing) parallel 

 to itself along BV until B coincides with V. The intersection L of 

 the original and the displaced curve enables us to find the direction 



