CONSTRUCTION OF CIRCLE DIAGRAM 199 



by it is not all absorbed by the core loss, a certain amount going 

 towards overcoming the frictional losses. If W A = core loss, and 

 W/ = frictional loss, we have 



W, = W - Wy 



This equation enables us to find WA if W/ is known. In order to 

 determine W/, we take a set of readings connecting the power 

 absorbed by the motor (when running light) with the line p.d. As 

 the p.d. is steadily decreased, the power decreases, on account of the 

 decrease in the hysteresis and eddy-current losses in the stator core. 

 By producing the curve connecting power with p.d. backwards until 

 it intersects the axis of power, we obtain the power corresponding to 

 zero p.d., i.e. the power when the core loss is vanishingly small. 

 This power will then simply represent W . 



The distance O'O is now given by 



V,.\/3 



and on laying this off horizontally from 0' we obtain the point in 

 our diagram. 



We next measure the resistance TI of one phase of the primary. 

 The effect due to eddy currents in the stator windings is equivalent 

 to an increase in the resistance of the windings. It is difficult to 

 estimate this equivalent increase of resistance, but as a rough value 

 we may, in order to be on the safe side, take it to amount to 30 per 

 cent. We shall suppose TI to be a resistance which is 30 per cent. 

 greater than the resistance of one stator phase as measured by using 

 a continuous current ( 88). Then rJo gives us the resistance drop 

 at no load. 



Now, on referring to the diagram of Fig. 128, we notice that OB 

 makes an angle a with the vertical such that 



_ TI __ rilo __ resistance drop at no load 



~ ~ WV-f ^Lr ~~ primary phase p.d. 



Returning now to Fig. 131, we erect a perpendicular at 0, 

 and draw OB making an angle a with the vertical, such that 



T 



sin o = 4*. Along OB lay off a length OQ = I sin equal to the 



v/3 



wattless component of I<j, and from Q draw a line QY making an 

 angle o with OB (or 2a with the vertical). On again referring to 

 Fig. 129, we see that the centre of the circle must lie on QY. 



