412 THE INDUCTION MOTOR. 



taken by the"statorl =J(7 o and the r power by means of a 

 watt-meter^as described in experiment L. 



. (6) Then make a fresh experiment with the shaft 

 clamped so that it cannot rotate and the rotor conductors 

 short-circuited. Again apply the normal voltage to 

 the stator, and measure the current and watts supplied. 

 If the current taken by the motor under these conditions 

 is too high for safety or convenient measurement, reduce 

 the primary voltage, and multiply the current observed 



normal volts . , . 



by the ratio r. r m order to obtain the true static 



volts applied 



current, and multiply the watts observed by the square 

 of this fraction in order to get the static watts. This 

 is done, because the no-load watts will be nearly all 

 copper losses, and proportional to C' 2 . 



(c) After these tests measure the resistance of the 

 stator winding while still hot by passing a measured 

 direct current through the windings, and observing the 

 drop of potential in them. 



These measurements are sufficient for the construction 

 of the complete diagram from which the performance of 

 the motor at all loads can be predetermined. 



Construction of Diagram from the Observations. Draw 

 vertical and horizontal lines E, A (see Fig. 198). 

 Calculate the angle of lag 0" at no load from the readings 

 (a) and set off the line K representing the no-load current 

 making an angle EOK = <t>". Make OK equal to 

 the no-load current to a scale ofjamperes, arid complete 

 the no-load triangle of currents K F by drawing K F 

 vertical. .Draw the horizontal dotted line a b through 

 K. From the readings (b) calculate the angle of lag 0' 

 for the stationary motor, and hence draw the line O B v 

 equal to the measured static current C s and making 

 the angle </>' with E. 



Draw the semi-circle A B F through the points B l 

 and K, the centre of the circle being a point p lying on 

 the line A. The point A is thus determined. Join 

 B 1 A and draw A f perpendicular to B 1 A. This cuts 

 the vertical through p at the point /, which is the centre 

 of the output circle A N F, drawn through the points 

 A and F. A f is drawn perpendicular to A B 1 because 

 A B 1 must be a tangent to the output circle, since the 



