408 THE INDUCTION MOTOR. 



no-load losses of the motor. The losses (b) increase in 

 proportion to the square of the current in each winding. 

 In the stator, the effect of the copper losses is a reduction 

 in the voltage made use of by the motor. Thus, effective 

 voltage is E C* R, instead of E. 



In the rotor, the effect of the copper loss is to dimmish 

 the current, and thus to diminish the output. 



The application of the diagram to the case of a complete 

 motor must now be briefly given, and will then be further 

 explained by an example from an actual motor. Referring 

 to Fig. 196, which is the diagram Fig. 195 extended to 

 apply to the more complete conditions, we can represent 

 the various quantities in the following manner : 



(1) Power Supplied to Motor. The apparent watts 

 supplied to a 3-phase motor are 1-73 x E x (7 S , where 

 E and C s are the line current and voltage supplied. 

 These can therefore be obtained from the diagram by 

 multiplying the length of the line O B measured on the 

 scale of amperes by 1-73 and by the terminal voltage. 

 The true power supplied will be due to the energy current, 

 i.e., the component of O B in phase with E x , represented 

 by the line B parallel to E. Thus power supplied = 

 B G x 1 . 73 x E, and the power supplied to motor will 

 always be proportional to the length B G. 



(2) Torque. This is proportional to the product of 

 the rotor flux Z r and the rotor current C r . Hence, 

 except for copper losses in the stator, the torque will be 

 proportional to A B x B F in the diagram. In the actual 

 motor, however, the rotor flux is diminished by the 

 voltage lost in the resistance of the stator. A B is the 

 direction of the rotor flux, hence in the triangle of fluxes 

 ABO (where A B = Z r , A O = Z,, B = z) we must 

 diminish the length A B by a portion B H equivalent 

 to the copper drop in the primary. Mark off a length 

 B H on the line A B to represent this drop at the par- 

 ticular value of the primary current for which A B 

 has been drawn. Draw a semi-circle through the 

 three points A H F, and call its centre /, which must 

 evidently lie on the line g q bisecting A F at right .angles. 

 Then the length H B will be the copper drop correspond- 

 ing to any position of A B and the rotor flux will be 

 represented by A H, and the length of B H will vary, 



