334 ELECTRICAL ENGINEERING 



= the e.m.f. of self-inductance generated in one phase of 



the rotor by the leakage flux $ 2 L EZL = sltfh. 

 E r = the e.m.f. generated in one phase of the rotor by the 

 flux $ r . E r is the difference between E 2g and EZL and is 

 equal to 7 2 r 2 . It is in phase with the rotor current 7 2 

 and lags 90 degrees behind the flux <l> r . 



As the induction motor is loaded the end d of the vector $1 fol- 

 lows a circle passing through / and having its centre on of pro- 

 duced. 



207. Proof that the Locus is a Circle. From d draw dk at 

 right angles to fd to cut of produced in k. Then the semicircle 

 / dk is the locus of d. 



In the triangles aob and / dk 



Z oab = Zfdk, being right angles, 

 < oba = Z dfk, 



fk fd fd fd 



therefore ^ = ^ = ^^ = ^^, 



and 



, r ob Xfd Vi'ofXvfoh f ViV 2 

 fk = r = =-*- T = of*-= = a constant 

 oh cb oh Viv 2 oh j 1 ViVz 



since of is constant. 



Therefore the locus of d is a circle described on the diameter fk. 



208. Magnetomotive Force Diagram. Since the magnetic 

 circuit of the machine is not saturated by the fluxes 4> a and 3> r , the 

 flux diagram, Fig. 315, may be replaced by the m.m.f. diagram, 

 Fig. 316. 



od = nJi = total m.m.f. of the stator per phase. 

 ol = nil' = m.m.f. of the load component of stator current. 

 of = n\I M = m.m.f. of the magnetizing current. 

 op n z h = m.m.f. of the rotor per phase; it is equal and opposite 

 to rci/'. 



209. Stator Current Diagram. The m. m. f. diagram, Fig. 

 316, may be replaced by the stator current diagram, Fig. 317. 



od = 1 1 total current m one phase of stator. /i = IM + !' 



of = IM = magnetizing current in one phase of stator. 



ol = F = load component of current in one phase of stator. 



O p = J 2 = current in one phase of rotor. 7 2 = /'. 



nz 



df = ol = I' = / 2 = load component of stator current per 

 HI 



phase and represents the rotor current per phase. 



