Consider a phasor diagram with normal excitation i.e. such a current through field winding which will produce flux that will adjust magnitude of E

_{bph}same as V_{ph}. Let δ be the load angle corresponding to the load on the motor. So from the exact opposing position of E

_{bph}with respect to V_{ph}. E_{bph}gets displaced by angle δ. Vector difference of E

_{bph}and V_{ph}, gives the phasor which represents I_{a}Z_{s}, called E_{Rph}. Now Z

_{s }= R_{a }+ j X_{s }Ω where R

_{a }= Resistance of stator per phase X

_{s }= Synchronous reactance of stator per phase i.e. θ = tan

^{-1}(X_{s}/R_{a}) and

**|**Z_{s}**|**_{}= √(R_{a}^{2}+ R_{s}^{2}) Ω This angle 'θ' is called internal machine angle or an impedance angle.

The significant of 'θ' is that it tells us that phasor I

_{aph }lags behind E_{Rph}i.e. I_{a}Z_{s}by angle θ. Current always lags in case of inductive impedance with respect to voltage drop across that impedance. So phasor I_{aph }can be shown lagging with respect to E_{Rph}by angle θ. Practically R_{a}is very small compared to X_{a}and hence θ tends to 90^{o}.**Note**: The power factor at which motor is running, gets decided by the angle between V

_{ph}and I

_{aph }shown. This angle is denoted as Φ and called power factor angle.

and cos Φ = Power factor at which motor is working.

The nature of this p.f. is lagging if Iaph lags V

_{ph}by angle Φ. While it is leading if Iaph leads V_{ph}by angle Φ. Phasor diagram indicating all the details is shown in the Fig.1.Fig. 1 Phasor diagram under normal working condition |

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