**Phasor Diagrams for Transformer on Load**Consider a transformer supplying the load as shown in the Fig. 1.

Fig. 1 |

The various transformer parameters are,

R

_{1 }= Primary winding resistance X

_{1 }= Primary leakage reactance R

_{2 }= Secondary winding resistance X

_{2 }= Secondary leakage reactance Z

_{L }= Load impedance I

_{1}= Primary current I

_{2 }= Secondary current = I_{L }= Load currentnow Ī

_{1}= Ī_{o}+ Ī_{2}'where I

_{o }= No load current I

_{2}'= Load component of current decided by the load = K I

_{2}where K is transformer component The primary voltage V

_{1 }has now three components,1. -E

2. I

3. I

The secondary induced e.m.f. has also three components,

1. V

2. I

3. I

As load power factor is unity, the voltage V

1. Consider flux Φ as reference

2. E

3. E

4. Assume V

5. I

6. Add I

7. Reverse I

8. Add I

9. Add I

_{1}, the induced e.m.f. which opposes V_{1}2. I

_{1}R_{1}, the drop across the resistance, in phase with I_{1}3. I

_{1}X_{1}, the drop across the reactance, leading I_{1}by 90^{o}The secondary induced e.m.f. has also three components,

1. V

_{2}, the terminal voltage across the load2. I

_{2}R_{2}, the drop across the resistance, in phase with I_{2}3. I

_{2}X_{2}, the drop across the reactance, leading I_{2}by 90^{o} The phasor diagram for the transformer on load depends on the nature of the load power factor. Let us consider the various cases of the load power factor.

__1.1 Unity power factor load, cosΦ__

_{2}= 1_{2 }and I_{2 }are in phase. Steps to draw the phasor diagram are,1. Consider flux Φ as reference

2. E

_{1 }lags Φ by 90^{o}. Reverse E_{1 }to get -E_{1}.3. E

_{1 }and E_{2 }are inphase4. Assume V

_{2 }in a particular direction5. I

_{2 }is in phase with V_{2}.6. Add I

_{2 }R_{2 }and I_{2 }X_{2 }to to get E_{2}.7. Reverse I

_{2 }_{ }to get I_{2}**'**.8. Add I

_{o }and I_{2}**'**_{}to get I_{1}.9. Add I

_{1 }R_{1 }and to -E_{1 }to get V_{1}. Angle between V

_{1 }and I_{1}is Φ_{1}and cosΦ_{1}is primary power factor. Remember that I_{1}X_{1 }leads I_{1 }direction by 90^{o}and I_{2}X_{2 }leads I_{2}by 90^{o}as current through inductance lags voltage across inductance by 90^{o}. The phasor diagram is shown in the Fig.2Fig. 2 Phasor diagram for unity power factor load |

__Lagging Power Factor Load, cos Φ__

_{2} As load power factor is lagging cosΦ

The complete phasor diagram is shown in the Fig. 3._{2}, the current I_{2 }_{ }lags V_{2 }_{ }by angle Φ_{2}. So only changes in drawing the phasor diagram is to draw I_{2 }_{ }lagging V_{2 }_{ }by Φ_{2}in step 5 discussed earlier. Accordingly direction of I_{2 }R_{2}, I_{2 }X_{2}, I_{2}**'**, I_{1}, I_{1 }R_{1 }and I_{1}X_{1 }will change. Remember that whatever may be the power factor of load, I_{2}X_{2 }leads I_{2 }by 90^{o }and I_{1}X_{1 }leads I_{1}by 90^{o}.Fig. 3 Phasor diagram for lagging power factor |

__Loading Power Factor Load, cos Φ__

_{2} As load power factor is leading, the current I

_{2}leads V_{2 }by angle Φ_{2}. So change is to draw I_{2 }leading I_{2 }by angle Φ_{2}. All other steps remain same as before. The complete phasor diagram is shown in the Fig. 4Fig. 4 Phasor diagram for leading power factor |

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Very good information!!!! Thank you, Erik Arckens

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ReplyDeletethere are two inductances on primary side one is that of Py winding and other due to leakage reactance. lenz's law is applicable on both inductors.i.e there should be -ve sign with E1 as well as I1X1 drop. but we have used with only E1 not with the other inductor.please explain with very basic concepts so that I can get the idea.further with applied signs both should be voltage drop across the inductor and should have same sign?

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