As primary parameters are referred to secondary, there are no voltage drops in primary.

When there is no load, I

_{2}= 0 and we get no load terminal voltage V_{20}as E_{2}.**.**V

^{.}._{20}= E

_{2}= No load terminal voltage

while V

_{2}= Terminal voltage on load Consider the phasor diagram for lagging p.f. load. The current I

_{2}lags V_{2}by angle Φ_{2}. Take V_{2}as reference phasor. I_{2 }R_{2e}is in phase with I_{2}while I_{2}X_{2e}leads I_{2}by 90^{o}. The phasor diagram is shown in the Fig.2.Fig. 2 |

To derive the expression for approximate voltage drop, draw the circle with O as centre and OC as redius, cutting extended OA at M. As OA = V

_{2}and now OM = E_{2}, the total voltage drop is AM = I_{2}Z_{2e}. But approximating this voltage drop is equal to AN instead of AM where N is intersection of perpendicular drawn from C on AM. This is because angle is practically very very small and in practice M and N are very close to each other.

Approximate voltage drop = AN

Draw perpendicular from B on AM intersecting it at D and draw parallel to DN from B to the point L shown in the Fig. 2.

**.**AD = AB cos Φ

^{.}._{2}= I

_{2 }R

_{2e }cos Φ

_{2}

and DN = BL = BC sin Φ

_{2}= I_{2 }X_{2e }sin Φ_{2}**.**AN = AD + DN = I

^{.}._{2 }R

_{2e }cos Φ

_{2}+ I

_{2 }X

_{2e }sin Φ

_{2}

Assuming Φ

_{2}= Φ_{1}= Φ**.**Approximate voltage drop = I

^{.}._{2 }R

_{2e }cos Φ + I

_{2 }X

_{2e }sin Φ

If all the parameters are referred to primary then we get,

Approximate voltage drop = I

_{1 }R_{1e }cos Φ + I_{1 }X_{1e }sin Φ If the load has leading p.f. then we get the phasor diagram as shown in the Fig. 3. The I

_{2 }leads V_{2}by angle Φ_{2}.Fig. 3 |

In this case, the expression for approximate voltage drop remains same but the sign of I

Approximate voltage drop = I

= I

_{2 }X_{2e }sin Φ_{ }reverses.Approximate voltage drop = I

_{2 }R_{2e }cos Φ - I_{2 }X_{2e }sin Φ ....... Using referred to secondary values= I

_{1 }R_{1e }cos Φ - I_{1 }X_{1e }sin Φ ...........Using referred to primary values It can be noticed that for leading power factor E

_{2 }< V_{2}. For the unity power factor, the phasor diagram is simple and is shown in the Fig. 4. For this case, as cos Φ = 1 and sin Φ = 0, the approximate voltage drop is I

_{2 }R_{2e }or I_{1}R_{1e}.Fig. 4 |

Thus the general expression for the total approximate voltage drop is,

Approximate voltage drop = E

= I

= I

Approximate voltage drop = E

_{2 }- V_{2 }= I

_{2e }R_{2e }cosΦ I_{2e }X_{2e }sin Φ ........Using referred to secondary values= I

_{1e }R_{1e }cos Φ I_{1e }X_{1e }sin Φ ........Using referred to primary values + sing for lagging power factor while - sign for leading power factor loads.

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