**1. Effect OF Winding Resistances**A practical transformer windings process some resistances which not only cause the power losses but also the voltage drops. Let us see what is the effect of winding resistance on the performance of the transformer.

Let R

_{1}= primary winding resistance in ohms R

_{2}= secondary winding resistance in ohms Now when current I

_{1}flows through primary, there is voltage drop I_{1}R_{1}across the winding. The supply voltage V_{1}has to supply this drop. Hence primary induced e.m.f. E_{1}is the vector difference between V_{1}and I_{1}R_{1}. Similarly the induced e.m.f. in secondary is E

_{2}. When load is connected, current I_{2}flows and there is voltage drop I_{2}R_{2}. The e.m.f. E_{2}has to supply this drop. The vector difference between E_{2}and I_{2}R_{2}is available to the load as a terminal voltage.The drops I

_{1}R_{1}and I_{2}R_{2}are purely resistive drops hence are always in phase with the respective currents I_{1}and I_{2}.__1.1 Equivalent Resistance__

The resistance of the two windings can be transferred to any one side either primary or secondary without affecting the performance of the transformer. The transfer of the resistances on any one side is advantageous as it makes the calculations very easy. Let us see how to transfer the resistances on any one side.

The total copper loss due to both the resistances can be obtained as,

total copper loss = I

_{1}^{2}R_{1}+ I_{2}^{2}R_{2} = I

_{1}^{2}{ R_{1}+(I_{2}^{2}/I_{1}^{2}) R_{2}} = I

_{1}^{2}{R_{1}+ (1/K^{2}) R_{2}} .......(3)Where I

_{2}/I_{1}= 1/K neglecting no load current. Now the expression (3) indicates that the total copper loss can be expressed as I

_{1}^{2}R_{1}+ I_{1}^{2}**.**R_{2/}K^{2}. This means R_{2}/K^{2 }is the resistance value of R_{2 }shifted to primary side which causes same copper loss with I_{1}as R_{2 }causes with. This value of resistance which R_{2 }/K^{2 }is the value of R_{2 }referred to primary is called equivalent resistance of secondary referred to primary. It is denoted as R_{2}**'**. R

_{2}**'**= R_{2 }/K^{2 }........(4) Hence the total resistance referred to primary is the addition of R

_{1}and R_{2}**'**called equivalent resistance of transformer referred to primary and denoted as R_{1e}. = R

_{1}+ R_{2}**'**= R_{1}+ R_{2 }/K^{2 }.........(5) This resistance R

_{1e }causes same copper loss with I_{1}as the total copper loss due to the individual windings. total copper loss = I

_{1}^{2}R_{1e }= I_{1}^{2}R_{1}+ I_{2}^{2}R_{2}......(6) So equivalent resistance simplifies the calculations as we have to calculate parameters on one side only.

Similarly it is possible to refer the equivalent resistance to secondary winding.

total copper loss = I

_{1}^{2}R_{1}+ I_{2}^{2}R_{2} = I

_{2}^{2}{(I_{1}^{2}/I_{2}^{2}) R_{1}+ R_{2}} = I

_{2}^{2}( K^{2 }R_{1}+ R_{2}) ........(7) Thus the resistance K

^{2 }R_{1}is primary resistance referred to secondary denoted as R_{1}**'**. R

_{1}**'**= K^{2 }R_{1}.......(8) Hence the total resistance referred to secondary is the addition of R

_{2 }and R_{1}**'**called equivalent resistance of transformer referred to secondary and denoted as R_{2e}. R

_{2e }= R_{2 }+ R_{1}**'**= R_{2 }+ K^{2 }R_{1}.........(9) total copper loss = I

_{2}^{2}R_{2e }........(10) The concept of equivalent resistance is shown in the Fig. 1(a), (b) and (c).

Fig. 1 Equivalent resistance |

**Key Point**: When resistance are transferred to primary, the secondary winding becomes zero resistance winding for calculation purpose. The entire copper loss occurs due to R

_{1e}. Similarly when resistances are referred to secondary, the primary becomes resistanceless for calculation purpose. The entire copper loss occurs due to R

_{2e}.

**Important**

**Note**: When a resistance is to be transferred from the primary to secondary, it must be multiplied by K

^{2}. When a resistance is to be transferred from the secondary to primary, it must be divided by K

^{2}. Remember that K is N

_{1 }/N

_{2}.

The result can be cross-checked by another approach. The high voltage winding is always low current winding and hence the resistance of high voltage side is high. The low voltage side is high current side and hence resistance of low voltage side is low. So while transferring resistance from low voltage side to high voltage side, its value must increase while transferring resistance from high voltage side to low voltage side, its value must decrease.

**Key point**:

**High voltage side →**

**Low current side → High resistance side**

**Low voltage side → High current side → Low resistance side**

**Example 1**: A 6600/400 V single phase transformer has primary resistance of 2.5 Ω and secondary resistance of 0.01 Ω calculate total equivalent resistance referred to primary and secondary.

**Solution**: The given values are,

R

_{1 }= 2.5 Ω R

_{2 }= 0.01 Ω

K = 400/6600 = 0.0606

While finding equivalent resistance referred to primary, transfer to primary as,

R

_{2}

**'**= R

_{2 }/K

^{2 }= 0.01/(0.0606)

^{2}= 2.7225 Ω

R

_{1e }= R

_{1 }+ R

_{2}

**'**= 2.5 + 2.7225 = 5.2225 Ω

It can be observed that primary is high voltage hence high resistance side hence while transferring from low voltage to on high voltage, its value increases.

To find total equivalent resistance referred to secondary, first calculate ,R

_{1}

**'**= K

^{2 }R

_{1 }= (0.0606)

^{2}x 25 = 0.00918 Ω

R

_{2e }= R

_{2 }+ R

_{1}

**'**= 0.01 + 0.00918 = 0.01918 Ω

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