USA: +1-585-535-1023

UK: +44-208-133-5697

AUS: +61-280-07-5697

Equivalent Resistance and Reactance

Equivalent Resistance. In Fig. 25 is shown a transformer whose primary and secondary windings have resistances of R1 and R2 respectively (shown external to the windings). The resistance of the two windings can be transferred to anyone of the two windings. If both the resistances are concentrated in one winding the calculations become simple since then we can work in one winding only.


In the secondary, copper loss is I22R2. This loss is supplied by primary which takes a current of I1. Hence, if R2‘ is the equivalent resistance referred to primary which would have caused the same loss as R2 in secondary, then

Similarly, equivalent primary resistance as referred to secondary is


In Fig. 26, secondary resistance has been transferred to primary side (leaving secondary circuit resistanceless),

Similarly, the equivalent resistance of the transformer as referred to the secondary (Fig. 27)

R02 =R2 +R1‘ =R2 +K2R1                                                           … (7)

The following points are worth remembering:

(i)                           When shifting resistance to the secondary, multiply it by K2.

(ii)                         When shifting resistance to the primary, divide by K2.

Leakage reactance. Leakage reactance can also be transferred from one winding to the other in the same way as resistance (Fig. 28 and Fig. 29).


Total Voltage Drop in a Transformer

(i) Approximate voltage drop. When there is no-load on the transformer, then,

V1 = E1 (approximately).

and E2 = KE1 = KV1

Also    E2 = 0V2, where 0V2 is secondary terminal voltage on no-loud

E2 = 0V2 =KV1

V2 = secondary voltage on load.

Refer Fig. 32. The procedure of finding the approximate voltage drop of the transformer as referred to secondary is given below:

  • Taking O as center, radius OC draw an arc cutting OA produced at E.

The total power voltage drop I1Z02 = AC = AE which is approximately equal to AG.

  • From B draw BF perpendicular on OA produced. Draw CG perpendicular to OE and draw BD parallel to OEApproximate voltage drop = AG = AF + FG = AF + BD            [= I2R02 cos ɸ + I0X02sin ɸThis is the value of approximate voltage drop for a lagging power factor. Figs. 33 and 34 refer to unity and leading power factor respectively..


  • The approximate voltage drop for a leading power factor becomes:

(I2R02 cos ɸ – I2X02 sin ɸ)

  • The approximate voltage drop for a transformer in general is given by :

(I2R02 cos ɸ ± I1X02 sin ɸ)                                               … (12)

  • The voltage drop as referred to primary is given by:

(I1R01 cos ɸ ± I1X02 sin ɸ

  • Percentage voltage drop in secondary 

ii) Exact voltage drop. When we refer to Fig. 32 we find that exact drop is AE and not AG. If quantity GE is added to AG the exact value of voltage drop will be obtained.

From right angled triangle OCG, we get

CG2 = OC2 – OG2 = (OC + OG) (DC – OG)

= (DC + OG) (DE – OG) = GE × 2OC