Equivalent resistance and reactance of transformer

The transformer has two independent circuits primary and secondary winding. The two circuits can be resolved into an equivalent circuit to make the calculations simple.

Consider a practical transformer as shown below, which has primary and secondary windings whose resistance and reactances are R₁ and R₂ and X₁ and X₂ respectively.

The voltage drops due to primary resistance and reactances are I₁R₁ and I₁X₁ respectively. Similarly, the voltage drops due to secondary resistance and reactances are I₂R₂ and I₂X₂ respectively.

The equivalent resistance and reactance of transformer is calculated by transferring the circuit parameters either from primary to secondary or from secondary to primary.

Equivalent resistance and reactance referred to Secondary

To make the calculations simple, let us transfer the circuit parameters(primary voltage, resistance, and reactances) from the primary side to the secondary side.

As you know, the transformation ratio is given by,

$K = \frac{V_2}{V_1} = \frac{I_1}{I_2}$

The voltage drops in the primary winding referred to secondary is K times, that is KV₁. So the primary resistive and reactive drops referred to secondary can be written as KI₁R₁ and KI₁X₁ respectively. If I₁ is replaced with KI₂(from the transformation ratio), the drops due to primary resistance and reactance will become K²I₂R₁ and K²I₂X₁ respectively.

Total resistive drop in the transformer = K²I₂R₁ + I₂R₂ = I₂(K²R₁ + R₂) = I₂R₀₂

Total reactive drop in the transformer = K²I₂X₁ + I₂X₂ = I₂(K²X₁ + X₂) = I₂X₀₂

In the above equations, R₀₂ and X₀₂ represent the equivalent resistance and reactances of the primary winding referred to secondary. Now the above circuit can be redrawn as,

The phasor diagram for the obtained circuit is drawn below. Consider the terminal voltage as the reference phasor. Since the load is inductive, the secondary current I₂ lags behind V₂ by Φ. The drops due to resistance(I₂R₀₂) and reactances(I₂X₀₂) are drawn parallel and perpendicular to the current phasor. Join OC, the input voltage referred to secondary.

Phasor diagram of transformer referred to Secondary

From the phasor diagram,

$KV_1 = V_2+I_2R_{02} +j I_2X_{02}$

$KV_1 = \sqrt{(V_2+I_2R_{02} cos \phi +I_2X_{02} sin\phi)^2 +(I_2X_{02} cos \phi -I_2R_{02} sin\phi)^2 }$

In this equation, the term (I₂ X₀₂ cosɸ – I₂ R₀₂ sinɸ) is very small compared to (V₂ + I₂R₀₂ cosɸ + I₂ X₀₂ sinɸ). Hence neglecting the smaller term, we get,

$KV_1 = \sqrt{(V_2+I_2R_{02} cos \phi +I_2X_{02} sin\phi)^2 }$

$KV_1 = V_2+I_2R_{02} cos \phi +I_2X_{02} sin\phi$

$V_2 = KV_1 - I_2R_{02} cos \phi - I_2X_{02} sin\phi --- (1)$

For pure resistive load, Φ = 0, hence the equation(1) becomes,

$V_2 = KV_1 - I_2R_{02}$

For pure capacitive load, current I₂ leads the voltage V₂ by Φ and so it is taken as negative(-), hence the equation(1) becomes

$V_2 = KV_1 - I_2R_{02} cos \phi + I_2X_{02} sin\phi$

Equivalent resistance and reactance referred to Primary

In the same way as before, the secondary winding parameters can be transferred to the primary to make the calculations simple.

With respect to the transformation ratio, the voltage drop due to secondary resistance and reactance referred to primary are $\frac{I_2R_2}{K}$ and $\frac{I_2X_2}{K}$ respectively. But from the transformation ratio, $I_2 = \frac{I_1}{K}$ . Thus the secondary resistive and reactive voltage drops referred to primary will become $\frac{I_1R_2}{K^2}$ and $\frac{I_1X_2}{K^2}$ respectively.

Total resistive drop in the transformer =

$I_1R_1 + \frac{I_1R_2}{K^2} = I_1(R_1 + \frac{R_2}{K^2}) = I_1R_{01}$

Total reactive drop in the transformer =

$I_1X_1 + \frac{I_1X_2}{K^2} = I_1(X_1 + \frac{X_2}{K^2}) = I_1X_{01}$

In the above equations, R₀₁ and X₀₁ represent the equivalent resistance and reactances of the secondary winding referred to primary. Now the above circuit can be redrawn as,

Equivalent resistance and reactance of transformer referred to primary

The phasor diagram for the above circuit is drawn below. Consider the terminal voltage V₁ as the reference phasor. Since the load is inductive, the primary current I₁ lags behind V₁ by Φ. The drops due to resistance(I₁R₀₁) and reactances(I₁X₀₁) are drawn parallel and perpendicular to the current phasor. Join OC, the input voltage referred to primary.

Phasor diagram of transformer referred to primary

From the phasor diagram,

$V_1 = \frac{V_2}{K}+I_1R_{01} +j I_1X_{01}$

$V_1 = \sqrt{(\frac{V_2}{K}+I_1R_{01} cos \phi +I_1X_{01} sin\phi)^2 +(I_1X_{01} cos \phi - I_1R_{01} sin\phi)^2 }$

In this equation, the term $(I_1X_{01} cos \phi - I_1R_{01} sin\phi)$ is very small compared to $(\frac{V_2}{K}+I_1R_{01} cos \phi +I_1X_{01} sin\phi)$ . Hence neglecting the smaller term, we get,

$V_1 = \sqrt{(\frac{V_2}{K}+I_1R_{01} cos \phi +I_1X_{01} sin\phi)^2 }$

$V_1 = \frac{V_2}{K}+I_1R_{01} cos \phi +I_1X_{01} sin\phi ---- (3)$

For pure resistive load, Φ = 0, hence the equation(3) becomes,

$V_1 = \frac{V_2}{K} + I_1R_{01}$

For pure capacitive load, current I₂ leads the voltage V₂ by Φ and so it is taken as negative(-), hence the equation(3) becomes

$V_1 = \frac{V_2}{K} - I_1R_{01} cos \phi + I_1X_{01} sin\phi$

Solved Problem 1

A 20 kVA, 2000/200 V, 50 Hz, single-phase transformer has the following parameters: R₁ = 1.25 Ω, R₂ = 0.0025 Ω, X₁ = 2.8 Ω, X₂ = 0.0065 Ω. Calculate: (i) Equivalent resistance referred to primary and secondary. (ii) Equivalent reactance referred to primary and secondary.

Solution: Given data: R₁ = 1.25 Ω, R₂ = 0.0025 Ω, X₁ = 2.8 Ω, X₂ = 0.0065 Ω

Transformation ratio,

$K = \frac{V_2}{V_1} = \frac{200}{2000} = 0.1$

(i) Equivalent resistance referred to primary,

$R_{01} = R_1 + \frac{R_2}{K^2} = 1.25 + \frac{0.0025}{0.1^2} = 1.5 \Omega$

Equivalent resistance referred to secondary,

$R_{02} = K^2 R_1 + R_2 = (0.1^2 \times 1.25)+ 0.0025 = 0.015 \Omega$

(ii) Equivalent reactance referred to primary,

$X_{01} = X_1 + \frac{X_2}{K^2} = 2.8 + \frac{0.0065}{0.1^2} = 3.45 \Omega$

Equivalent reactance referred to secondary,

$X_{02} = K^2 X_1 + X_2 = (0.1^2 \times 2.8)+ 0.0065 = 0.0345 \Omega$

Solved Problem 2

A 10 kVA, 2000/400 V, 50 Hz single-phase transformer at no load has R₁=4.5 Ω, R₂=0.3 Ω, X₁=11 Ω, X₂=0.50 Ω. Determine the approximate value of the secondary voltage at full load, 0.8 pf (lagging).

Solution: Given data: R₁ = 4.5 Ω, R₂ = 0.3 Ω, X₁ = 11 Ω, X₂ = 0.50 Ω

Transformation ratio,

$K = \frac{V_2}{V_1} = \frac{400}{2000} = 0.2$