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Transformer equivalent circuit referred to primary

Short answer

Secondary parameters are referred to primary by dividing voltage by K, multiplying current by K, and dividing impedances (R, X, Z) by K^2, where K=N2/N1. Equivalent series resistance R01 = R1 + R2/K^2 and reactance X01 = X1 + X2/K^2. Shunt branch remains unchanged.

Draw the equivalent circuit of a transformer with their parameters as it is in primary side and secondary side. How all parameters can be transferred to primary side - explain with mathematical derivation.

The equivalent circuit of a transformer is a circuit of resistances and reactances that behaves exactly like the real transformer, so its performance can be calculated by simple circuit analysis.

Exact equivalent circuit

  I1   R1    X1          I2'     R2'   X2'
 o--->-/\/\--mmm--+--------+-->--/\/\--mmm--o
 +                |   I0   |                +
               [ R0 ]   [ X0 ]
 V1   E1          |        |            V2'
 -                |        |                -
 o----------------+--------+----------------o
     primary side      |  secondary referred

Parameters:

  • R1,X1R_1, X_1: primary winding resistance and leakage reactance.
  • R2,X2R_2, X_2: secondary resistance and leakage reactance.
  • R0R_0: core-loss resistance (carries IwI_w, the working component of I0I_0).
  • X0X_0: magnetising reactance (carries IμI_\mu).

In the actual transformer, R1,X1,R0,X0R_1, X_1, R_0, X_0 are on the primary side and R2,X2R_2, X_2 and the load ZLZ_L are on the secondary side, joined by an ideal transformer of ratio K=N2/N1=E2/E1K = N_2/N_1 = E_2/E_1.

Transferring secondary quantities to primary

The rule: the referred quantity must give the same power, losses and VA as the original.

Voltage: since E2=KE1E_2 = KE_1,

E2′=E2K=E1,V2′=V2KE_2' = \frac{E_2}{K} = E_1, \qquad V_2' = \frac{V_2}{K}

Current: by ampere-turn balance N1I2′=N2I2N_1I_2' = N_2I_2,

I2′=KI2I_2' = KI_2

Resistance: copper loss must be unchanged:

I2′2R2′=I22R2  ⇒  R2′=R2(I2I2′)2=R2K2I_2'^2R_2' = I_2^2R_2 \;\Rightarrow\; R_2' = R_2\left(\frac{I_2}{I_2'}\right)^2 = \frac{R_2}{K^2}

Reactance: reactive power I2XI^2X must be unchanged:

I2′2X2′=I22X2  ⇒  X2′=X2K2I_2'^2X_2' = I_2^2X_2 \;\Rightarrow\; X_2' = \frac{X_2}{K^2}

Load impedance: similarly ZL′=ZLK2Z_L' = \frac{Z_L}{K^2}.

Check with impedance: Z2′=V2′I2′=V2/KKI2=Z2K2Z_2' = \frac{V_2'}{I_2'} = \frac{V_2/K}{KI_2} = \frac{Z_2}{K^2}.

Approximate equivalent circuit referred to primary

Since I0I_0 is only 2-5% of rated current, the shunt branch can be moved to the input terminals. Then the series parts combine:

R01=R1+R2′=R1+R2K2X01=X1+X2′=X1+X2K2Z01=R012+X012\begin{aligned} R_{01} &= R_1 + R_2' = R_1 + \frac{R_2}{K^2} \\ X_{01} &= X_1 + X_2' = X_1 + \frac{X_2}{K^2} \\ Z_{01} &= \sqrt{R_{01}^2 + X_{01}^2} \end{aligned}
 o---+--------+----[ R01 ]----[ jX01 ]----o
     |        |                           +
  [ R0 ]   [ jX0 ]                   V2' = V2/K
     |        |                     Z_L' = Z_L/K^2
 o---+--------+---------------------------o

Similarly, referred to the secondary: R02=R2+K2R1R_{02} = R_2 + K^2R_1, X02=X2+K2X1X_{02} = X_2 + K^2X_1, and R0,X0R_0, X_0 are multiplied by K2K^2.

Rule to remember: going from the low-voltage side to the high-voltage side, impedances are multiplied by the square of the voltage ratio (HV/LV)2^2; voltages scale by the ratio, currents by its inverse.

✓ Answer checked by Claude Opus 5.5 (high effort)

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