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Equivalent circuit of 8 kVA 400/120 V transformer

Short answer

OC test on LV gives R0=2133.3 ohm and X0=337.5 ohm referred to HV. SC test on HV gives R01=0.275 ohm, X01=0.3873 ohm, and Z01=0.475 ohm.

A 8 kVA, 400/120 V, 50 Hz single phase transformer on test gives the following results: O.C. test: 120 V, 4 A, 75 W (carried on LV side) S.C. test: 9.5 V, 20 A, 110 W (carried on HV side) Calculate the equivalent circuit parameters referred to HV side.

Method. The no-load (OC) test gives the shunt branch: cos⁡ϕ0=P0V0I0\cos\phi_0 = \frac{P_0}{V_0 I_0}, Iw=I0cos⁡ϕ0I_w = I_0\cos\phi_0, Iμ=I0sin⁡ϕ0I_\mu = I_0\sin\phi_0, R0=V0IwR_0 = \frac{V_0}{I_w}, X0=V0IμX_0 = \frac{V_0}{I_\mu}. The SC test gives the series branch: Z=VscIscZ = \frac{V_{sc}}{I_{sc}}, R=PscIsc2R = \frac{P_{sc}}{I_{sc}^2}, X=Z2−R2X = \sqrt{Z^2 - R^2}. Values found on one side are moved to the other side by multiplying by the square of the voltage ratio.

Ratio: HV = 400 V, LV = 120 V, so to refer an LV value to HV multiply by (400120)2=11.11\left(\frac{400}{120}\right)^2 = 11.11.

From OC test (on LV side): 120 V, 4 A, 75 W

cos⁡ϕ0=75120×4=0.1563,sin⁡ϕ0=0.9877Iw=4×0.1563=0.625 A,Iμ=4×0.9877=3.951 AR0(LV)=1200.625=192 Ω,X0(LV)=1203.951=30.37 Ω\begin{aligned} \cos\phi_0 &= \frac{75}{120 \times 4} = 0.1563, \quad \sin\phi_0 = 0.9877 \\ I_w &= 4 \times 0.1563 = 0.625\ \text{A}, \quad I_\mu = 4 \times 0.9877 = 3.951\ \text{A} \\ R_0 (\text{LV}) &= \frac{120}{0.625} = 192\ \Omega, \quad X_0 (\text{LV}) = \frac{120}{3.951} = 30.37\ \Omega \end{aligned}

Referred to HV:

R0(HV)=192×11.11=2133.3 ΩX0(HV)=30.37×11.11=337.5 Ω\begin{aligned} R_0 (\text{HV}) &= 192 \times 11.11 = 2133.3\ \Omega \\ X_0 (\text{HV}) &= 30.37 \times 11.11 = 337.5\ \Omega \end{aligned}

From SC test (on HV side): 9.5 V, 20 A, 110 W

The test is on the HV side, so the results are already referred to HV. (Rated HV current = 8000/400 = 20 A, so this is a full-load test.)

Z01=9.520=0.475 ΩR01=110202=0.275 ΩX01=0.4752−0.2752=0.3873 Ω\begin{aligned} Z_{01} &= \frac{9.5}{20} = 0.475\ \Omega \\ R_{01} &= \frac{110}{20^2} = 0.275\ \Omega \\ X_{01} &= \sqrt{0.475^2 - 0.275^2} = 0.3873\ \Omega \end{aligned}

Equivalent circuit referred to HV side

 I1 ->        I0                 I2'
 o-----+--------+---[ R01 ]---[ jX01 ]---o
 +     |        |                        +
     [ R0 ]  [ jX0 ]                   load
 V1    |        |                       V2'
 -     |        |                        -
 o-----+--------+------------------------o
 All values referred to HV side:
 R0 = 2133.3 ohm, X0 = 337.5 ohm
 R01 = 0.275 ohm, X01 = 0.387 ohm

Answer: R0=2133.3 ΩR_0 = 2133.3\ \Omega, X0=337.5 ΩX_0 = 337.5\ \Omega, R01=0.275 ΩR_{01} = 0.275\ \Omega, X01=0.387 ΩX_{01} = 0.387\ \Omega, Z01=0.475 ΩZ_{01} = 0.475\ \Omega (all referred to HV).

✓ Answer checked by Claude Opus 5.5 (high effort)

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