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250/500 V transformer, parameters both sides

Short answer

Using the printed test-side assignment, LV side: Req=0.694 Ω, Xeq=1.515 Ω, R0=195.3 Ω, X0=65.97 Ω; HV side: Req=2.778 Ω, Xeq=6.06 Ω, R0=781.25 Ω, X0=263.9 Ω. If the OC and SC side labels are interchanged, LV values become R0=781 Ω, X0=264 Ω, Req=0.174 Ω, Xeq=0.379 Ω.

A 1 phase 250/500 V, 50 Hz transformer gave the following test results: Open circuit test: 250 V, 1 A, 80 W on H.V. side Short-circuit test: 20 V, 12 A, 100 W on L.V. side Calculate the equivalent circuit parameters and draw the equivalent circuit referred to low voltage side and high voltage side.

Data as printed: the OC test is on the HV winding (250 V, 1 A, 80 W) and the SC test is on the LV winding (20 V, 12 A, 100 W). The test values are used exactly as given. The transformer is 250/500 V, so the LV winding is 250 V, the HV winding is 500 V, and K=2K = 2, K2=4K^2 = 4.

Open-circuit test (HV side)

cos⁡ϕ0=80250×1=0.32,Iw=0.32 A,Iμ=1−0.322=0.9474 AR0H=2500.32=781.25 Ω,X0H=2500.9474=263.9 Ω\begin{aligned} \cos\phi_0 &= \frac{80}{250\times1} = 0.32,\quad I_w = 0.32\ \text{A},\quad I_\mu = \sqrt{1 - 0.32^2} = 0.9474\ \text{A} \\ R_{0H} &= \frac{250}{0.32} = 781.25\ \Omega,\qquad X_{0H} = \frac{250}{0.9474} = 263.9\ \Omega \end{aligned}

Short-circuit test (LV side)

Zeq,L=2012=1.667 Ω,Req,L=100122=0.6944 ΩXeq,L=1.6672−0.69442=1.515 Ω\begin{aligned} Z_{eq,L} &= \frac{20}{12} = 1.667\ \Omega,\qquad R_{eq,L} = \frac{100}{12^2} = 0.6944\ \Omega \\ X_{eq,L} &= \sqrt{1.667^2 - 0.6944^2} = 1.515\ \Omega \end{aligned}

Referred to the high-voltage (HV) side

Impedance moves from LV to HV by multiplying by K2=4K^2 = 4:

Req,H=4×0.6944=2.778 Ω,Xeq,H=4×1.515=6.060 ΩR_{eq,H} = 4\times0.6944 = 2.778\ \Omega,\qquad X_{eq,H} = 4\times1.515 = 6.060\ \Omega

with shunt branch R0H=781.25 ΩR_{0H} = 781.25\ \Omega, X0H=263.9 ΩX_{0H} = 263.9\ \Omega.

  HV side:  Req=2.778 ohm  Xeq=6.06 ohm
 o--+-----/\/\/\-----UUUU------o
    |
   [R0=781.25] [X0=263.9]   Load
    |
 o--+---------------------------o

Referred to the low-voltage (LV) side

Shunt branch is divided by K2K^2:

R0L=781.254=195.3 Ω,X0L=263.94=65.97 ΩR_{0L} = \frac{781.25}{4} = 195.3\ \Omega,\qquad X_{0L} = \frac{263.9}{4} = 65.97\ \Omega

with series branch Req,L=0.6944 ΩR_{eq,L} = 0.6944\ \Omega, Xeq,L=1.515 ΩX_{eq,L} = 1.515\ \Omega.

  LV side:  Req=0.694 ohm  Xeq=1.515 ohm
 o--+-----/\/\/\-----UUUU------o
    |
   [R0=195.3] [X0=65.97]   Load
    |
 o--+---------------------------o

Answer: HV side: R0=781.25 ΩR_0 = 781.25\ \Omega, X0=263.9 ΩX_0 = 263.9\ \Omega, Req=2.778 ΩR_{eq} = 2.778\ \Omega, Xeq=6.060 ΩX_{eq} = 6.060\ \Omega. LV side: R0=195.3 ΩR_0 = 195.3\ \Omega, X0=65.97 ΩX_0 = 65.97\ \Omega, Req=0.694 ΩR_{eq} = 0.694\ \Omega, Xeq=1.515 ΩX_{eq} = 1.515\ \Omega

Note: 250 V is the LV rating, so the side labels in the paper are probably swapped (normally the OC test is done on the LV side and the SC test on the HV side). With that reading the results are: LV side R0=781.25 ΩR_0 = 781.25\ \Omega, X0=263.9 ΩX_0 = 263.9\ \Omega, Req=0.1736 ΩR_{eq} = 0.1736\ \Omega, Xeq=0.3788 ΩX_{eq} = 0.3788\ \Omega; HV side R0=3125 ΩR_0 = 3125\ \Omega, X0=1055.5 ΩX_0 = 1055.5\ \Omega, Req=0.6944 ΩR_{eq} = 0.6944\ \Omega, Xeq=1.515 ΩX_{eq} = 1.515\ \Omega.

✓ Answer checked by Claude Opus 5.5 (high effort)

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