250/500 V transformer test, equivalent circuit

Given: 250/500 V, 50 Hz. The primary is the 250 V (LV) winding, the secondary is the 500 V (HV) winding, $K = 500/250 = 2$.

Open-circuit test (on the LV side)

The no-load power is almost entirely core loss, so the shunt branch is found from it.

$$\begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{80}{250\times1} = 0.32 \\ I_w &= I_0\cos\phi_0 = 0.32\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 0.9474\ \text{A} \\ R_0 &= \frac{V_0}{I_w} = 781.25\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 263.9\ \Omega\quad(\text{on the LV side}) \end{aligned}$$

Short-circuit test (on the HV side)

At the reduced test voltage the core loss is negligible, so the wattmeter reading is the copper loss.

$$\begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{20}{12} = 1.667\ \Omega \\ R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{100}{12^2} = 0.6944\ \Omega \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 1.515\ \Omega\quad(\text{on the HV side}) \end{aligned}$$

Referring to the LV side

The turns ratio is $K = 500/250 = 2$, so $K^2 = 4$. Impedances are multiplied by $K^2$ when moved from LV to HV, and divided by $K^2$ when moved from HV to LV.

$$\begin{aligned} R_{eq} &= 0.6944 \div 4 = 0.1736\ \Omega \\ X_{eq} &= 1.515 \div 4 = 0.3788\ \Omega \end{aligned}$$

Equivalent circuit referred to the LV side

   Req=0.174 ohm   Xeq=0.379 ohm
 o--+-----/\/\/\-----UUUU------o
    |                          
   [R0=781.25] [X0=263.9]   Load
    |      (in parallel)
 o--+---------------------------o

$R_{eq} = 0.1736\ \Omega$, $X_{eq} = 0.3788\ \Omega$ ($Z_{eq} = 0.4167\ \Omega$), shunt branch $R_0 = 781.25\ \Omega$ in parallel with $X_0 = 263.9\ \Omega$.

Answer (referred to primary): $R_0 = 781.25\ \Omega$, $X_0 = 263.9\ \Omega$, $R_{01} = 0.1736\ \Omega$, $X_{01} = 0.3788\ \Omega$