10 kVA 200/400 V tests, parameters on LV side

Reading of the tests: "HV open" means the OC test is done on the LV winding at its rated 200 V. "LV short" means the LV winding is short-circuited while the supply is applied to the HV (400 V) winding, so the SC readings are on the HV side. Hence $K = 400/200 = 2$, $K^2 = 4$.

Open-circuit test (on the LV side)

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

$$\begin{aligned} \cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{120}{200\times1.3} = 0.4615 \\ I_w &= I_0\cos\phi_0 = 0.6\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 1.153\ \text{A} \\ R_0 &= \frac{V_0}{I_w} = 333.3\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 173.4\ \Omega\quad(\text{on the LV side}) \end{aligned}$$

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.

$$\begin{aligned} Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{22}{30} = 0.7333\ \Omega \\ R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{200}{30^2} = 0.2222\ \Omega \\ X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 0.6989\ \Omega\quad(\text{on the HV side}) \end{aligned}$$

Referring to the LV side

The turns ratio is $K = 400/200 = 2$, 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.

$$\begin{aligned} R_{eq} &= 0.2222 \div 4 = 0.05556\ \Omega \\ X_{eq} &= 0.6989 \div 4 = 0.1747\ \Omega \end{aligned}$$

Equivalent circuit referred to the LV side

   Req=0.0556 ohm   Xeq=0.175 ohm
 o--+-----/\/\/\-----UUUU------o
    |                          
   [R0=333.3] [X0=173.4]   Load
    |      (in parallel)
 o--+---------------------------o

$R_{eq} = 0.05556\ \Omega$, $X_{eq} = 0.1747\ \Omega$ ($Z_{eq} = 0.1833\ \Omega$), shunt branch $R_0 = 333.3\ \Omega$ in parallel with $X_0 = 173.4\ \Omega$.

Answer (referred to LV side): shunt branch $R_0 = 333.3\ \Omega$, $X_0 = 173.4\ \Omega$; series branch $R_{eq} = 0.0556\ \Omega$, $X_{eq} = 0.1747\ \Omega$