To determine the parameters of the equivalent circuit of a single phase transformer and predetermine the performance characteristics, by performing the OC and SC test on single phase transformer
NAME PLATE DETAILS OF SINGLE PHASE TRANSFORMER 

KVA Rating 
2 KVA 
HV Volts 
230 VAC 
LV Volts 
115 VAC 
HV current 
8.6 A 
LV current 
17.3 A 
Winding type 
Core 
S.No. 
Name of the Equipment 
Range 
Quantity 
Type 
1. 
Single Phase Verica 
0 to 230V/15 A

1 No

Core 
2. 
Voltmeter 
0 to 150 V 
1 No 
Digital 
3. 
Voltmeter 
0 to 50 V 
1 No 
Digital 
4. 
Ammeter 
5 A 
1 No 
Digital 
5. 
Ammeter 
10 A 
1 No 
Digital 
6. 
Wattmeter ( LPF ) 
150/300V/10A 
1 No 
Digital 
7. 
Wattmeter ( UPF ) 
75/150/20A 
1 No 
Digital 
A transformer is a static device comprising of two coils coupled through a magnetic medium connecting two ports at different (sometime same) voltage levels in an electric system allowing the interchange of electrical energy in either direction via magnetic field. The schematic diagram of a transformer is shown in Figure.
An ideal transformer obeys the following relations:
$$\begin{aligned} \dfrac{E_1}{E_2}&=\dfrac{V_1}{V_2}=\dfrac{N_1}{N_2}\\ \dfrac{I_1}{I_2}&=\dfrac{N_2}{N_1} \end{aligned}$$
Impedance seen on primary (impedancetransformation property)
$$\begin{aligned} Z_2^{\prime}&=Z_2\cdot \left(\dfrac{N_1}{N_2}\right)^{2} \end{aligned}$$
S.No. 
V_{oc}(V) 
I_{oc}(A)

W_{oc}= W× M.F.(w)





S.No. 
V_{sc}(V) 
I_{sc}(A)

W_{sc}= W× M.F.(w)





where,
M. F. = Multiplication factor = \((VI\cos\phi)/\mathrm{FSD}\) FSD = Full scale divisions
By the equivalent circuit parameters \( R_0, X_0, R_{01}, R_{02}, X_{01} \) and \( X_{02} \) are found from the O.C. and S.C. test results, draw the equivalent circuit referred to L.V. side as well as H.V. side.
With transformer parameter values obtained OC and SC tests, fill in the values in the equivalent circuits
At 0.8 Pf lag
Output P_{0} = K × (Rated KVA) × (0.80)
Iron loss (Constant loss) = P_{i} = W_{0}
Copper loss = K^{2} P_{C} = K^{2} W_{sc}
Where W_{sc} at rated current = copper loss at rated = I_{rated}^{2} × R_{equ}
Total loss = P_{L} = (P_{i} + K^{2} P_{C})
Input power (P_{input}) = P_{0} + P_{L} = K × (rated KVA) × (0.8) + P_{i} + K^{2} P_{C}
Efficiency = η = output / input
S.No  1  2  3  4  5  6  7  8 

Different Transformer Loading (K)  0.15  0.30  0.45  0.60  0.75  0.90  1.05  1.20 
Copper Losses (K^{2}P_{C})  
Iron Losses (P_{i}) / Constant Losses  
Total loss = P_{L} = (P_{i} + K^{2}P_{C})  
Output Power (P_{0}) in KW 
Determine the following either on LV or HV side
Rated voltage V = R=
Rated current I = X =
Power Factor  1.0  0.8  0.6  0.4  0.2 

R cos ɸ  
X sin ɸ  
R cos ɸ + X sin ɸ  
R cos ɸ  X sin ɸ  
Regulation (LAG)  
Regulation (LEAD) 
Mark the power factor corresponding to zero voltage regulation.
Plots should be drawn between
 \( Y_0 = \frac{I_0}{V_0} \)
 \( G_i = \frac{W_0}{(V_0)^2} \)
 \( B_m = \sqrt{Y_0^2  G_i^2} \)
 \( Z_{eq} = \frac{V_{sc}}{I_{sc}} \)
 \( R_{eq} = \frac{W_{sc}}{(I_{sc})^2} \)
 \( X_{equ} = \sqrt{Z_{eq}^2  R_{eq}^2} \)
Draw the Equivalent circuit referred to the LV side & HV side.
LOAD  EFFICIENCY 
1/4  
1/2  
3/4  
FULL 
PF at which regulation is zero =