1. Example Consider the following details from a 2 kVA 240V/ 240 V transformer. Winding Resistances Primary winding resistance - 0.5 Ω Secondary winding resistance - 0.6 Ω No-Load Test Primary Applied Voltage - 240 V No-load current - 0.13 A ∠ -57  Core losses - 17 W Secondary no-load Voltage - 250 V Short Circuit Test Impedance Voltage - 10 V Primary Full load Current - 8.7 A Secondary Full load Current - 8.3 A Copper Losses - 130 W Phase Angle - 16.8  lagging Using the above information draw the equivalent circuit of the transformer. Answer The Primary No-load Impedance The magnetising current and iron loss currents can be determined from the no-load current. No-load current = 0.13 ∠ -57  = (0.0708 - J 0.109) A That is, 0.0708 A in phase with the supply voltage and 0.109 A lagging the supply voltage by 90  E. For the equivalent diagram then and

2. Checking As the above resistance and reactance are connected in parallel, their combined impedance should be equal to Z 0 that was calculated above. The total equivalent impedance, referred to the primary can be determine from the short-circuit test. The resistance and reactance referred to the primary are a 1.158 Ω resistor in series with a 0.332 Ω inductive reactance. Because the transformer in this case has a ratio of not quite 1:1, (240:250) the resistances can be seen to be made up of: Primary resistance = 0.5 Ω Secondary Resistance = 0.6 Ω Note that these value should agree with the value from the d.c. resistance tests. In a similar manner, the total leakage reactance can be divided by 2 to give the primary and secondary leakage reactance. Primary Leakage Reactance = 0.33321 / 2 = 0.166 Ω Secondary Leakage Reactance = 0.33321 / 2 = 0.166 Ω