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
