2- A 10 kVA, 2500/250 V, single phase transformer gave the following results: V (Volt) I (Amp) Power (Watt) Side O.C. Test 250 0.8 S.C. Test 60 3 50 45 L.V. H.V. Find the equivalent circuit parameters referred to LV side? = Ans:- (R'o 1250 S2, X'o=323 2, R02 = 0.05 2, X02 = 0. 1936 2)
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- Answer only please The resistance of a transformer winding is 0.25 ohm at 25oC. When operating at full load, the temperature of the winding is 75oC. The temperature coefficient of resistance for copper at 0oC is 0.00427 per oC. What is the winding resistance at full load? a. 0.512 ohm b. 0.315 ohm c. 0.298 ohm d. 0.271 ohmI need the answer quicklyNone
- Can you solve D and E1. A 2.4-KVA, 2400/240-V, 50-Hz, the step-down transformer has the following parameters: : RH = 1.59, :XH = 2.50, R₂ = 0.020, X₁ = 0.030, : Rc = 6k2, and: Xm = 8kQ. It is operating at 50% of its load at 0.5 power factor leading. Using the exact equivalent circuit embodying the ideal transformer, determine the efficiency of the transformer. Also, sketch its phasor diagram.dc tansformd
- Q3/A/ The equivalent circuit impedances of a 20-kVA, 8000/240 V, 60-Hz transformer are to be determined. The open-circuit test was performed on the secondary side of the transformer and the short circuit test was performed on the primary side of the transformer .The following data was taken: Open-circuit Short-circuit test test (on secondary) Voc 240 v (on primary) Vsc 489 v Isc 2.5 A Psc 240 W loc 7.133 A Poc 400 WA voltage transformer in Figure 8 has 1,500 turns of wire on its primary winding and 500 turns of wire for its secondary winding. What will be the turns ratio, a of the transformer. If 240 volts is applied to the primary winding, what will be the resulting secondary voltage. And if a 400 n resistor is connected to the secondary winding, what are the values of current Is and Ip. primary Vp Transformer Core NF Ns Figure 8 secondary VsPlease explain each step
- i need ans within 15 minutes my best wishes tonA Buck converter has an output power of 1950 Watts, and a source voltage of 1000V (the constant DC voltage). Using the information provided and the "“transformer-like" relationships derived in class, estimate the corresponding load currents for the following duty ratios: a) D=0.35 b) D=0.9 c) D=0.5A 100/S00 V. 1 KVA, 50 Hz transformer was put through open and short-circuit tests to find its internal parameters. The test conditions and derived parameters are as follows Open circut test: The high voltage side terminals were open. The low voltäge side was connected to 10020ª, 50 Hz supply. The low voltage side current was measured as 50245º MA. Short-circuit test The high voltage side terminals were short-circuited. The low voltage side was connected to a variable voltage supply (50 Hz). A full load current of 102-30° A was achieved when the variable voltage supply was at 820° V. The transformer was then put to operation by connecting a load of 400230° 2 to the high voltage side (secondary) while applying a primary voltage of 10020º V. Calculate the primary current (the current drawn from the power supply).