# Question: 4-1) A 5 KVA 200/1000 V, 50 Hz single-phase transformer gave the follow test results: V2 0.C. test (L.V.S: 200V, 1.2 A, and 90 w. S.C. test (H.V.S: 50V, 5 A, and 110 w. Using the transformer as step-up transformer to delivering 3 KW at a power factor of 0.8 lagging. I. Calculate the parameters of the equivalent circuit referred to low voltage – Free Chegg Question Answer

`Transcribed text From Image: 4-1) A 5 KVA 200/1000 V, 50 Hz single-phase transformer gave the follow test results: V2 0.C. test (L.V.S: 200V, 1.2 A, and 90 w. S.C. test (H.V.S: 50V, 5 A, and 110 w. Using the transformer as step-up transformer to delivering 3 KW at a power factor of 0.8 lagging. I. Calculate the parameters of the equivalent circuit referred to low voltage 19. (Q; side, Rase 22 Fuerzac PI. II. Compute the value of the output secondary voltage, also find percentage regulation. VR = 3.252 III. Max. Efficiency. - 04.34 % Pout Ve=9325 4-2) The following test data were taken on a 110 - KVA, 4400/440 - volt, 60 - Hz Transformer: Short-circuit test: P= 2000 w, I = 200 amps, V = 18 v Open - circuit test: P = 120 w, 1=2 amps , V = 440 v a) Calculate the voltage regulation of this transformer when it supplies Vidal full - load at 0.8 pf lagging. Neglect the magnetizing current .b) Find the efficiency at 0.8 full - load at 0.85 pf lagging. 1 93-24 2 V, 237 Set up =-2 / 4-3) A 50-KVA, 2300/230 - volt, 60 Hz transformer has a high-voltage winding resistance of 0.065 ohm and a low-voltage winding resistance of 0.0065 ohm. Laboratory tests showed the following results: Open - circuit test: V = 230 v, 1= 5.7 amps, P= 190w Y Short-circuit test: V = 41.5 v, I = 21.7 amps , P = No WM used B . Compute the value of primary voltage needed to give rated secondary voltage when the transformer is connected step-up and is delivering 40KVA at a power factor of 0.8 lagging. Vi = 232- SV • Compute the efficiency and the voltage regulation under the conditions of part a. 17975% 3) VR=1238% 12 h. But`

`Answer:`