Question

please solve it clearly

A 1000-VA, 230/115-V has the following equivalent circuit parameters Referred to high voltage side Referred to low voltage side Req=0.503 Ohm Rc=3383 Ohm Xeq=1.905 Ohm XM=1099 Ohm (Find the expected values for the open and short circuit tests and fill the following table (show all steps Short Circuit test (High voltage side) =VSC - Isc-rated current Pse Open Circuit test (Low voltage side) -Voc -Ioc Poc Then .a) Find the equivalent circuit of this transformer referred to low voltage side) .b) Find the transformer's voltage regulation at rated conditions with 0.8 PF lagging) ..) Draw the phasor diagram of the transformer based on the conditions of part b)

Answer 1

A 1000 VA 230/115 u has the following equivalent Cituit parameters. Reffered to HU side Refered to tu sidle Req = 0.5032 Re = 33832 xeq=1.9052 XM = 10gg. solution: opai ciruit test i This test is performed on Ly side primary Voc = rated primary voltage on au side = 11h Hoc =115 ull parametes 115 - - - Ioc = Cussant through Shant Joe V XMXRc XM+ Rc Ioc = 0.1386 Ampere 1099x3383 1099 +3383 Poc = Voc² = 115² Ro 3383 Poc = 3.9 watt ! Short ciruit test: This test is performed on Hu side as Primary. Ise = Rated cussent of Hu side 1000 VA 230 V Ise = 4.3478 Ampere/

Page = Isa Req = (4.3478)’x 0.503 Pigc = 9.5 watt Poc Vo Tse-4 Psc = Usex Isc 4. Ise = Psc = Vse. Use a Zsc Use Zse Psc = use Vse = / Pse • Zse Zsc = √ Req² + Xeq? = 10.5032+.095 Zse = 1097 Vse = √9058 1997 Usc = 4.324 Short ciruit test (High voltage side) Vse = 4.32u open Ciruit test I c loni noltage side) Voc = 115 volt Ise= 4.34 A Joc = 0.1386 Ampere Poc a to 3.g watt Psc = 9.5W I a) Equivalent Ciruit sefessed to low voltage side. Eres transformation fetio EK) = k =0.5

* Req Refered to Lv side Req (Luside) = Reqxk² = 0.503 (0.5)2 = 0.125 r * Xeq Refered to cu side xeq Zcu side) = Xeqxk? . = 1.905 x(0.5)2 [Xeq = 0.47 2 equiualent Ciruit Req=0.125 a nteq=0.47 1150 EXM=3383 2304 of voltage Regulation at rated power with 0.8 pf lagging. Vili Iz = VA - = 4,3478 Aupee - = 1000 Uz 230 Molterge Regulation - int Iz X2 sing2 x 100 I2 R2 Cos 02 + Iz X, sino - V₂ - Iz Reg cosø + Iz Xeq sin oz ve - x100 - U₂ = 4.34 x 0.503 X 0.8 +434 x 1.905 x 0.6 -X100 Holtage Regulation = 2.91% 230

Phaset diagram of transformer for about constition. V2-230V 9.20.8 keq Iz = 4.34 A