A 2200/220V, single phase transformer has the following resistance and leakage reactance referred to primary: R5.6 ohms, X22 ohms1. calculate V2, I2 and cos(2) when the primary is supplied with 2200V, 9A at a lagging power factor of 0.8. Neglect Io.
you may get the comparable voltage out of one output as in case you had related all 3. the different 2 might have decrease voltages for the period of them. case in point, say the transformer replace into regularly a 240 to 480 3-section delta. in case you have been to place 240 for the period of legs A and B, you may get 480 for the period of A and B of the output. i've got self assurance the B/C and C/A outputs might have 240 each and each, because of the fact the unconnected C section might seem as a midsection-faucet to the A and B legs
I2 I1[V1/V2] 9[2200/V2] I29[2200/20] 9[10] I2 90 amperes. V2 220 volts
This is not really a very hard problem but I'm afraid artsupre's solution of using only the turns ratio is a little too simplistic. According to the problem definition, we need to represent this transformer in an equivalent circuit, as a perfect or ideal transformer of turns ratio 10 preceded by a resistance and leakage reactance in series and equal respectively to the R and X values quoted. This shows us that the current will be totally transformed (therefore I2 10 times I1 90 A) but there will be some voltage drop in the preceding impedance. So how to calculate the voltage drop? We need to take the voltage applied to the terminals and subtract from it the total voltage drop across R and X. This is a vector sum since resistance drops are in phase with current but reactive drops are 90 degrees lagging. We normally use complex representation to accomplish these vector summations. In this case the procedure is: Take the current vector as real. Then the applied voltage is 2200(0.8 + j0.6) and the voltage drop across X+R is 9(5.6 + j22) Hence the resultant voltage at the terminals of the ideal transformer (1710 + j1122) 2045 at a leading angle of arc cos (0.836), i.e. at a power factor of 0.836. This voltage of course has to be transformed by the turns ratio giving V2 204.5 at the above power factor. If you need further help or clarification, please edit your question and I will try to respond. Please take a moment to award one of your answers Best Answer. That's partly why we do this!!
