Question

Phasor17

364012 7.620°kV | |-jX. 2j1280 22 Given: A 7.6 kV circuit has a load consisting of 640 22 of resistance in series with 1280 2 of inductive reactance. Required: How much parallel capacitive reactance is needed to correct the power factor to 0.96 lagging? Solution: Xc = 's Submit this value If you are not ready to submit the answer now, you can print this, work it off-line, then log in again to submit the answer. As long as you log in under the same user name, you will get the same parameters when you re-select this problem. D... . .. ..1.1..... .

Answer 1

So! 41 7.60 7-640 Ejl2cor I 401)280 1640 +j1280) and jxare in porillel So equivalent admittance. Yes azónical Yega *°40 - j) 26. 64. +3/280 640 -j/ze. Yes = 640-jizeo j 2048000 Yes = 0.0003125 - 30.000625 +3 Mes = 0.0003125 + v/ € - 0.000625) Izvyes since power factor is given as coop=0.96 $ = Coslys.96) = 16.26 So. I is legging by v by 16.26 and tang = -0.000G25 0-00g/25 tan(16-26) = * 0.0003125 - 9-1145*15= -0.0082. 445 * . / 5 L. 0.00025 + 9.1/ 45 < = 7.16145.X.157 * = 1394- 34 -