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# soln2 - EE 212 Assignment 2 Solutions Problem 1 10 120 V A...

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EE 212 – Assignment 2 Solutions Problem # 1 Figure 1 Voltage = 120 0 0 Frequency = 60 Hz Angular frequency, ϖ = 2 ×π× f = 377 rad/s X L1 = ϖ L 1 = 377 * 50 * 10 -3 = 18.85 X L2 = ϖ L 2 = 377 *100 *10 -3 = 37.70 Z = (10 + j18.85) + (100 || j37.7) = (10 + j18.85) + (35.3 69.3 0 ) = (10 + j18.85) + (12.4 + j33.0) = 22.4 + j51.9 = 56.5 66.6 0 ohms i) Calculate the current drawn from the source. Ι = V/Z = 120 0 0 V/ 56.5 66.6 0 Ω = 2.12 -66.6 0 A ii) Value of the capacitor required to correct power factor to unity. Current p.f. =cos -66.6 0 = 0.397 lagging S = V Ι * = (120 0 0 ) (2.12 -66.6 0 )* VA = (120 0 0 ) (2.12 66.6 0 ) VA = 255 66.6 0 VA = (101 + j234) VA 120 V ~ j18.85 100 10 + - A B j37.7

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Q = 234 VAr Capacitor C must generate the reactive power, Q c = 234 VAr Figure 1a Q c = |V| 2 / X c and X c = 1 / ϖ C C = Q / ϖ |V| 2 = 234 / (377) * (120) 2 = 43.1 µF iii) With the power factor corrected to unity, the current drawn from the source. Current drawn from source, Ι new S = V Ι * new With unity power factor, S = 101 + j0 (i.e. no reactive power) Ι * new = S / V = 101 0 0 / 120 0 0 = 0.843 0 0 A Alternate Method: X c = (120) 2 V 2 / 234 VAr = 61.6 -90 0 since X c lags V by 90 0 Z new = Z || X c = ( 56.507 66.5972 0 )Ω || (61.5724 -90 0 = 142 0 0 Ω Ι * new = V / Z new = (120 0 0 ) V/ (142 0 0 ) Ω = 0.843 0 0 A
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soln2 - EE 212 Assignment 2 Solutions Problem 1 10 120 V A...

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