hw1soln - Problem 1. (Problem 2.1 on page 47 of the text.)...

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6 ohms 10.39 ohms Z 6 10.392i + ohm = Z V I := To find the series circuit, first find the equivalent impedance. The series circuit will be two elements, a resistor to model the real part of the impedance and a reactance (in this case, an inductive reactance, to model the reactive portion. Q 1.039 10 3 × VAr = QI m S () := P 600W = PR e S := S 600 1039i + volt amp = SV I := VAr volt amp := Calculate the power quantities I1 0 e j 30 deg A := V 120 e j30 deg volt := Use RMS Phasors j1 := pt ( ) 600 1200 cos 2 ω t + = ( ) 1200 cos 60 deg ( ) cos 2 ω t + = ( ) 2400 cos ω t 30 deg + cos ω t 30 deg = vt ()it = a. Find p(t), S, P, and Q into the network. b. Find a simple, two element, series circuit consistent with the prescribed terminal behavior as described in this problem. it 2 10 cos ω t 30 deg = 2 120 cos ω t 30 deg + = Problem 1. (Problem 2.1 on page 47 of the text.) In Figure 2.1, 1
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Problem 2. (problem 2.4 on page 48). A 3phase load draws 200 kW at a PF of 0.707 lagging froma 440-V line. in parallel is a 3phase capacitor bank which suppoies 50 kVAr. Find the resultant power factor and current into the parallel combination. P i 200 kW . PF i 0.707 V LLi 440 volt . Q cap 50 kV . A . Find the initial values of apparant power S and reactive power Q S i P i PF i Q i S i S i . P i P i . Q i 200.06 kV A . = Real power P is unchanged by addition or deletion of reactive power Q. P f P i P f 210 5 . W = Add the two reactive power components together. Capacitive reactive power is negative under the convention assumed in the text. Q f Q i Q cap Q f 150.06 kV A . = Calculate the new apparent power. The power factor is the ratio of real power to apparent power.
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hw1soln - Problem 1. (Problem 2.1 on page 47 of the text.)...

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