Question

2) For the circuit shown, draw the phasor equivalent circuit for the described circuit, and answer the questions below. then find the voltage across the current source (v(t)). Note for questions b and c you need to determine the instantaneous power associated with the source using values for P and Q(P and Q are determined from the phasor analysis results for V and Ig) and then use the trig identity Acos(x)Bsin(x) = V42 + B2 cos(x-Tan(B/A to convert Pcos(x) - Qsin (x) to the form Dcos(x- theta). ig(t)= 0.1cos(1000) Amps vt) 300 mH 400 Ohms a) Find the phasors i, and V b) Find the instantaneous power associated with the source in the form p(t) = P + Pcos(2wt) - Qsin(2wt). Hint: Find P and Q using the phasor values from part a. c) What is the instantaneous power associated with the source written in terms of a single cosine function? d) What is the peak power delivered by the source? e) What is the peak absorbed power by the source?

Answer 1

1(1- 0.1 cos 1000+ A ( 1 ) Vio your 200m H stanice (cm) - [ too * Szo A 3 = 0.1 X 1200 Ltant 4 [3= 24 L 53.13² v V = 2 4 2

© igly 0.1 Cos 1000 A, vit) - 24 Cos (10004 +53-139) v PH) = iga vit) PCH = Cool Cos 1000 €) (24 Cos (1000+ +53.13')) PTH = 2.4 Cos (1000 4) Cos (10004 +53-13°) POH = 2.4 Cos (1000 t) (COS 1000 cos 53.139 - Sin 1000+ Sin 53.13) Ples - 2.4 (Cos 1000 6) COS (53.3") - 2.4 (Cos 10004) (Sin 1000+) (Sim 5313) Plt) = (2-4) (5+ Cos 2(104(06) - 2.4 (Sin (2004)) (0-) P(t) = 0.72 + 0.72 Cos 2000+ - 0.96 Sin 20006 0.96 P(E)= 0.12 + √(0.12)" +(-0.96) Cos (2000k - tant (-0.15) PCH = 0.72 + 1.2 Cos (20001 + 53.13%) w

a pak proti do delivered by the Peak power source 21.92 = 0.72 +1+2 = 1.92 walts. @ Peak source = 0.72 - 1-2 = -0-48 watts Peak power power by the obsorbed obsorbed by the