ECE 201 Spring 2010Homework 13 SolutionsProblem 9(a)2R and 6R in series gives 8R. 8R and 8R in parallel gives 4R. Thus 12R is inseries with Vsin the simplified circuit. Thus the Thevenin voltage is givenbyVoc=Vs12R×12× 6R=Vs4= 30 VTo find the Thevenin resistance, we short circuit Vs. Thus 8R and 8R arein parallel, which gives 4R, which in turn is in series with 2R. This gives 6Rand 6R in parallel. Thus Rth= 3R = 900 Ω.(b)Using the Thevenin equivalent circuit to simplify our calculations,PRL=30900 + RL2RL= 0.1875 W, 0.24 W, 0.244898 W (RL= 300 Ω, 600 Ω, 1200 Ω)The Thevenin equivalent circuit analysis allows us to analyze the circuitwithout the load once and then plug in the various load resistor values tocompute the relevant quantities. However, using earlier techniques wouldrequire 3 separate analyses to compute these values. Hence the use of aThevenin equivalent reduces the effort needed to obtain the answers.1(c)Using Maximum Power Transfer Theorem, fo r maximum power trans-fer, RL= 900 Ω and resultant power delivered to the load is given byPmax= (30)2/(4 × 900) = 0.25 W .(d)If Vsis doubled, power becomes four times, as Pmax∝ V2s. Thus powerdelivered to the load becomes Pload= 0.25 × 4 = 1 W .Problem 17We proceed to find the Thevenin voltage Vocand Thevenin resistance Rth.Norton current is then given by Isc= Voc/Rthand Norton resistance is sameas Thevenin resistance. Since this problem involves a dependent source, wecannot apply the conventional method. We write KVL equations to find therequired values.vAB− 300iA− vx= 0vx− 200(iA− vx/18000) −43vx= 0⇒ vx= −1800029iAvAB= −320.689655iAvAB= RthiA+ Voc⇒ Voc= 0Rth= −320.69 Ω⇒ Isc= 0Rnorton= −320.69 ΩProblem 21(a)Again, we write KVL equations to find the Thevenin and Norton equivalentcircuits.vAB− 40iA− 200(iA+ kvx+ vx/50) = 0Vs− vx+ 40iA− vAB= 02⇒ vAB[1 + 200(k + 0.02)] = iA[240 + 8000(k + 0.02)] + 2 00Vs(k + 0.02)vAB= 60iA+ 18vAB= iARth+ Voc⇒ Voc= 18 VRth= 60 ΩIsc= 0.3 A(b)Clearly, using the previous set of equations, Vocis zero f or k = −0.02 and forthis value o f k Rth= 240 Ω.Problem 37(a)vAB= iARth+ Voc54 = 0.01Rth+ Voc66 = 0.04Rth+ Voc⇒ Rth= 400 ΩVoc= 50 VIsc= 0.125 A(b)RL= 400 Ω, Pmax= V2oc/4RL= 2500/(4 × 400) = 1.56 25 W
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