ECE 201 Spring 2010Homework 5 SolutionsProblem 26Since V2= 60 V , the current through 60 Ω branch is 1 A. The resistors 90 Ωand 180 Ω are in parallel. Their equivalent resistance isReq=90 × 18090 + 180= 60Now 60 Ω and 60 Ω a r e in series. Thus the volta ge drop across the parallelcombination of 40 Ω and 120 Ω is 120 V. Thus current thro ug h 40 Ω resistoris 120/40=3 A. Hence,Is= 3 + 1= 4 AApplying K VL around the loop containing the source and 40 Ω resistor,Vs− 1 80 × 4 − 120 = 0⇒ Vs= 840 VPower delivered by the source is given by,Ps= Vs× Is= 840 × 4= 3360 WProblem 32(a)Using voltage division, the following equations can be written,R1+ R2+ Rs= 2400 (1)R1+ R2R1+ R2+ Rs= 0.75 (2)R2R1+ R2+ Rs= 0.25 (3)1Using (1) and (3), R2= 600 Ω. Using (2), R1= 1200 Ω. Using (1), Rs=600 Ω.Problem 40The resistors 9 kΩ and 18 kΩ are in parallel. Thus their equivalent resistanceof 6 kΩ is in series with 6 kΩ. Now the combination 4 kΩ and 12 kΩ are inparallel. The current Iindivides into I1and I2through the para llel resistorsin inverse ratio of t heir resistances. Similarly, the current I2divides in inverseratio of resistances 9 kΩ and 18 kΩ in parallel. Thus we get,I1= 120 ×1212 + 4= 90 mAI2= 120 − 90 = 30 mAI3= 30 ×99 + 18= 10 mAVin= 0.09 × 4 × 103= 360 VPsource= VinIin= 43.2
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