ECE 201 Spring 2010 Homework 5 Solutions Problem 26 Since V2 60 V the current through 60 branch is 1 A The resistors 90 and 180 are in parallel Their equivalent resistance is 90 180 90 180 60 Req Now 60 and 60 are in series Thus the voltage drop across the parallel combination of 40 and 120 is 120 V Thus current through 40 resistor is 120 40 3 A Hence Is 3 1 4A Applying KVL around the loop containing the source and 40 resistor Vs 180 4 120 0 Vs 840 V Power delivered by the source is given by Ps Vs Is 840 4 3360 W Problem 32 a Using voltage division the following equations can be written R1 R2 Rs 2400 R1 R2 0 75 R1 R2 Rs R2 0 25 R1 R2 Rs 1 1 2 3 Using 1 and 3 R2 600 Using 2 R1 1200 Using 1 Rs 600 Problem 40 The resistors 9 k and 18 k are in parallel Thus their equivalent resistance of 6 k is in series with 6 k Now the combination 4 k and 12 k are in parallel The current Iin divides into I1 and I2 through the parallel resistors in inverse ratio of their resistances Similarly the current I2 divides in inverse ratio of resistances 9 k and 18 k in parallel Thus we get 12 12 4 90 mA I1 120 I2 120 90 30 mA 9 9 18 10 mA I3 30 Vin 0 09 4 103 360 V Psource Vin Iin 43 2 W 2
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