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Problem 2: Determine the power absorbed by the 47 kO2 resistor in the circuit below: 47k2 3 kN v, > 2k2 Ans: P472 = 18.117 uW
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Answer #1

Given circit 47 kr V12 VI uur I V 2 20kr 12K2 a SMAL 0.5 look as This first we need to find out vx - I = Sx03 2x10² 34103 +7+voltage acurors = 47kr in 1.25 x 472103 474103 + 50x103 = 0.92277 Volts. absorbed by 47 kr resistant is - power 2 47K (0.9227

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