Solution Manual for Electrical Engineering: Principles and Applications, 7th Edition

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1CHAPTER 1ExercisesE1.1Charge = CurrentTime = (2 A)(10 s) = 20 CE1.2A)2cos(200)200cos(2000.010t)0.01sin(20()()(ttdtddttdqtiE1.3Becausei2has a positive value, positive charge moves in the samedirection as the reference.Thus, positive charge moves downward inelementC.Becausei3has a negative value, positive charge moves in the oppositedirection to the reference.Thus positive charge moves upward inelementE.E1.4Energy = ChargeVoltage = (2 C)(20 V) = 40 JBecausevabis positive, the positive terminal isa and the negativeterminal isb. Thus the charge moves from the negative terminal to thepositive terminal, and energy is removed from the circuit element.E1.5iabenters terminala.Furthermore,vabis positive at terminala. Thusthe current enters the positive reference, and we have the passivereference configuration.E1.6(a)220)()()(ttitvtpaaaJ666732032020)(310031001002ttdttdttpwaa(b)Notice that the references are opposite to the passive signconvention. Thus we have:20020)()()(ttitvtpbbbJ100020010)20020()(1002100100ttdttdttpwbb

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