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Unformatted text preview: 1 CHAPTER 1 Exercises E1.1 Charge = Current × Time = (2 A) × (10 s) = 20 C E1.2 A ) 2cos(200 ) 200cos(200 0.01 0t) 0.01sin(20 ( ) ( ) ( t dt d dq i = × = = = E1.3 Because 2 has a positive value, positive charge moves in the same direction as the reference. Thus, positive charge moves downward in element C . Because 3 has a negative value, positive charge moves in the opposite direction to the reference. Thus positive charge moves upward in element E . E1.4 Energy = Charge × Voltage = (2 C) × (20 V) = 40 J Because v ab is positive, the positive terminal is a and the negative terminal is b . Thus the charge moves from the negative terminal to the positive terminal, and energy is removed from the circuit element. E1.5 enters terminal . Furthermore, is positive at terminal . Thus the current enters the positive reference, and we have the passive reference configuration. E1.6 (a) 2 20 ) ( ) ( ) ( p = = J 6667 3 20 3 20 20 ) ( 3 10 3 10 10 2 = = = = = ∫ ∫ w (b) Notice that the references are opposite to the passive sign convention. Thus we have: 200 20 ) ( ) ( ) ( − = − = J 1000 200 10 ) 200 20 ( ) ( 10 2 10 10 − = − = − = = ∫ ∫ 2 E1.7 (a) Sum of currents leaving = Sum of currents entering i a = 1 + 3 = 4 A (b) 2 = 1 + 3 + b ⇒ = 2 A (c) 0 = 1 + c + 4 + 3 ⇒ = 8 A E1.8 Elements A and B are in series. Also, elements E , F , and G are in series. E1.9 Go clockwise around the loop consisting of elements , , and C : 3  5 + v = 0 ⇒ = 8 V Then go clockwise around the loop composed of elements , D and :  (10) + e = 0 ⇒ = 2 V E1.10 Elements and are in parallel; elements and are in series....
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This note was uploaded on 11/02/2010 for the course ECE 3710 taught by Professor Haris during the Spring '07 term at Georgia Tech.
 Spring '07
 Haris

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