0132130084_ism01

0132130084_ism01 - 1 CHAPTER 1 Exercises E1.1 Charge =...

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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 t dt d dt t dq t i = × = = = E1.3 Because i 2 has a positive value, positive charge moves in the same direction as the reference. Thus, positive charge moves downward in element C . Because i 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 i ab enters terminal a . Furthermore, v ab is positive at terminal a . Thus the current enters the positive reference, and we have the passive reference configuration. E1.6 (a) 2 20 ) ( ) ( ) ( t t i t v t p a a a = = J 6667 3 20 3 20 20 ) ( 3 10 3 10 10 2 = = = = = ∫ ∫ t t dt t dt t p w a a (b) Notice that the references are opposite to the passive sign convention. Thus we have: 200 20 ) ( ) ( ) ( − = − = t t i t v t p b b b J 1000 200 10 ) 200 20 ( ) ( 10 2 10 10 − = − = − = = ∫ ∫ t t dt t dt t p w b b 2 E1.7 (a) Sum of currents leaving = Sum of currents entering i a = 1 + 3 = 4 A (b) 2 = 1 + 3 + i b ⇒ i b = -2 A (c) 0 = 1 + i c + 4 + 3 ⇒ i c = -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 A , B , and C : -3 - 5 + v c = 0 ⇒ v c = 8 V Then go clockwise around the loop composed of elements C , D and E : - v c- (-10) + v e = 0 ⇒ v e = -2 V E1.10 Elements E and F are in parallel; elements A and B are in series. E1.11 The resistance of a wire is given by A L R ρ = . Using 4 / 2 d A π = and substituting values, we have: 4 / ) 10 6 . 1 ( 10 12 . 1 6 . 9 2 3 6 − − × × × = π L ⇒ L = 17.2 m E1.12 R V P 2 = ⇒ Ω = = 144 / 2 P V R ⇒ A 833 . 144 / 120 / = = = R V I E1.13 R V P 2 = ⇒ V 8 . 15 1000 25 . = × = = PR V mA 8 . 15 1000 / 8 . 15 / = = = R V I E1.14 Using KCL at the top node of the circuit, we have i 1 = i 2 . Then, using KVL going clockwise, we have - v 1- v 2 = 0; but v 1 = 25 V, so we have v 2 = -25 V. Next we have i 1 = i 2 = v 2 / R = -1 A. Finally, we have W 25 ) 1 ( ) 25 ( 2 2 = − × − = = i v P R and W. 25 ) 1 ( ) 25 ( 1 1 − = − × = = i v P s E1.15 At the top node we have i R = i s = 2A. By Ohm’s law we have v R = Ri R = 80 V. By KVL we have v s = v R = 80 V. Then p s = -v s i s = -160 W (the minus sign is due to the fact that the references for v s and i s are opposite to the passive sign configuration). Also we have W. 160 = = R R R i v P 3 Problems P1.1 Four reasons that non-electrical engineering majors need to learn the fundamentals of EE are: 1. To pass the Fundamentals of Engineering Exam....
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This note was uploaded on 11/20/2011 for the course ELE 220 taught by Professor Anas during the Spring '11 term at American Dubai.

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0132130084_ism01 - 1 CHAPTER 1 Exercises E1.1 Charge =...

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