Chapter 01

Chapter 01 - CHAPTER 1 Solutions for Exercises E1.1 E1.2...

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1 CHAPTER 1 Solutions for 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 0 3 10 0 10 0 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 0 2 10 0 10 0 !
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This note was uploaded on 10/28/2008 for the course ECE 331 taught by Professor Baliga during the Spring '08 term at N.C. State.

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Chapter 01 - CHAPTER 1 Solutions for Exercises E1.1 E1.2...

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