This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 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...
View Full
Document
 Fall '08
 Styblinski

Click to edit the document details