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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...
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This note was uploaded on 02/23/2009 for the course ECEN 215 taught by Professor Styblinski during the Fall '08 term at Texas A&M.
- Fall '08