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Unformatted text preview: 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 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....
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This homework help was uploaded on 03/19/2008 for the course EE 331 taught by Professor Preston during the Fall '06 term at University of Texas.
- Fall '06