EE 2950 Chapter : Chapter 01
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Charge = current time = (2 a) (10 s) = 20 c ti tdq dt d dt. Because i2 has a positive value, positive charge moves in the same direction as the reference. Thus, positive charge moves downward in element c. Because i3 has a negative value, positive charge moves in the opposite direction to the reference. Thus positive charge moves upward in element e. Energy = charge voltage = (2 c) (20 v) = 40 j. Because vab 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. iab enters terminal a. Thus the current enters the positive reference, and we have the passive reference configuration. (a) titv. 220 t tp a a a w a. J (b) notice that the references are opposite to the passive sign convention.
