PracticeExam1_solution - EE 310 Midterm 2 Sample 1...

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EE 310 Midterm 2 - Sample # 1 Instructions: This examination is closed book, closed notes. By signing this document, you are pledging to adhere to the honor system stating that all work performed here is strictly your own effort. Limited partial credit given for some problems. Check your answers carefully for arithmetic mistakes. Be sure to include appropriate units . Only one answer will be considered; more than one answer is automatically graded incorrect. Work on top of an erasure will be not be reconsidered for any reason after the exam is returned; circle and draw a line through work to be discarded. List all assumptions, show all work and clearly indicate your answer. Only answers placed where specified are considered. You have up to 50 minutes to take this examination. Good luck! Name __________________________________________________________ Signature ____________________________________________________ Problem 1: Basics of Power in Sinusoidal Circuits (Ch 11, Problem 2, hw 5, modified) Given the circuit in the figure on the right, find a) the average power supplied or absorbed by each element, b) the instantaneous power supplied by the source, assuming ω = 100 rad/s. Solution: a) This part is solved in the hw solution, but I don’t like it very much. As we saw in class today, i) the average power absorbed by the reactive elements (capacitor and inductor) is zero, because the voltage and current through them are out of phase by 90, so that cos θ = 0. For the resistor, by the current divider rule . 34 6 3 5 6 | | 3 5 6 5 4 1 ) 4 1 ( 2 2 2 = + = - - = + - - = res res I j j j j j j I Therefore . 5 . 2 17 45 5 34 36 2 1 | | 2 1 2 W R I P res res = = = This must also be the power supplied by the current source (by power conservation). b) From lecture 9, the instantaneous power ) 2 cos 2 1 ) ( I V m m res t ( I V P t p θ θ ω + + + = , where . 0 , 2 = = I m I θ Note this would simply change to ) 2 cos 2 1 ) ( I V m m res t ( I V P t p θ θ ω - - + = if we used sin() as reference instead of cos. The first of these 2 terms is already computed. Note that since the current
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