W08 314 ex2 all-s - EECS 314 Winter 2008 Midterm exam 2 The...

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EECS 314 Winter 2008 Midterm exam 2 Instructor: Alexander Ganago The Key A B C D E 1 3 6 2 4 5 10 12 11 8 7 9 Problem # Correct answer 1 A 2 D 3 B 4 E 5 A 6 C 7 A 8 E 9 E 10 B 11 D 12 C
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EECS 314 Winter 2008 Midterm exam 2 © 2008 Alexander Ganago Problem 1 In the circuit shown on this diagram, the resonant frequency equals " RES = 10,000 rad /sec At the resonant frequency, the ratio of magnitudes of the capacitor voltage and the resistor voltage V C V R is equal to… A. 10 B. 100 C. 1,000 D. Not enough information E. None of the above. Sample Solution The resonant frequency of a series RLC circuit shown on the diagram equals RES = 1 L # C substitute the given values and obtain: 1 10 " 3 # C = 10 8 thus C = 10 " 5 F The magnitude of capacitor’s impedance in ohms equals, at resonance: Z C = 1 C = 1 10 4 # 10 $ 5 = 10 1 = 10 % Since the same current flows through all circuit elements, the ratio of the magnitudes of voltages equals the ratio of the magnitudes of impedances V C V R = Z C R = 10 " 1 " = 10 Answer: A
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EECS 314 Winter 2008 Midterm exam 2 © 2008 Alexander Ganago Problem 2 In the circuit shown on this diagram, assume DC steady- state conditions at t = 0- At time t = 0, the input voltage V IN abruptly changes from 10 V to 4 V. At t = 1 ms, the input voltage abruptly changes from 4 V to 10 V, as shown on the first sketch. On the second sketch, the capacitor voltage V C is shown with the solid line while the input voltage V IN is shown with the dotted line (for reference). Evidently, V C has not reached a new DC steady state by t = 1 ms. Indeed, V C t = 1 ms ( ) = 5 V Determine the value of the output voltage at t = 1 ms ( ) + right after the input voltage changed to 10 V. A. –6 V B. –5 V C. +6 V D. +5 V E. None of the above.
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EECS 314 Winter 2008 Midterm exam 2 © 2008 Alexander Ganago Problem 2 Sample Solution KVL, valid at any time: " V IN + V C + V OUT = 0 . Due to the continuity demand, V C t = 1 ms ( ) = 5 V is the same before and after the abrupt change of V IN . Thus at t = 1 ms ( ) + , we obtain from KVL: " V IN + V C + V OUT = 0 " 10 V ( ) + 5 V ( ) + V OUT = 0 V OUT = 5 V Answer: D
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EECS 314 Winter 2008 Midterm exam 2 © 2008 Alexander Ganago Problem 3 In the circuit shown on this diagram, terminals b and c are connected with a wire that has zero resistance and zero capacitance. Determine the equivalent capacitance C ad between terminals a and d . A. 1500 nF B. 150 nF C. 138.5 nF D. 635.7 nF E. None of the above Sample Solution Note that the circuit above can be replaced with this one: Then, it becomes evident which capacitors are connected in parallel: Replace parallel combinations with their equivalents and obtain: Replace the series combination with their equivalent and obtain: C ad = 150 nF Answer: B
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EECS 314 Winter 2008 Midterm exam 2 © 2008 Alexander Ganago Problem 4 In the circuit shown on this diagram, you are required to obtain the transfer function magnitudes H ( " ) = V OUT
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This homework help was uploaded on 04/10/2008 for the course EECS 314 taught by Professor Ganago during the Winter '07 term at University of Michigan.

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W08 314 ex2 all-s - EECS 314 Winter 2008 Midterm exam 2 The...

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