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F07%20314%20ex2%20all-s

# F07%20314%20ex2%20all-s - EECS 314 Fall 2007 Exam 2 Sample...

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Alexander Ganago The Key A B C D E 2 5 4 3 1 10 8 7 6 9 20 12 13 11 16 18 17 14 19 15 Problem # Correct answer 1 E 2 A 3 D 4 C 5 B 6 D 7 C 8 B 9 E 10 A 11 D 12 B 13 C 14 D 15 D 16 E 17 C 18 B 19 B 20 A EECS 314 Fall 2007 Exam 2 Sample solutions

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Instructor: Alexander Ganago Problem 1 In the circuit shown on this diagram, under DC steady- state conditions, the energy stored in the 200-mF capacitor equals … A. Zero B. 60 J C. 311 J D. 720 J E. None of the above. Solution Under DC steady-state conditions, the capacitors act as open circuits thus there is no current through the 40-ohm resistor. The current in the circuit equals 480 V (4 + 10 + 2) " = 30 A Thus the voltage drop across the 2-ohm resistor equals 60 V, and the energy stored in the 200-mF capacitor equals 1 2 " 60 V ( ) 2 " (200 mF ) = 360 J . Answer: E
Instructor: Alexander Ganago Problem 2 (submitted by Barry Mullins, modified) In the circuit shown on this diagram, what is the energy stored in the 2-mH inductor under DC, steady-state conditions? A. 250 μJ B. 125 μJ C. 500 μJ D. 27.78 μJ E. 0 μJ Solution by Barry Mullins, modified Under DC, steady state, the inductors act as shorts, and the capacitor acts as an open. We can therefore redraw the circuit to reflect this. Therefore, the current through the resistor is 20V/40 Ω = 0.5A. This current is the same current that flows through the 2 mH inductor, so we can calculate the energy stored. P 2 mH = 1 2 L " I 2 = 1 2 (0.002 H ) " (0.5 A ) 2 = 250 μ J The answer is A

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Instructor: Alexander Ganago Problem 3 In the circuit shown on this diagram, the equivalent capacitance C ad between the terminals a and d (when nothing is connected to terminals b and c ) equals … A. 3 " C B. 4 9 " C C. 1 2 " C D. 1 3 " C E. C Solution Using the rules for calculation of the effective capacitance for capacitors connected in series and in parallel, reduce the circuit step-by-step as shown on the diagrams, and obtain: Answer: D
Instructor: Alexander Ganago Problem 4 In the circuit shown on this diagram, terminals b and c are connected with a wire (short circuit). The equivalent inductance between terminals a and d equals… A. 10 mH B. 20 mH C. 30 mH D. 25 mH E. None of the above. Solution The circuit can be redrawn: Evidently, no current ever flows through the 40-mH inductor thus it can be deleted from the equivalent diagram. The circuit reduces to this one, whose effective inductance equals: 10 mH + (20 mH ) " (20 mH ) (20 mH ) + (20 mH ) + 10 mH = = 10 mH + 10 mH + 10 mH = 30 mH Answer: C

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Instructor: Alexander Ganago Problem 5 In the circuit shown on this diagram, the input voltage is shown on the sketch. Assume DC steady-state conditions at time 0- and determine the output voltage at time = 1 msec. The nominal values are: R = 2 k Ω C = 1 μ F V OUT (t = 1 msec) equals … A. 9 V B. 8 V C. 7 V D. 6 V E. 5 V (Choose the nearest value) Solution The output voltage is measured across the capacitor; at time 0 < t 1 msec, it equals V OUT ( t ) = K 1 " e # t \$ + K 2 , where " = R # C = (2 k \$ ) # (1 μ F ) = 2 m sec .
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F07%20314%20ex2%20all-s - EECS 314 Fall 2007 Exam 2 Sample...

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