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Unformatted text preview: EECS 314 Fall 2008 Exam 2 Instructor: Alexander Ganago The Key A B C D E 4 2 3 5 1 9 8 7 6 10 14 15 13 12 11 Problem # Correct answer 1 E 2 B 3 C 4 A 5 D 6 D 7 C 8 B 9 A 10 E 11 E 12 D 13 C 14 A 15 B EECS 314 Fall 2008 Exam 2 Instructor: Alexander Ganago Question 1 In the circuit shown on the diagram, an EECS 314 student observed unwanted oscillations. Furthermore, the student’s calculations showed that in this circuit " = # 10 . Which of the following measures can help the student to get rid of oscillations: A. Increase L by a factor of 20; keep R and C unchanged B. Increase R by a factor of 20; keep L and C unchanged C. Increase C by a factor of 20; keep R and L unchanged D. Increase L by a factor of 20 and increase C by a factor of 20; keep R unchanged E. None of the above helps get rid of oscillations. Solution In a parallel RLC circuit, " = 1 2 # RC and " = 1 L # C . To get rid of oscillations, achieve " > # , that is L R ¡ " C > 4 . Note that initially, at " = # 10 , the value of L R 2 " C = 1 25 . The goal is to increase it to 4. None of the suggestions A – D achieves this goal. Even the most promising A increases L R 2 " C from 0.04 to only 0.8, which still corresponds to underdamped response, with oscillations. Answer: E. Solutions EECS 314 Fall 2008 Exam 2 Instructor: Alexander Ganago Question 2 In the circuit shown on the diagram, use the offset diode model, although V D0 is not given. When the source voltage equals 3 V, the source supplies 60 mW. When the source voltage increases to 12 V, the diode absorbs… A. 80 mW B. 110 mW C. 140 mW D. Cannot be found because V D0 is not given E. None of the above. Solution The 3 V source supplies 60 mW, thus the current through the circuit equals 60 mW 3 V = 20 mA . From Ohm’s law, the voltage across 100 Ω resistor equals 2 V, thus V D0 = 1 V. When the source voltage equals 12 V, the current through the circuit equals I = V S " V D R = 12 V ( ) " 1 V ( ) 100 # = 110 mA , thus the diode absorbs I " V D = 110 mA ( ) " 1 V ( ) = 110 mW . Answer: B 100 Ω Solutions EECS 314 Fall 2008 Exam 2 Instructor: Alexander Ganago Question 3 (Submitted by James Kim; modified) In the circuit shown on the diagram, what is the nominal value of capacitance C that ensures the total capacitance equal 2 F ? A. 9 F B. 6 F C. 3 F D. 1 F E. No solution is possible, because any capacitance C produces C TOTAL > 2 F. Solution C 2 F ( ) 4 F ( ) = 6 F ( in parallel , add ) C TOTAL = C " 6 F ( ) C + 6 F ( ) = 2 F C " 6 F ( ) = C " 2 F ( ) + 12 F 2 ( ) 4 " C = 12 F C = 3 F Answer: C Solutions EECS 314 Fall 2008 Exam 2 Instructor: Alexander Ganago Question 4 (Contributed by Philip Choi) In the circuit shown on the diagram, at what time will the resistor voltage equal v R t ( ) = v C t = ( ) 2 A. 1.386ms B. 0.6021ms C....
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 Spring '09
 Ganago
 Rectifier, Alexander Ganago

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