F08 EECS314 EXAM1 sol - EECS 314 Fall 2008 Exam 1 A 10 13...

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EECS 314 Fall 2008 Exam 1 Instructor: Alexander Ganago The Key A B C D E 10 4 1 9 2 13 7 3 12 5 15 8 6 14 11 Problem # Correct answer 1 C 2 E 3 C 4 B 5 E 6 C 7 B 8 B 9 D 10 A 11 E 12 D 13 A 14 D 15 A
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EECS 314 Fall 2008 Exam 1 Instructor: Alexander Ganago Question 1 In the circuit shown on the diagram, the 1 A source absorbs 10 W. Thus the voltage source absorbs … A. 57 W B. 30 W C. 3 W D. Cannot be found because the source voltage is not given E. None of the above. Solution From the given power and the Passive Sign Convention, V 1 = +10 V By KCL, I 3 = 3 A Thus V 3 = 9 V From KVL around the left mesh, " V S " 9 V ( ) + 10 V ( ) = 0 thus V S = + 1 V From the Passive Sign Convention, the voltage source absorbs V S " I 3 = 1 V ( ) " 3 A ( ) = 3 W Answer: C Comment From KVL around the right mesh, " V 1 " 20 V ( ) " V 4 = 0 thus V 4 = " 30 V and the 4 A source absorbs –120 W, or supplies 120 W; the power balance is fine.
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EECS 314 Fall 2008 Exam 1 Instructor: Alexander Ganago Question 2 In the circuit shown on the diagram, the 8 Ω resistor absorbs 8 W. Thus all resistors combined absorb… A. 12 W B. 32 W C. 52 W D. 104 W E. 140 W Solution: Since the power absorbed by a resistor P R = I R 2 " R is determined by the current through the resistor and the resistance, for 8 Ω we obtain P 8 = I 8 2 " (8 # ) = 8 W thus I 8 = 1 A Due to parallel connection, the voltage drop across 2 Ω resistor must be equal to that across 8 Ω , which means that the current I 2 = 4 A thus P 2 = I 2 2 " (2 # ) = 4 2 " 2 = 32 W By KCL, the current through 4 Ω resistor equals I 4 = I 2 + I 8 = 5 A thus P 4 = I 4 2 " (4 # ) = 5 2 " 4 = 100 W The total current equals 140 W Answer: E
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EECS 314 Fall 2008 Exam 1 Instructor: Alexander Ganago Question 3 In the circuit shown on the diagram, terminals A and B are connected with a wire (zero resistance); other resistances are in ohms. The equivalent resistance between terminals D and C equals (in ohms): A. 5.714 B. 7.5 C. 10 D. 13.103 E. 20 Solution: see the diagrams Answer: C
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EECS 314 Fall 2008 Exam 1 Instructor: Alexander Ganago Question 4 The circuit on this diagram is used to measure the amount of liquid in a cylindrical tank. The cross-section area of the tank equals S = 300 cm 2 ; the length of the linear potentiometer (from A to B) equals L = 50 cm; the resistance is linearly proportional to the displacement of the tap; the middle position of the potentiometer’s tap corresponds to the liquid level H = 80 cm, the source voltage equals V S = 25 V. Neglect the volume of the float. Calculate in liters ( l ) the volume of liquid in the tank at V OUT = 10 V. Recall that 1 l = 1,000 cm 3 . A. 16.5 l B. 22.5 l C. 27.5 l D. Not enough information E. None of the above.
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This note was uploaded on 10/20/2009 for the course EECS EECS314 taught by Professor Ganago during the Spring '09 term at University of Michigan.

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F08 EECS314 EXAM1 sol - EECS 314 Fall 2008 Exam 1 A 10 13...

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