F08 EECS314 EXAM1 sol

# 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 … 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 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.
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… 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

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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):
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. Solution Due to the linear dependence of the resistance on the position of the tap, displacement of the tap by Δ h causes the voltage change of " V = " h # " V " h \$ % & ( ) = " h # V S L \$ % & ( ) = " h # 25 V 50 cm \$ % & ( ) = " h # 0.5 V cm \$ % & ( ) The middle position of the potentiometer corresponds to the output voltage of 25:2 = 12.5 V. Thus, at the output voltage of 10 V, the tap is below H by 5 cm, or the liquid level equals 75 cm. The liquid volume equals (75 cm) (300 cm 2 ) = 22.5 l. Answer: B

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• Spring '09
• Ganago
• Electrical resistance, Thévenin's theorem, Voltage source, Norton's theorem, Voltage drop, Alexander Ganago

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