W08 314 ex1 all-s

# W08 314 ex1 all-s - EECS 314 Winter 2008 Midterm exam 1 The...

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EECS 314 Winter 2008 Midterm exam 1 Instructor: Alexander Ganago The Key A B C D E 6 2 1 4 3 7 8 5 10 9 12 11 Problem # Correct answer 1 C 2 B 3 E 4 D 5 C 6 A 7 A 8 B 9 E 10 D 11 D 12 C

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EECS 314 Winter 2008 Midterm exam 1 © 2008 Alexander Ganago Problem 1 In the circuit shown on this diagram, the current source I S is variable. Determine the value of its current I S such that the 1-A current source absorbs zero power. A. Zero B. +2 A C. –2 A D. Not enough information E. None of the above Sample Solutions 1. Solution by node voltage Ground is chosen on the bottom node; V 1 is the voltage at the node connecting both current sources. V 1 = 0 V, to ensure zero power consumption. Thus the node voltage equation for V 1 is: V 1 " 6 V ( ) 2 # + 1 A ( ) " I S ( ) = 0 0 V ( ) " 6 V ( ) 2 # + 1 A ( ) " I S ( ) = 0 I S = " 2 A Answer: C 2. Solution by KCL, KVL, and Ohm’s law Due to KVL, the voltage at the node that connects both current sources is zero; thus the voltage drop across the 2- Ω resistor equals 6 V; from Ohm’s law, the current through this resistor must be 3 A, to the right. From KCL at the node that connects both current sources, 3 A comes from the left, 1 A goes down; thus 2 A must flow to the right. From KCL at the unknown current source, 2 A flows in the direction opposite to the arrow symbol thus I S = –2 A . Both solutions yield the same answer, as expected.
EECS 314 Winter 2008 Midterm exam 1 © 2008 Alexander Ganago Problem 2 In the circuit shown on this diagram, the resistances are in ohms. Terminals A and B are connected with a wire (zero resistance). Determine of the equivalent resistance in ohms between terminals C and D. A. 13.75 B. 17.50 C. 19.375 D. 21.25 E. 25.00 Sample Solution Redrawing the circuit is the key tool to successful solution of this problem. Here, we simply get rid of the unnecessary details such as the casing of the IC chip and terminals A and B, which are not used in the measurement of the equivalent resistance. First of all, note that the two 20- Ω resistors are connected in parallel. Replace them with a single 10- Ω resistor. Thus we can replace the 5- Ω and 10- Ω resistors in series with their equivalent, as shown here. Continued on the next page…

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EECS 314 Winter 2008 Midterm exam 1 © 2008 Alexander Ganago Problem 2 Sample Solution, continued The two 15- Ω resistors in parallel can be replaced with their equivalent: Now, the entire circuit can be reduced to a single resistor, as shown below. Answer: B
EECS 314 Winter 2008 Midterm exam 1 © 2008 Alexander Ganago Problem 3 When the circuit shown on this diagram is connected to the source at terminals a and b, the 1- Ω resistor absorbs 1 W. Determine the power absorbed by the 3-

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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 ex1 all-s - EECS 314 Winter 2008 Midterm exam 1 The...

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