1 ECE 201 – Fall 2009 Exam #1 September 21, 2009 Division 0101: Tan (11:30am) Division 0201: Clark (7:30 am) Division 0301: Elliott (1:30 pm) Instructions 1. DO NOT START UNTIL TOLD TO DO SO. 2. Write your Name, division, professor, and student ID# (PUID) on your scantron sheet. 3. This is a CLOSED BOOKS and CLOSED NOTES exam. 4. There is only one correct answer to each question. 5. Calculators are allowed (but not necessary). 6. If extra paper is needed, use back of test pages. 7. Cheating will not be tolerated. Cheating in this exam will result in an F in the course. 8. If you cannot solve a question, be sure to look at the other ones and come back to it if time permits. 9. As described in the course syllabus, we must certify that every student who receives a passing grade in this course has satisfied each of the course outcomes. On this exam, you have the opportunity to satisfy outcomes i, ii, and iii. (See the course syllabus for a complete description of each outcome.) On the chart below, we list the criteria we use for determining whether you have satisfied these course outcomes. Course Outcome Exam Questions Total Points Possible Minimum Points required to satisfy course outcome i 1-8 56 28 ii 9-12 28 14 iii 13-14 14 7 If you fail to satisfy any of the course outcomes, don’t panic. There will be more opportunities for you to do so.2 1. The current i(t) conducted along a wire is shown in the plot. Choose the plot that best represents the charge q(t) that has passed along the wire during the interval from t=0 to t.3 2. The current i(t) through an element, and the power it absorbs, p(t), are as shown in the plots. Which of the following plots best represents the energy absorbed by the element in the interval from t=0 to t?4 3. Determine the equivalent resistance eqR. (1) 1Ω (2) 2 Ω (3) 3 Ω (4) 4 Ω (5) 5 Ω (6) 6 Ω (7) 7 Ω (8) 8 Ω 4. Determine the expression of ( )outVt. (1) 11234=+++out inRVVRRRR (2) 121234+=+++out inRRVVRRRR (3) 1234234+++=++out inRRRRVVRRR (4) 231234=+++out inRRVVRRRR (5) 234123++=++out inRRRVVRRR (6) 2341++=out inRRRVVR (7) 1234=++out inRVVRRR (8) 2341234++=+++out inRRRVVRRRR5 5. If 4W of power is absorbed by the 16 Ω resistor, then determine the amount of power delivered by the 12V voltage source. All that is known about the third rectangular element is that it absorbs power. (1) 1W (2) 2W (3) 3W (4) 4W (5) 5W (6) 6W (7) 7W (8) 8W 6. Determine the current xi. (1) 1A (2) 2 A (3) 3 A (4) 4 A (5) 5 A (6) 6 A (7) 7 A (8) 8 A6 7. Determine the value of the voltage xV across the 3A current source. (1) 0V (2) 4V (3) 8V− (4) 6V (5) 10V− (6) 5V (7) 4V− (8) 12V 8. Determine the current 2i passing through the 5V independent voltage source. (1) −1A (2) 2A (3) −3A (4) 4A (5) −5A (6) 6A (7) −7A (8) 8A7 9. The voltage at node A in the circuit below is: (1) 1 V (2) 2 V (3) 3 V (4) 4 V (5) 5 V (6) 6 V (7) 7 V 10. The equivalent resistance Req of the network (consisting of the dependent current source and the resistor) shown below is: (1) 1 Ω (2) 2 Ω (3) 3 Ω (4) 4 Ω (5) 5 Ω (6) 6 Ω (7) 7 Ω8 11. The matrix form of the loop equations of the circuit shown below is as follows. 1o2I30 10 VI21 0 = − What is the value of RL? (1) 10 Ω (2) 20 Ω (3) 30 Ω (4) 40 Ω (5) 50 Ω (6) 60 Ω (7) 0 Ω 12. The power absorbed by the dependent source in the circuit shown below is: (1) 0 W (2) −8 W (3) −3.2 W (4) −2 W (5) −1.6 W (6) −1.28 W (7) 1.28 W9 13. In the figure below, the rectangular box represents a linear network containing no independent sources. The table to the right gives the value of Vout measured for various values of Is and Vs. Determine the voltage sV′ that, when Is = 2A, corresponds to Vout = 10V. Is Vs Vout 1A 2V 5V 0A 5V 20V 2A sV′ 10V (1) −6V (2) −4V (3) −2V (4) 0V (5) +2V (6) +4V (7) +6V 14. Using superposition, determine the current I1 in the following network. (1) -3A (2) -2A (3) -1A (4) 0 (5) +1A (6) +2A (7)
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