ECE 201 – Fall 2008 Exam #1 September 22, 2008 Division 0101: Clark (7:30am) Division 0201: Elliott (10:30 pm) Division 0301: Capano (3:30 pm) Division 0401: Qi (4:30 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 1-6, 11-14 70 35 ii 7-10 28 14 If you fail to satisfy any of the course outcomes, don’t panic. There will be more opportunities for you to do so. 11. The current i(t) through an element is shown in the plot. Determine the total charge that has passed through this element in the interval between t = 0 and t = 4s. (1) 0 C (2) +1 C (3) -1 C (4) +2 C (5) -2 C (6) +3 C (7) -3 C 2. Consider a circuit element shown below with two nodes A and B. When a series of voltages VAB are applied, the current IAB is listed in the following table. What could this element be? IAB VAB B A Unknown element Applied voltage, VABMeasured current, IAB2 V 1 A 0 V 1 A –2 V 1 A (1) 2V voltage source (2) 2Ω resistor (3) – 2 Ω resistor (4) 1A current source (5) None of the above 23. The current flow through the 5Ω resistor, I, and the power absorbed by the 1A current source, P, are: (1) I = 2A, P = 10 W (2) I = 2A, P = -10 W (3) I = 1A, P = 10 W (4) I = 1A, P = -10 W (5) I = -1A, P = 10 W (6) I = -1A, P = -10 W 4. The equivalent resistance between nodes A & B, Req, is: (1) 5Ω (2) 6Ω (3) 10Ω (4) 15Ω (5) 25Ω (6) 30Ω (7) 45Ω 35. The voltage drop from node A to node B is: (1) 1V (2) 2V (3) 3V (4) 4V (5) 5V (6) 6V (7) 7V (8) 8V (9) 9V (10) 10V 6. Find the output voltage, Vout, for the circuit below (in V): (1) cannot be determined (2) 2 (3) 3 (4) 4 (5) 8 (6) 10 (7) 12 47. In the circuit below, find nodal voltage VB when nodal voltage VA equals 16 V. (1) 1V (2) 2V (3) 3V (4) 4V (5) 8V (6) 12V (7) 16V 8. The following circuit consists of a floating voltage source and a current source. Determine the node voltage . aV (1) (2) 0aV= V 1aVV= (3) 2aVV=− (4) 4=aVV (5) (6) 3=aVV V5aV= (7) 6aVV=− (8) 7aVV= 59. For the circuit shown with nodal voltages V1, V2 and V3, as labeled, which of the following equations is correct? (The node at which each equation might be derived is given in parentheses.) (1) 112VVV5V 010 15−+−ΩΩ= (node 1) (2) 131VVV5V010 40 10−++=ΩΩΩ(node 1) (3) 112VVV5V010 15 10−+−ΩΩΩ= (node 1) (4) 2321 2VVVV V015 50 25−−++ΩΩΩ= (node 2) (5) 2321VVVV015 25−−+ΩΩ= (node 2) (6) 2321 xVVVV V015 50 25−−−+ΩΩΩ= (node 2) (7) 3321VVVV020 25 10−++ΩΩΩ= (node 3 + 1) 610. For the circuit shown, with mesh currents I1, I2, and I3, as labeled, which one of the following equations is correct? (The mesh at which each equation might be derived is given in parentheses.) (1) (mesh 1) ()()112 135V I I I 25 I I 15 0+− Ω+− Ω= (2) (mesh 1) ()()112 135V I I I 25 I I 15 0−+− Ω+−Ω= (3) () (mesh 2) 21 2II25 I20 0.1A−Ω−Ω−=00000 (4) () (mesh 2) 21 2II25 I20 0.1A−Ω−Ω+= (5) () (mesh 2) ()21 2II25 I20−Ω− Ω= (6) () (mesh 2) ()21 2 AII25 I20 V−Ω− Ω−= (7) () (mesh 2) ()21 2 AII25 I20 V−Ω+ Ω−= 711. The constant g in the circuit below is 1/5 S. Determine the equivalent resistance Req of the dependent source – resistor combination in the circuit below. (1) 1.25Ω (2) 8Ω (3) 1Ω (4) 10Ω (5) 0.5Ω (6) 2Ω (7) 0.67Ω 12. Find the current that flows through the 24 ohm resistor I24 (in A): (1) 0 (2) 12/5 (3) 2 (4) 4 (5) 5/2 (6) 6/5 (7) 5 813. In the circuit shown, determine VR. (1) 5V (2) -5V (3) 4V (4) -4V (5) 1V (6) -1V (7) 0V 14. In the following circuit, the current of the VCCS (Voltage Controlled Current Source) depends on the voltage xV across the “diamond-shaped” Wheatstone Bridge. The transconductance of the VCCS is 6S . What is the gain out inVV of this amplifier? (1) 1=out inVV (2) 2=out inVV (3) 3=out inVV (4) 4=out inVV (5) 5=out inVV (6) 6=out inVV (7) 7=out inVV (8) 8=out inVV
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