ECE 201 – Spring 2008 Exam #1 February 5, 2008 Division 0101: Qi (7:30am) Division 0201: Elliott (12:30 pm) Division 0301: Capano (2:30 pm) Division 0401: Jung (3:30 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 1-10 70 35 ii 11,12 14 7 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. 11. Fig. 1(a) illustrates the charge flow in a conductor. Assume you are sitting on the right side of the cross-section boundary and are recording the total amount of net charges that have passed across the boundary. Your measurement is shown in Fig. 1(b) for time 0<t<4 seconds. q(t) (Coulombs) 21 Assume the reference current direction is shown in Fig. 1(a). Which of the following figures shows the correct current flow? i(t) (Amperes) 1 -1 2 1) 2) 3) 4) 5) 6) 1 2 3 1 4 t (sec) i(t) (Amperes) 1-1 212314 t (sec) i(t) (Amperes) 1 -1 2 1 2 3 1 4 t (sec) 12i(t) (Amperes) 1 -1 2 1 2 3 1 4 t (sec) 1t (sec) 123Fig. 1(a) Fig. 1(b) 4 i(t) (Amperes) 21-1 134 t (sec) i(t) (Amperes) 1-1 22314 t (sec) 22. An unknown two-lead element, as shown, has a current I = -5mA when the voltage V is +10V. Based on this limited information, which of the following statements is correct? (1) This element might be a resistor, of resistance R = 2 kΩ. (2) This element might be an ideal 10V voltage source. (3) This element cannot be a current source. (4) This element might be an independent source, but cannot be a dependent source. 3. What is the resistance of the resistor, R, and the power absorbed by the resistor, P? I= 0.5 A - + V= – 4 V 1) R=8 Ω , P= -2 W 2) R=2 Ω, P=8W 3) R=8 Ω , P= 2W 4) R=2 Ω, P=- 8W 5) R= – 8 Ω , P= -2 W 6) R= – 2 Ω , P= -2 W 7) R= – 8 Ω , P= 2 W 8) R= – 2 Ω , P= 2 W 34. In the network shown below, find the current I. (1) 1A (2) -1A (3) 2A (4) -2A (5) 3A (6) -3A (7) 0A 1A 5A I 3A 1A3A 5. In the following circuit determine Vx. (1) -11V (2) -5V (3) -1V (4) 0V (5) +1V (6) +5V (7) 11V 46. The equivalent resistances seen between the inputs of circuits 1and 2 are REQ1 and REQ2, respectively. The two circuits are connected as shown below. Find the equivalent resistance, REQ, seen between the inputs. REQ1Circuit 1 REQ2Circuit 2REQ Circuit 2 Circuit 2Circuit 1 Circuit 1 (1) 2⋅REQ1+REQ2/2 (2) REQ1/2+2⋅REQ2 (3) (REQ1+REQ2)⋅2 (4) (REQ1+REQ2)/2 (5) REQ1+REQ2 (6) 2⋅REQ1+REQ2 (7) REQ1+2⋅REQ2 57. Determine the voltage vx in the following circuit. (1) 2.0V (2) 3.0V (3) 4.5V (4) 6.0V (5) 6.67V (6) 9.0V (7) 13.5V 8. All resistors in the circuit shown below have the same resistance of 8 Ω. Find the output voltage, VOUT. (1) 1V (2) 3V (3) 5V (4) 7V (5) 8V (6) 9V (7) 11V (8) 13V 15V VOUT 69. Find the current I, in the circuit shown below. (1) 1A (2) 2A (3) 3A (4) 4A (5) 5A (6) 6A (7) 7A 10. In order to calculate the equivalent resistance of the network shown to the left below, we can add a 1A current source and calculate the voltage drop across the current source, as we show on the right side. With this hint, determine the equivalent resistance of the network on the left. 1) -4 Ω 2) -3 Ω 3) -2 Ω 4) -1 Ω 5) 1 Ω 6) 2 Ω 7) 3 Ω 8) 4 Ω + - V1 – + 1 A Req 2 Ω + - V1 – + 2 Ω 3 V1 3 V1 7A R R/2 I R/4 R 711. For the following circuit, select the correct equation relating V1, V2, and/or ix. (1) xi2= A (2) ()12xxVV2A i 2i 01−−++ + =Ω (3) ()x12 x2A i V V 1 2i 0++ − Ω+ = (4) ()12xVV2A i 2i 01−++ +=Ω (5) () ()x21 22i V V 1 V 4 0−+ − Ω+ Ω= (6) 21 2VVV014−+=ΩΩ 812. In the circuit shown, which one of the following equations relating to currents IA, IB, IC and voltage Vd is correct? (1) ()()AAB AC4A 4 I 1 I I 2 I I 0+Ω +Ω − +Ω − = (2) A1I4A=−I00 (3) () AB B B dII1I3I70.3V−Ω+Ω+Ω− = (4) () ()CA BA B B CI I 2 I I 1 I3 I4 I2 0− Ω+ − Ω+ Ω+ Ω+ Ω= (5) ()d0.3V 4A 2=Ω (6) ()CA CII2I2−Ω+Ω= 913. Use superposition, or any other methods that you feel appropriate, in the circuit below to find the voltage V1 (in volts): (1) 8 (2) 12 (3) 20 (4) 24 (5) 36 (6) 60 (7) 100 1014. Using Source Transformations, simplify the following network to the following form: What are Vs and R in this simplified equivalent network? VsR 1) 4.0V 20Ω 2) 7.0V 20Ω 3) 10.0V 20Ω 4) 4.0V 40Ω 5) 7.0V 40Ω 6) 10.0V 40Ω 7) 4.0V 80Ω 8) 7.0V 80Ω 9) 10.0V 80Ω
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