ECE201 -- Exam 2 October 19, 2004 INSTRUCTIONS: ♦ There are eleven (11) multiple choice problems worth 9 points each. ♦ Students who properly identify themselves with name and id number on scantron sheet will receive one extra point. ♦ To maximize our assessment of your knowledge and understanding, do NOT dwell on a single problem. If you get stuck, move on to the next problem and return later, time permitting. ♦ This is a closed book, closed notes exam. All students are expected to abide by the usual ethical standards of the university, i.e., your answers must reflect only your own knowledge and reasoning ability. Students caught cheating will receive a grade of ‘F’ for the course. POSSIBLY USEFUL EQUATION: ∞∞o-(t-t )/τ+ox(t)=x()+[x(t)-x()]e1 1. For the op. amp circuit shown below, the output voltage vout (in V) is (1) −2 (2) 2 (3) 3 (4) 4 (5) 5 (6) −3 (7) − 4 * vout+−−+400Ω2 V+_200Ω100Ω300Ω2 2. In the above circuit, R3 =10 Ω. In order for the output Vout = −3V1 − 0.2V2, the resistance R1 and R2 should be chosen as (1) R1 = 10 Ω, R2 = 10/3 Ω (2) R1 = 30 Ω, R2 = 2 Ω (3) R1 = 30 Ω, R2 = 50 Ω (4) R1 = 10/3 Ω, R2 = 50 Ω * (5) R1 = 50/3 Ω, R2 = 2 Ω (6) R1 = 50 Ω, R2 = 30 Ω (7) none of the above V1 V2 Vout R2R3R13 3. Suppose that two experiments are performed on the above circuits, with results Vs Is Iout Experiment 1 2 V 3 A 2 A Experiment 2 3 V 5 A 3 A Then in the third experiment, if we choose Vs=1 V and Is=1 A, the output Iout will be (1) 1 V * (2) 2 V (3) 3 V (4) −1 V (5) 0 V (6) −2 V (7) none of the above Linear network with no independent sources4 4. The current I in the above circuit is (1) 1.6 A (2) −0.4 A (3) 1.4 A (4) 0 A (5) 0.2 A * (6) 0.4 A (7) none of the above5 5. The Norton equivalent circuit of the circuit shown above is +_2Ω2Ω2Vx2A+_abVxab(5)1.5 A8Ωab(6)1.5 A8Ωab(4)1.5 A1/8Ωab(3)1.5 A1/8Ωab(2)12 A8Ωab(1)12 A8Ω(7) none of the above6 6. The load resistance RL is chosen to maximize power delivered to the load. The maximum power, in W, delivered to the load is (1) 2.5 (2) 5.0 (3) 12.5 (4) 25 (5) 62.5 (6) 125 * (7) 250 7. In the circuit shown, the voltage source is vs(t) = 2 cos (t) V. Then vL(t) is (1) − 3 cos (t) V * (2) − 3 sin (t) V (3) 3 cos (t) V (4) 3 sin (t) V (5) − 12 sin (t) V (6) 12 sin (t) V (7) −12 cos (t) V 20A10ΩRL5Ω5 ΩLoadvL(t)−+2 F0.25 H+_vs(t)ix3 ix7 8. When simplified, the circuit below becomes (1) a series combination of 1 H inductance and 1 F capacitance * (2) a parallel combination of 1 H inductance and 1 F capacitance (3) a series combination 6/41 H inductance and 0.4 F capacitance (4) a parallel combination of 6/41 H inductance and 0.4 F capacitance (5) a series combination 41/6 H inductance and 2.5 F capacitance (6) a parallel combination of 41/6 H inductance and 2.5 F capacitance (7) none of the above 9. The switch in the RL circuit below has been at position A for a long time and it turns to position B at t = 0. For t > 0, the resistor voltage vR(t) (in V) is (1) −3 + 3e – 50t (2) 3 − 3e – 50t * (3) 3e – 50t (4) − 3 + 3e – 0.02 t (5) 3 − 3e – 0.02 t (6) 3e – 0.02 t (7) 0.2 − 0.2e – 0.02 t 15Ωt = 03 V+_AB0.3 H−+vR(t)1 H2 F1/6 H1 H1 H1F1F1 H1 H8 10. Suppose vC(0−)=2 V in the circuit shown above. Then the current iC(t) is (t>0) (1) 10−8e-t/8 A (2) 2e-t/6 A (3) 2e-t/8 A * (4) 4−2e-t/12 A (5) −2 e-t/6 A (6) e-t/8 A (7) 5−3e-t/6 A 11. In the above circuit the switch is open for a long time before it is closed at time t=0. Then the energy stored in the 4 H inductor at time t>0 is (in W) (1) 8e-2t (2) 0 (3) 8(4−3e-t)2 (4) 4−3e-t (5) 8(8−7e-t)2 (6) 8 * (7) none of the aboveAnswers to Test #2, Oct. 19, 2004 1. (7) 2. (1) 3. (5) 4. (5) 5. (6) 6. (1) 7. (1) 8. (6) 9. (2) 10. (3) 11.
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