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 x t x x to x e t to 1 For the op amp circuit shown below the output voltage vout in V is 1 2 2 2 3 3 5 5 6 3 7 4 4 4 200 100 2 V 300 400 vout 1 R1 V1 R3 Vout R2 V2 2 In the above circuit R3 10 In order for the output Vout 3V1 0 2V2 the resistance R1 and R2 should be chosen as 1 R1 10 R2 10 3 2 R1 30 R2 2 3 R1 30 R2 50 4 R1 10 3 R2 50 5 R1 50 3 R2 2 6 R1 50 R2 30 7 none of the above 2 Linear network with no independent sources 3 Suppose that two experiments are performed on the above circuits with results Experiment 1 Experiment 2 Vs 2V 3V Is 3A 5A Iout 2A 3A 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 3 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 above 4 2Vx 2 a Vx 2 2A b 5 The Norton equivalent circuit of the circuit shown above is 1 12 A 2 a 8 12 A a 8 b b 3 1 5 A 4 a 1 8 1 5 A a 1 8 b 5 1 5 A b 6 a 8 1 5 A a 8 b b 7 none of the above 5 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 4 25 7 250 2 5 0 5 62 5 3 12 5 6 125 5 5 RL 10 20A Load 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 ix vs t 2F 0 25 H 3 ix 6 vL t 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 1F 1H 1H 1F 1H 1H 1H 1 6 H 2F 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 B t 0 0 3 H 3V A 15 vR t 7 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 above 8 Answers 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 6
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