EE 201 Final Exam December 17, 1997General Instructions:• The exam is closed book, closed notes.• Do not open the exam until you are told to begin.• Fill in your name, student identification number, and section number in the appropriateplaces on the computer scan forms.Time Instructor Section Number7:30 am DeCarlo 00019:30 am Krogmeier 000212:30 pm Krogmeier 00033:30 pm Chen 0004• The exam consists of 30 equally weighted multiple choice questions.• Please keep your computer scan sheets covered while you are working on the exam.1Problem 1. For the circuit shown below, Vsis:(1) 60 V (2) 40 V (3) 20 V(4) -46 V (5) -20 V (6) -40 V(7) 100 V+--+--2 A5 Ω20 V3 Ω8 AVsProblem 2. The circuit below contains a floating voltage source. It is convenient toincorporate the supernode concept in setting up the nodal equation. Pick the correct nodalequation involving the supernode.(1) 2V1+7V2=80 (2) 2V1+7V2=50 (3) 3V1− V2=40 (4) 3V1+ V2=40(5) 3V1+ V2=10 (6) 7V1− 2V2=70 (7) 7V1− 2V2=40+--+--+--1/2 Ω 1/4 Ω1/3 Ω5 V2 V10 V30 Areference nodeV1V22Problem 3. Consider the circuit below with voltage sources Vs1and Vs2andmeshcurrentsI1and I2. Find the correct mesh equations for the circuit.(1)(6I1− 4I2= Vs14I1− 7I2= Vs2(2)(6I1− 3I2= Vs1−2I1+7I2= Vs2(3)(6I1+4I2= Vs14I1+7I2= Vs2(4)(6I1− 3I2= Vs12I1+7I2= Vs2(5)(9I1− 4I2= Vs1−4I1+9I2= Vs2(6)(9I1+4I2= −Vs14I1+9I2= Vs2(7)(6I1− 4I2= Vs14I1− 7I2= −Vs2+--+--2 Ω 3 Ω4 ΩVs1Vs2I1I23Problem 4. Compute Vafor the circuit drawn below. Answers (in Volts):(1) 1 (2) 2 (3) 3 (4) 0(5) -3 (6) -2 (7) -1+--+--+--+--3 Ω3 VVa2 Ω4 Ω2 Ω1 Ω2 V1 VProblem 5. In the circuit below, the current source is given byis(t)=2+5u(t)A.Find iC(0+) in Amps.(1) 1 (2) 2 (3) 3(4) 4 (5) 5 (6) 6(7) 7i (t)Ci (t)s15 Ω0.3 F4Problem 6. Find the output Voof the ideal operational amplifier below.(1) 1V (2) 2V (3) 3V(4) 4V (5) 5V (6) 6V(7) 7V+--+--+--ideal+--2 kΩ2 kΩ1 kΩ2 V- 6 VVoProblem 7. Find the differential equation for vC(t).(1)dvc(t)dt+2vC(t)=vs(t) − 0.5is(t) (2)dvc(t)dt+2vC(t)=vs(t)+0.5is(t)(3)dvc(t)dt+ vC(t)=vs(t)+0.5is(t) (4)dvc(t)dt+ vC(t)=vs(t) − 0.5is(t)(5)dvc(t)dt− vC(t)=vs(t) − 0.5is(t) (6)dvc(t)dt− 2vC(t)=vs(t)+0.5is(t)(7)dvc(t)dt+2vC(t)=−vs(t)+0.5is(t)+--+--2 F0.5 Ωv (t)si (t)sv (t)C5Problem 8. Choose the correct Norton equivalent circuit for the network shown below.+--+--8 Ω 8 Ω4 Ω2 V 2 V2 Aab4 Ω2.5 Aab(1)4 Ω2 Aab(3)20 Ω2 Aab(5)2 Ω1.5 Aab(2)2 Ω2.5 Aab(4)2 Ω2 Aab(6)8 Ω1.5 Aab(7)6Problem 9. In the circuit below, a capacitor of value C7> 0 is connected across terminalsA–B. The new equivalent capacitance Cneweqsatisfies:(1) Cneweq= Ceq(2) Cneweq>Ceq(3) Cneweq<Ceq(4) Cneweq= C7Ceq/(C7+ Ceq) (5) Cneweq= C7Ceq(6) Cneweq= Ceq/C7(7) none of theseC1C2C3C4C5C6CeqABProblem 10. If vin(t) is given by the graph to the left of the circuit below, the value ofvout(t)att = 2 seconds assuming vC(0) = 0 is:(1) 1V (2) 2V (3) 3V (4) 4V(5) -4 V (6) -2 V (7) 8V++----+--+--ideal2.5 kΩ0.1 mFv (t)outv (t)Cinv (t)inv (t)t-2 V1 sec.2 sec.7Problem 11. In the RC circuit below, vout(∞)is:(1) 1V (2) 2V (3) 3V (4) 4V(5) 8V (6) 12 V (7) 15 V++----+--+--ideal1 u(t) V2.5 Ω7.5 Ω0.1 FvCvoutProblem 12. The voltage Vxin the circuit below is of the formVx= AIs+ BVs.Find A and B.(1) A =2R, B = −0.25 (2) A =0.5, B = R (3) A = −2R, B =0.5(4) A = −R, B =0.25 (5) A =0.5R, B = −0.25 (6) A =1,B =1(7) A = −0.5R, B = −0.25+--+--R2RIsVsVxR8Problem 13. In the circuit below, the switch has been closed for a long time when it isopened at t =0. Fort ≥ 0, find the inductor voltage vL(t) in Volts.(1)3224e−8t/3(2) −3224e−8t/3(3) 16e−4t(4) −16e−4t(5) 2e−t/4(6) −2e−t/4(7) none of these+--+--32 V16 Ω8 Ω2 Ht = 0v (t)LProblem 14. Find the time constant of the circuit shown below.(1) 0.1 s (2) 0.2 s (3) 0.3 s(4) 0.4 s (5) 0.5 s (6) 0.6 s(7) 0.7 s+--+--0.01 F10 Ωv14 v19Problem 15. The RL circuit shown on the left has the inductor current shown in the graphon the right. The value of L is:(1) 0.08 H (2) 0.10 H (3) 0.15 H (4) 0.20 H(5) 0.30 H (6) 0.60 H (7) 0.03 H50 ΩLiL0 1 2 3 4 5 612345678910Response of Undriven RL CircuitTime in milli−secondsInductor Current in Amps0.751.510 e-210 e-110 e-0.510 e-0.25Problem 16. In the circuit shown below, the source voltage is given byvs(t) = 20 + 10u(t)V.Find the current iL(t)fort ≥ 0 (in Amps).(1) 7.5 − 2.5e−2t(2) 7.5 − 2.5e−0.5t(3) 3 − e−2t(4) 5e−0.5t(5) 5e−2t(6) 6 − 2e−2t(7) 6 − 2e−0.5t+--3 Ω2 Ω2.5 Hv (t)si (t)L10Problem 17. If vC(0−) = 0 V, find the value of R in Ohms which will producevC(0.1) = 10(1 − e−1)V.(1) 1 (2) 2 (3) 10 (4) 20(5) 50 (6) 100 (7) 200+--+--10 u(t) VR0.1 F v (t)CProblem 18. For what range of values of R (in Ohms) is the circuit shown below under-damped?(1) 0 ≤ R ≤ 5 (2) 0 ≤ R<2 (3) 2 <R≤∞ (4) R =3(5) 5 <R≤∞ (6) 0 ≤ R<10 (7) 10 <R≤∞+--15 u(t) V10 ΩR0.01 H0.01 F11Problem 19. In the circuit below, the switch has been open for a long time when it closesat t =0. FindvC(t)fort ≥ 0 (in Volts).(1) 6 − 6(1 + t)e−2t(2) 6+(−6 cos 8t +8sin8t)e−2t(3) 6 − 6(1 + t)e−8t(4) 6+(−8 cos 8t +6sin8t)e−2t(5) 6+2e−8t− 8e−2t(6) 8e−8t− 8e−2t(7) none of these+--+--6 V0.25 H0.25 F2.5 Ωt = 0vCProblem 20. Represented below is a resistive circuit with dependent sources that satisfiesthe linear relationshipiout=0.5vs1− 0.5is2.If vs1(t) = 2 cos(10t) V and is2(t)=2sin(10t) A, choose the correct expression for iout(t)(inAmps).(1) 0 (2)√2 cos(10t − 45◦)(3)√2 cos(10t +45◦) (4) 2sin(10t +45◦)(5) 2 cos(10t − 45◦) (6) cos(10t − 45◦)(7) cos(10t +45◦)+--resistive circuitwith dependentsourcesvs1is2iout12Problem 21. A capacitor has admittance YC= j0.34 Mhos at frequency ω = 100π rad/sec.Find the value of the capacitance in mF.(1) 7.7 mF (2) 4.2 mF (3) 301.8 mF (4) 9.4 mF(5) 1.1 mF (6) 594.4 mF (7) none of theseProblem 22. For the phasor circuit shown below, find the Thevenin equivalent impedanceZth(j1) in Ohms.(1) 1 (2) 1/(1 + j) (3) 1/(1 − j) (4) 1+j(5) 1/(1 − j2) (6) 0 (7) −j2+--1 Ω- j1 Ω1 I1I1Z (j1)th.13Problem 23. In the circuit below, if vs(t) = 10 cos(25t) V, the sinusoidal steady statecapacitor voltage vC(t) is (in Volts):(1) 0 (2) 2 cos(25t +90◦) (3) 2sin(25t)(4) 0.2 cos(25t +90◦) (5) 0.2sin(25t) (6) 0.1857 cos(25t − 111.8◦)(7) none of these+--+--10 Ω 0.08 H0.02 F v (t)Cv (t)sProblem 24. In the phasor circuit below, find the
View Full Document
Unlocking...