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exam2_review_session - EE-201 Review Problems Test 2 page-1...

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Unformatted text preview: EE-201 Review Problems Test 2 page-1 Fall 98 EE-201 REVIEW PROBLEMS TEST 2 (AND THEREABOUTS) 1. Consider the circuit below. in a Thevenin equivalent of the circuit at the terminals A-B, the open circuit voltage is (in V): (1) 24 (2) 8(3) 4(4) -4 (5) 0(6) -4 IOOQ 2. For the circuit of problem 1, the Thevenin equivalent resistance is RTH = (in Q): (1) 200 (2) 400 (3) 150 (4) 137.5 (5) 100 (6) 50 (7) 250 EE-201 Review Problems Test 2 page-3 Fall 98 5. in the circuit below, the value of RL which leads to‘ maximum power transfer to the load (BL) is (in Q): (1) 0.4 (2) 2 (3) 2.5 (4) 1 (5) 44 (6) cannot be determined (7) none of above 6. in the circuit of problem 5, suppose RL = 59. The value of VL is (in V): (1)1 (2) 2 (3) 2.5 (4) 3.5 (5) 5.5 (6) 7 (7) 7.5 EE-201 Review Problems Test 2 page-4 Fall 98 7. For the circuit below, Leq = (in H): (1) 0.3837 (2) 2 (3) 2.606 (4) 3.343 (5) 6 (6) 8.029 (7) none of above V 8. For the circuit below, Ceq=(in F): (1) 3 (2) 0.733 (3) 1.25 (4) 7 (5) 1.34 (6) 1.714 (7) 3.5 1.517 217 C A 1F EE-201 Review Problems Test 2 page—5 Fall 98 For problems 9 and 10, consider the linear resistive circuit below which contains only resistors and dependent sources. lf is1(t) = 2cos(2t) and V320) = 0, then vout(t) = 14cos(2t). lf v52(t) = locos(2t) and is1(t) = 0, then vout(t) = -5003(2t). 9. if is1(t) = 200$(2t) and v52(t) = -10cos(2t), then vout(t) = (in V): (1) 9cos(2t) (2) 14cos(2t) (3) 19cos(2t) (4) 24cos(2t) (5) 4cos(2t) (6) cannot be determined (7) none of above 10. if is1(t) = -4cos(2t) and vsg(t) = ZOcos(2t) then vout(t) = (in V): (1) -38cos(2t) (2) 18cos(2t) (3) -18003(2t) (4) lecos(2t) (5) -4800s(2t) (6) cannot be determined (7) none of above EE-201 Review Problems Test 2 page-7 Fall 98 K The value of H in the circuit below is (in A): (1) 5 (2) 1 (3) 15/13 ( ) -5 (5) 2.5 (6) -2.5 (7) none of the above 29 3v + v1 - 14. Consider the capacitive circuit belowv Suppose vc(0) = -1OV. The voltage, vC(t), at t = 1 second is vc(t) = (in V): (1) -44.59 (2) 27.29 (3) -24.59 (4) 17.29 (5) 7.29 (6) 0.432 (7) none of above 29 + 15a) = 10 e'Z‘um) vc(t) EE-201 Review Problems Test 2 page-8 Fall 98 15. Consider the circuit sketched below. Suppose vs(t) = Sin(7ct)u(’() and suppose vout(0) = 0. Then vout(t) at t = 0.5 seconds is vout(.5) = (in V): (1) 4 (2) 2 (3) O (4) '4 (5) 8 (6) 8n (7) none of above 1mm .._9 21mm) '+ vs“) V (t) 2F 0.517 011‘ 16. Referring again to the circuit of problem 15, the energy stored in the 2F—capacitor (assuming that the capacitor voltage is zero at t = 0) at t = 0.25 seconds is W(.25) = (in J): (1) 0.5 (2) 0.707 (3) 1 (4) 0.25 (5) 4 (6) 47: (7) 1.414 EE-201 Review Problems Test 2 page—9 Fa|i 98 at A ztiss Aids/:5 17. For the circuit below, if is(t) = ' then iC2(t) = (in A): (1) (2) (3) 6t (4) 12b (5) 9+, (6) 24+: (7) at etA ztis 18. In the circuit of problem 17, again suppose is(t) = Then vout(t) = (in V): (1) ’ (2) w«we— (3) 6t (4) 12%: (5) 9t (6) 24+: (7) 8t 4t(5 EE-201 Review Problems Test 2 page-1O ‘ Fall 98 19. In the RC circuit below, vC(O‘) = 10V. The value of iC(t=1sec) is (in A): (1) -0.0736 (2) -0.368 (3) 0.736 (4) -0.736 (5) 0 (6) 3.68 (7) none of above + SQ Vc(t) 0.217 20. In the circuit below, iL(O') = 25A. The value of iL(t=O.55ec) is (in A): (1) 0.458 (2) 3.38 (3) 4.72 (4) 0.892 (5) 51.5x10'9 (6) 9.197 (7) none of above 1115‘) 29 ' 21 0.5H EE-201 Review Problems Test 2 page-11 Fall 98 21. Consider the cirouit below. The value of the response, M0, at t = oo is iL(oo) = (in A): (1) 0 (2) 2 (3) 3 (4) 4 (5) 5 (6) 7 (7) none of above 22. In reference to problem 21, vL(0+) = (in V): (1) 8 (2) 2O (3) 12 (4) 24 (5) -8 (6) 0 (7) none of above EE—201 Review Problems Test 2 page-12 Fall 98 23. Consider the circuit below for which the response vc(t) == 20 - 1O e'o'4t for t 2 0. Suppose further that ic(t) = 0.4e'0-4t for t 2 o. The value of R is (in o): (1) 25 (2) 1o (3) -7.5 (4) 2.5 (5) 1 - (6) 15 (7) none of above R + 131m) 0 109 C V0“) x In the circuit below, vc(0‘) = 0. The value of vout(t) at t = 1 second is vout(1) = (in V) 1) 5.31 (2) 11.98 (3) 8 (4) 24 (5) 13.31 (6) 23.17 (7) none of above - EE-201 Review Problems Test 2 page-14 Fall 98 27. For the circuit below, if iL(O‘) 2A, then iL(t) at t = 1 second is M1) = (in A): (1) -1.472 (2) 2 (3) 4 (4) 0.528 (5) 2.944 (6) 0713(0 (7) 1.472 1 t A) V51 = 411(t) A o 6V. The value of vou((t) at t = 0.1 second is vou((0.1) = (in V): (1)12 (2) 9 (3) —3 (4) -2 (5) -2.33 (6) 5 (7) none of above EE-201 Review Problems Test 2 page-16 Fall 98 31. For the circuit shown below, the capacitor voltage at t = 0' is zero. The current in the 49 resistor at t = 3 seconds is iR(t=3) = (in A): (1) 0.19 (2) 0.57 (3) 0.79 (4) 1.0 (5) 1.26 (6) 1.73 (7) 1.9 t=0 I 120 1112“) 32V :— 4Q 117 32. In the circuit below; suppose M0") = 0. For t > O, iL(t) is (in A): 13. 1 iLfit) 6u{t) v o EE-201 Review Problems Test 2 page-17 Fall 98 33. The value of the capacitor voltage in the circuit below at t = 1 second is vC(t=1) = (in V): (1) 42.13 (2) 17.87 (3) 6.32 (4) 13.68 (5) 16.32 (6) 37.36 (7) 22.64 29 + 10+20u(t) volts v00) 0.517 34. For the circuit shown below, the value of v1 (t) at t = is is (in V): (1) 3.89 (2) 3.68 (3) 2.71 (4) 1.84 (5) 1.35 (6) 0.677 (7) none of above 49 49 40v EE-201 Review Problems Test 2 page-18 Fall 98 Workout Problem: (40 pts) The switches in the RC circuit below have been in the indicated positions for a long time before time t = 0. At t = 0 switch S1 is opened and 2 milliseconds later switch S2 is closed. The object of the problem is to compute the capacitor; voltage for t 2 O by completing the indicated steps. (a) (6 pts) Using a steady state analysis, find the capacitor voltage vC(O-) just before 81 is opened. What is vC(0+) and why? (b) (5 pts) Find the time constant of the capacitor voltage response valid for O < t < 0.002 seconds. (0) (5 pts) Now give the capacitor voltage vC(t) for «$430002 sec. (d) (10 pts) At t = 0.002 sec switch 82 is closed. ,Find the correct initial condition vC(O.002-) and the new time constant for the response valid for t > 0.002 sec. (e) (4 pts) Give the expression for the capacitor voltage vC(t) valid for t > 0.002 sec. (f) (4 pts) Sketch the capacitor voltage vC(t) for t 2 0 showing relevant times and voltages. (g) (6 pts) Compute the energy lost by the capacitor over the time interval 0 < t < 0.002 sec. ...
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