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Purdue ECE 20100 - ex4sp06

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DCBAECE 201 – Spring 2006Final ExamMay 5, 20068:00am-10:00amInstructionsThere are 20 multiple choice questions. Each is worth 10 points.Our goal is to assess what you know, not what you do not know. To maximizeour assessment of your knowledge and understanding, do NOT dwell on a singleproblem. If you get stuck, move on to the next problem and return later, timepermitting. It is important to work patiently, efficiently, and in an organizedmanner.This is a closed book, closed notes exam. No scrap paper is permitted. You arepermitted only a calculator.All students are expected to abide by the usual ethical standards of theuniversity, i.e., your answers must reflect only your own knowledge andreasoning ability.1Possible useful equations:(1) Trigonometry identities2)cos()cos(coscos2)cos()cos(sinsin(2) First-order circuitso(t t )/τ+ox(t)= x( )+ [x(t ) x( )] e- --� �(3) Second-order circuits2Question 1. Find node voltage VC (in V): (1) 1 (2) 2 (3) 3(4) 4 (5) 5 (6) 6(7) 7310 S3 S 12 A5 S1 SABCD+_8 V1 SQuestion 2. Find the current I in the circuit below (in A): (1) 1 (2) 2 (3) 3 (4) 4 (5) 5 (6) 6(7) 644 A4 3 1 4 2 2 +_Vd0.3VdIQuestion 3. The nodal and mesh equations of circuits containing inductors and capacitors aredifferential equations or integral-differential equations. Shown below is a circuit having threeresistors and an inductor. The mesh equations in terms of the designated currents i1(t) and i2(t) are(1)dt)t(dvdt)t(di2)t(i7)t(i2)t(v)t(i2)t(i6s221s21(2) dt)t(dvdt)t(di2)t(i7)t(i2)t(v)t(i2)t(i6s221s21(3) 0dt)t(di2)t(i7)t(i2)t(v)t(i2)t(i6221s21 (4)0dt)t(di2)t(i7)t(i2)t(v)t(i2)t(i6221s21(5) 0dt)t(di2)t(i3)t(i2)t(v)t(i2)t(i6221s21(6)0dt)t(di2)t(i3)t(i2)t(v)t(i2)t(i6221s21(7) none of the above52 Hixvs( t )+245i1i2Question 4. For the ideal op. amp. circuit below, the output voltage Vout is (in Volts): (1) 26 (2) 15 (3) 14(4) 16 (5) 25 (6) 24(7) 246+_++___+Vout1 V4 V2 10 4 Question 5.The Thevenin equivalent circuit of the ideal op. amp. circuit shown above is 73 0 0+1 0 02 Vb2 0 0a82 0 01 Vab( 4 )+_2 0 01 Vab( 3 )+_1 0 01 Vab( 1 )+_1 0 01 Vab( 2 )+_3 0 01 Vab( 5 )+_3 0 01 Vab( 6 )+_7 51 Vab( 7 )+_Question 6. A linear circuit is driven by a constant voltage source Vs1 and a constant currentsource Is2. The output voltage Vout is 3 V when Vs1 = 3 V and Is2 = 8 A. On the other hand, if Vs1 = 6V and Is2 = 4 A, Vout is again 3 V. If Vs1 = 0 V and Vout = 4 V, Is2 , in A, is (1) 16 (2) 2 (3)  (4) 4 (5) 8 (6)  (7) 64 9Vs 1+Is 2Vo u t_+L i n e a r c i r c u i tQuestion 7. Find the Thevenin equivalent resistance, RTH, for the circuit below (in ):(1) 10 (2) 20 (3) 7.5(4) 40 (5) 5 (6) 15(7) 30 10iy20 ixix+_20 20 0.5 iyQuestion 8. The Thevenin equivalent of the circuit below is:(1) a single 3 resistor(2) a single 3 resistor(3) a short circuit(4) an open circuit(5) a single 3V independent voltage source(6) a 3V independent voltage source in series with a 1 resistor (7) none of the above11Question 9. In the circuit below, the maximum power that can be transferred to the variableresistance R is (in W)(1) 2.5 (2) 5 (3) 6.25 (4) 10 (5) 12.5 (6) 25(7) none of the aboveV =10 Vs1+_+_V =5 Vs2222R12Question 10. The switch in the circuit below opens at t 0 after having been closed for a longtime. The sum of energy stored in the inductor and energy stored in the capacitor is 1J at t 0.The value of the inductance L is (in H): (1) 0 (2) 0.01 (3) 0.02 (4) 0.1 (5) 0.2 (6) 1 (7) none of the above 13Question 11. In the ideal op. amp. circuit shown below, the capacitor voltage, vc(t), is 0 V for t <0. The input voltage is vin(t) = 7 e2t u(t) V. The output voltage, vout(t), in V, for t > 0 is e2t) (2) e2t (3) e2t (4) e2t (5)  7e2t (6) –7(1  e2t ) (7) 7(1  e2t ) 14vo u t( t )++_1 0vi n( t )0 . 0 5 Fvc( t )+Question 12. The switch in the RL circuit shown below has been opened for a long time and itcloses at t = 0. Then the inductor current iL(t), in A, for t > 0 is(1) 2  2 e10t (2) 2 + 2 e10t (3) 4  4 e10t (4) 4 + 4 e10t (5) 4  2 e10t (6) 4 + 2 e10t (7) 2  4 e10t 154 A2 Ht = 0iL1 01 0Question 13. For the circuit below, the response for the capacitor voltage, vC(t), is desired. If thegeneral solution is of the form, x(t) = xn(t) + XF, select the final value, XF, in appropriate units:(1) 75 V (2) 0 V (3) 3 A (4) 4 A (5) 5 V (6) 6 A (7) 0 A16_+75 V25 15 4 mF20 mHiL(t)+_vc(t)Question 14. The switch in the circuit below has been opened for a long time. It closes at t = 0 s.The initial capacitor voltage prior to the switch closing is 0 V. The roots to the characteristicsequation are s1 =  2 s1 and s2 =  3 s1. The complete response of the circuit is(1) vc(t) = 8 e– 2t – 8 e– 3t V (2) vc(t) = – 12 e– 2t + 8 e– 3t V(3) vc(t) = 12 e– 2t – 8 e– 3t V (4) vc(t) = 4 – 12 e– 2t + 8 e– 3t V(5) vc(t) = 4 + 8 e– 2t – 12 e– 3t V (6) vc(t) = 4 – 12 e– 2t + 12 e– 3t V(7) vc(t) = 12 e– 2t + 12 e– 3t V170 . 5 H+2 . 5t = 04 V+_vc( t )1 / 3 FQuestion 15. In the circuit below, the current source is is(t) = 10 cos (10t) A. In the sinusoidalsteady state, the phasor voltage VC , in V, is(1) 1 + j 1 (2) 2 + j 2 (3) 4  j 3 (4)  4 (5) 10 (6) 6+ j 3 (7) j 7 184is0 . 0 5 F0 . 2 HvC+_Question 16. In the circuit shown below, the voltage lags behind the current by 30°. The circuitcontains: (1) R only (2) L only (3) C only (4) R and L (5) R and C (6) L and C (7) none of the above19Question 17. The current in the circuit shown below


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Purdue ECE 20100 - ex4sp06

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