EE 201 Final Exam May 5, 2000 General Instructions: Time ... · EE-201 Final page 1 Spring 00 EE 201 Final Exam May 5, 2000 General Instructions: • The exam is closed book, closed

EE-201 Final page 1 Spring 00

EE 201 Final Exam May 5, 2000 General 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

appropriate places on the computer scan forms.

Time Instructor Section Number

7:30 am DeCarlo 0001

11:30 am Capano 0002

3:30 pm Tan 0003

• The exam consists of 30 equally weighted multiple-choice questions.

• Please keep your computer scan sheets beneath this exam while you work.

• Our goal is to assess what you know, not what you do not know. 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. It is

important to work patiently, efficiently, and in an organized manner.

• 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.


Problem 1. Req = (in Ω): (1) 0.25 (2) 0.75 (3) 2.25 (4) 4 (5) 0.571 (6) 1.333 (7) 1.75

Problem 2. The power absorbed by the 5 Ω resistor = (in watts): (1) 10 (2) 20 (3) 8 (4) 50 (5) 500 (6) 80 (7) none of the above.


Problem 3. In the circuit below R2 = 4R1 and the power dissipated in R1 is 1 watt. Suppose the voltage Vs is doubled. Then the power absorbed by the resistor, R2, is (in watts): (1) 1 (2) 2 (3) 3 (4) 4 (5) 8 (6) 16 (7) 32

Problem 4. The current Ix in the circuit below is (in A): (1) 1 (2) 2 (3) 2.5 (4) 4 (5) 5 (6) 1.2 (7) none of above

Problem 5. In the circuit below, Vout = 20 V. Then Vin = (in V): (1) –4 (2) –5 (3) –16 (4) 4 (5) 5 (6) 80 (7) 16



Problem 6. For the circuit below, Rth = (in Ω): (1) 8 (2) 2 (3) 12 (4) 4 (5) 16 (6) 20 (7) 32

Problem 7. For the circuit below, the maximum power transferred to the load resistor for an appropriate load resistance RL is (in watts): (1) 25 (2) 200 (3) 12.5 (4) 40 (5) 50 (6) 800 (7) none of above

Problem 8. The current is(t) in amps given in the plot below excites the RL circuit to its right. The value of vs(t) at t = 0.5 sec is (in V): (1) 1 (2) 2 (3) 2.5 (4) 4 (5) 5 (6) 6 (7) none of the above.


Problem 9. What is vC(t) at t = 3 sec (approximately) assuming vC(0- ) = 10 V, and assuming the switch closes at t = 2 sec? (1) 1.35 (2) 2.23 (3) 3.68 (4) 4.72 (5) 0.5 (6) 8.45 (7) none of the above.

Problem 10. Using source transformations, the voltage across the 12 Ω resistor is (in volts): (1) 1 (2) 2 (3) 3 (4) 18 (5) 0.5 (6) 6 (7) none of above


Problem 11. The Norton equivalent circuit seen by the variable load resistor at the terminals A-B is: (1) Isc = 4 A, Rth = 4 Ω (2) Isc = 4 A, Rth = 5 Ω (3) Isc = 4 A, Rth = -4 Ω (4) Isc = -4 A, Rth = 4 Ω (5) Isc = -4 A, Rth = 5 Ω (6) Isc = -4 A, Rth = -4 Ω (7) none of these

Problem 12. The Thevenin equivalent resistance seen at the terminals A-B is (in Ω): (1) 1.5 (2) 2 (3) 3 (4) 3.5 (5) 0.2 (6) 2/3 (7) none of these


Problem 13. The inductor voltage vL for t ≥ 0 is an shown below. The part for t < 0 is not shown. It is also known that iL(∞) = 0 A. The initial condition iL(0+) is (in A): (1) 1 (2) –1 (3) 3 (4) –3 (5) 0.5 (6) –0.5 (7) 0

Problem 14. The RLC circuit below has an underdamped response of the form vC (t) = e−σ t Acos(ωdt) + Bsin(ωdt)[ ] + Vf (V) for t > 0. The range of R leading to this response is: (1) R < 0.25 Ω (2) R > 0.25 Ω (3) R = 0.25 Ω (4) R < 1 Ω (5) R > 1 Ω (6) R = 1 Ω (7) none of these


Problem 15. In the op amp circuit below, vs(t) = 2u(t) V and vC(0) = 2 V. The response vout(t) is (in V): (1) – 4t u(t) (2) 2 – t u(t) (3) 2 – 4t u(t) (4) –8t u(t) (5) 2 – 8t u(t) (6) –8t u(t) – 2 (7) 2 + 8t u(t)

Problem 16. In the circuit below, vC(0+) = – 5 V. The value of vC(t) at t = 0.05 sec is (in V): (1) – 1.065 (2) 1.32 (3) 1.839 (4) –5 (5) 5 (6) 8.678 (7) none of above

Problem 17. In the circuit below, vin(t) = 2 20 cos(10t + 63.435o) V. The phasor voltage VC = Vmax∠θ is (in V): (1) 2 10 ∠63.435o (2) 2∠0o (3) 2∠126.87o (4) 0.5∠–63.435o (5) 0.5∠63.435o (6) 2 10 ∠0o (7) none of above



Problem 18. The circuit below operates at ω = 1 rad/sec. The value of the phasor voltage V1 is (in V): (1) 2.236∠–116.57o (2) 2.236∠–63.435o (3) 2.236∠63.435o (4) –5∠0o (5) 5∠–63.435o (6) 11.18∠63.435o (7) 5∠0o

Problem 19. The average power absorbed by the 6Ω resistor in the circuit below is (in watts): (1) 1 (2) 2 (3) 3 (4) 0.5 (5) 1/3 (6) 1/6 (7) none of the above


Problem 20. The effective or rms value for the voltage waveform shown below is (in V). (1) 1 (2) 2 (3) 2 (4) 4 (5) 6 (6) 6 (7) none of the above.

Problem 21. The average power absorbed by the inductor with phasor voltage VL = 10∠60o and phasor current, IL, shown below is (in watts): (1) – 0.3 (2) 0.3 (3) 0 (4) 10 2 (5) 5π (6) – 10 (7) none of the above

Problem 22. For the capacitor shown below, vC(t) = 50 2 sin (10 t) V. The reactive power is (in VAR): (1) –25 (2) 25 (3) –100 (4) 50 (5) –50 (6) 100 (7) –2500


Problem 23. The circuit below operates in the sinusoidal steady state at ω = 10 rad/sec. The phasor source voltage is an effective value. The average power supplied by the source to the circuit is (in watts): (1) 100 (2) 200 (3) 320 (4) 400 (5) 50 (6) 282.84 (7) none of above

Problem 24. Each of the boxes in the circuit below is a device whose absorbed complex powers are: S1 = 100 + j 70 S2 = 100 + j 250 S3 = 75 – j 100 S4 = 65 + j 140 S5 = 60 – j 60 The apparent power delivered by the source is (in VA): (1) 100 (2) 200 (3) 300 (4) 400 (5) 500 (6) 600 (7) none of the above.


Problem 25. The circuit below operates in the sinusoidal steady state with ω = 3 rad/sec. The phasor source voltage is an effective value. The value of C that results in the unity power factor is (in F): (1) 1 (2) 2 (3) 3 (4) 0.1 (5) 0.5 (6) 0.01 (7) none of the above

Problem 26. The voltage phasor, V, leads the current phasor, I, by 25˚ in the circuit below. The circuit probably contains: (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 above

Problem 27. The value of | Vout | as ω → ∞ is (in V): (1) 1 (2) 2 (3) 2 (4) 0.4 (5) 0.5 (6) 0 (7) none of the above.


Problem 28. The circuit below operates in the sinusoidal steady state at ω = 1 rad/sec. The phasor gain, Vout/Vin =: (1) 1 (2) 2 (3) –2 (4) –1 (5) 2j (6) –2j (7) none of the above.


Problem 29. The motor in the circuit below consumes 30kW of average power at a power factor of 0.6 lagging (without the capacitor). A capacitor is connected in parallel with the motor to increase the power factor to 0.894 lagging. The voltage across the motor is kept at 250 Vrms. It is already known that without compensation, Qm

old = 40.0 kVAR, and with compensation, QmCnew = 15.0 kVAR. The

impedance of the capacitor (in Ω) is: (1) 1.25 (2) j1.25 (3) –j1.25 (4) 2.5 (5) j2.5 (6) –j2.5 (7) none of the above


Problem 30. The load circuit that maximizes power transfer to the load at ω = 100 rad/sec is:

(7) none of above

Solutions to 201 Final Spring 2000 1. Is = 0.5Vs – 0.25VR = (0.5 – 0.25)Vs. Req = Vs/Is = 4 Ω. Answer: (4).

2. V5Ω = 510

(50 – 10) = 20 V. P5Ω = (20)2/5 = 80 watts. Answer: (6).

3. Is

2R1 = 1 watt. Is2R2 = Is

24R1 = 4(Is2R1) = 4 watts. If Vs is doubled, then 2Vs = Req(2Is).

Hence, power absorbed by R2 is (2Is)2R2 = 4Is

2R2 = 16 watts. Answer: (6).

4. Ix = 11 +1 + 0.5

5 = 2 A. Answer: (2).


5. V– = 1

1 + 4 (20) = 4 V. Vin = –V– = –4 V. Answer: (1).

6. Rth = 4 ||8|| 8 = 2 Ω. Answer: (2). 7. Given RL = Rth = 2 Ω, the voltage across RL is 10 V. Hence, PRL = (10)2/2 = 50 watts. Answer:

(5). 8. vs(0.5) = 4(0.5) + 2 × 1 = 4 V. Answer: (4). 9. vC(2) = 10 exp(–2/2) = 10 exp(–1) = 3.6788 V. vC(3) = 3.6788 exp(–(3 – 2)/1) = 3.6788 × exp(–

1) = 1.35 V. Answer: (1). 10. The Norton equivalent seen by the 12-Ω resistor consists of a 2A source (upwards) in parallel

with a 4-Ω resistor. 4||12 = 3 Ω. Therefore, V12Ω = 2 × 3 = 6 V. Answer: (6). 11. By inspection, ISC = 4 A and VOC = 20 V. Thus, Rth = VOC/ISC = 5 Ω. Answer: (2). 12. Excite with a 1-A source so that Ix = 1 A. Then VAB = 2 + V1. To find V1, observe that

1 ⋅ (V1 – 2) + V1 – 1 = 0. Hence, 2V1 = 3 or V1 = 1.5 V and Rth = VAB = 3.5 Ω. Answer: (4).

13. 0 = iL(∞) = iL(0+) + 1L

vL (τ) dτ0

∞∫ = iL(0+) + 2(1 – 0.5). Hence, iL(0+) = –1 A. Answer: (2).

14. Characteristic poly for parallel RLC is

s2 + 1RC

s + 1LC

= s2 + 4Rs + 4 = 0.

Undamped requires that


4R

⎛ ⎝

⎞ ⎠

2− 16 < 0 .

Hence, 4/R < 4 requires that R > 1. Answer: (5).

15. vout(t) = vC(0+) – 1RC

vs(τ)dτ = 2 − 4 2u(τ)dτ = 2 − 8t u(t)0

t∫0

t∫ . Answer: (5).

16. vC(t) = vC(∞) + (vC(0+) – vC(∞))e–t/τ where vC(0+) = –5, vC(∞) = 5, and τ = 0.5 × 0.1 = 0.05 sec.

Then, vC(0.05) = 5 – 10e–1 = 1.32 V. Answer: (2).

17. VC =

11 + jωc

1 + 11 + jωc

Vin =1

2 + 4jVin =

2 20∠63.435°20∠63.435°

= 2 . Answer: (2).

18. j(V1 – 10j) + 1j

V1 + (V1 – 5) = 0. Hence, V1 = 5 – 10 = –5. Answer: (4).

19. Pave =12

i2 (t)Rdt0

1∫ =

62

t2 dt0

1∫ = t3]0

1=1 . Answer: (1).

20. Veff2 =

13

v2 (τ) dτ0

3∫ =

132 + 4[ ] = 2 . Therefore, Veff = 2 V. Answer: (3).

21. Ideal inductors do not absorb average power. Answer: (3). 22. VC,eff = 50 ∠90° = 50j. YC(j10) = j0.01. SC = VC,eff (VC,eff j0.01)* = –j0.01 |VC,eff|

2 = –j0.01(2500) = –j25. Answer: (1).

23. Zin (j10) = j1+1

0.8 + j0.6= 0.8 + j0.4 . Zin

* (j10) = 0.8 − j0.4 . Yin* (j10) = 1 + j0.5. It follows that

Ss =V sV s*

Zin* (j10)= V s

2Yin* (j10) = 400 + j200. Pave = 400 watts. Answer: (4).


24. By conservation of power, the complex power delivered by the source is Ss = 400 +j300. The apparent power is |Ss| = 500 VA. Answer: (5).

25. Unity power factor means |S| = Pave which means the input impedance and admittance are real.

Yin(j3) = j3C +1

1+ j3= j3C + 0.1 − j0.3. Hence, C = 0.1 F. Answer: (1).

26. Answer: (4). R and L. 27. The circuit is lowpass. For large ω capacitor looks like a short and the inductor as open. Hence,

|Vout| = 0. Answer: (6) 28. Vout

Vin= −

YinYf

= −1+ j.5 + .5j

= −2 . Answer: (3)

29. Answer: (6). –j2.5 Ω.

30. Zth = 30 + 1jωc

= 30 – j10. For maximum power transfer ZL = Z th* = 30 + j10. 10 = ωL = 100L

implies L = 0.1 H. Answer: (2)

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EE 201 Final Exam May 5, 2000 General Instructions: Time ... · EE-201 Final page 1 Spring 00 EE 201 Final Exam May 5, 2000 General Instructions: • The exam is closed book, closed