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ECE 201 – Spring 2008 Exam #3

April 15, 2008

Division 0101: Qi (7:30am)

Division 0201: Elliott (12:30 pm) Division 0301: Capano (2:30 pm)

Division 0401: Jung (3:30 pm) Instructions

1. DO NOT START UNTIL TOLD TO DO SO. 2. Write your Name, division, professor, and student ID# (PUID) on your scantron sheet. 3. This is a CLOSED BOOKS and CLOSED NOTES exam. 4. There is only one correct answer to each question. 5. Calculators are allowed (but not necessary). Please clear any formulas, text, or other information

from your calculator memory prior to the exam. 6. If extra paper is needed, use back of test pages. 7. Formulas are given on the final page of this exam. 8. Cheating will not be tolerated. Cheating in this exam will result in an F in the course. 9. If you cannot solve a question, be sure to look at the other ones and come back to it if time permits. 10. As described in the course syllabus, we must certify that every student who receives a passing grade

in this course has satisfied each of the course outcomes. On this exam, you have the opportunity to satisfy outcomes i, iii, iv, v, and ix. (See the course syllabus for a complete description of each outcome.) On the chart below, we list the criteria we use for determining whether you have satisfied these course outcomes. Outcomes i and iii were covered on Exam #1. You only need to satisfy the outcomes once during the course.

Course

Outcome Exam

Questions Total Points

Possible Minimum Points required to satisfy course outcome

i 1-14 100 50 iii 10, 12, 13 21 7 iv 1-4 28 14 v 8-14 49 21 ix 5-7 21 7

If you fail to satisfy any of the course outcomes, don’t panic. There will be more opportunities for you to do so.

1. For the circuit shown below, choose the correct υC vs. time plot. Assume ( )0 0.5 + =C Vυ and

( )04 /

+

=CdV s

dtυ

. [Hint: Find solution for the characteristic equation.]

2

1F1H0.2Ω

0 10 200

0.2

0.4

0.6

υ5 t( )

t

0 10 200.5

0

0.5

υ4 t( )

t

0 10 205

0

5

υ3 t( )

t

0 10 205

0

5

υ t( )

t

0 10 2050

0

50

υ2 t( )

t

(1) (2)

(3) (4)

(5)

υC(t) υC(t)

υC(t) υC(t)

υC(t)

2. For the circuit shown, the two resistors have the same value R. Find R (in Ω) that yields characteristic roots s1 = -2s-1 and s2 = -3s-1.

1) 1.2 2) 0.2 3) 1.6 4) 0.4 5) 0.8 6) 0.6 7) 2.4

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3. Suppose R in the previous problem is changed to 67

Ω . Find vc(t) (in V). (In the answer

selections, K1, K2, A and B are appropriate constants.) 1) ( ) ( )7t

d d12e A cos t Bsin t− ⎡ ⎤ω + ω⎣ ⎦ 2) t 6

1 2K e K e− −+ t

t

t

ω

3) t 6

1 26 K e K e− −+ + 4) t 6

1 212 K e K e− −+ + 5) ( ) ( )( )3.5t

d d12e A cos t Bsin t− ω + ω

6) ( ) ( )( )3.5t

d d6 6e A cos t Bsin t−+ ω +

7) Not enough information is given to determine vc(t).

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4. In the following circuit, the switch, which was open for a long time prior to t = 0, closes at t = 0. Determine the current iR at t = 0+.

1) -90 mA 2) -60 mA 3) -30 mA 4) 0 5) 30 mA 6) 60 mA 7) 90 mA

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5. For the circuit shown below, find υout(t) assuming ideal OP-amp behavior.

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(1) υout(t) = 2cos(ωt) [V] (2) υout(t) = -2cos(ωt) [V] (3) υout(t) = 5cos(ωt) [V] (4) υout(t) = -5cos(ωt) [V] (5) υout(t) = 13cos(ωt) [V] (6) υout(t) = -13cos(ωt) [V] (7) υout(t) = 16cos(ωt) [V]

+

2R2R

υs1(t)= 3cos(ωt) [V] υs2(t)= 8cos(ωt) [V] + R + R

6. Find the output voltage Vo (in V), assuming ideal op amp behavior:

1) 1 2) 2 3) 3 4) 4 5) 5 6) 10 7) 25

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7. For the circuit shown below, find υout(t) when υin(t)=2t+3 [V] assuming ideal OP-amp behavior.

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(1) υout(t) = 2t [V]

R1 = 1 kΩ R2 = 1 kΩ C = 1 μF

(2) υout(t) = -2t [V] (3) υout(t) = 2×10-3 [V] (4) υout(t) = -2×10-3 [V] (5) υout(t) = 2t+3 [V] (6) υout(t) = -2t-3 [V] (7) υout(t) = -4×10-3 [V]

8. The following figure shows the voltage and current of a two-terminal device. The dashed line is the waveform of the current i(t), and the solid line is the waveform of the voltage v(t). The x-axis is time in seconds.

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Which of the following could be the device? (1) a 0.5 Ω resistor (2) a 2F capacitor (3) a 2H inductor (4) a 0.5F capacitor (5) a 0.5H inductor (6) a 2 Ω resistor

2 A

1 V

t (s)π 0 2π 3π 4π 5π

current

v(t)

i(t)

voltage

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°9. For the element shown, the time-dependent voltage and current are and i( .

( )v(t) 5V cos t 30= ω −t) 200mAsin t= ω

The impedance of this element is 1) ( )2 cos t 30 sin tΩ ω − ° ω

°

2) ( )25 cos t 30 / sin tΩ ω − ° ω 3) 2 tan( t 120 )Ω ω − 4) 5 3 0∠ − °Ω 5) 0.2 90∠ − °Ω 6) 25 60∠ °Ω

10. Find the equivalent impedance (Zeq) for the circuit shown below. Assume ω = 100 [rad/sec].

2 H

0.1mF

2 H

0.1 mFZeq

100Ω

(1) Zeq = 100 + j100 [Ω] (2) Zeq = 100 – j100 [Ω] (3) Zeq = 200 + j100 [Ω] (4) Zeq = 200 – j100 [Ω] (5) Zeq = 100 + j200 [Ω] (6) Zeq = 100 – j200 [Ω] (7) Zeq = 200 + j200 [Ω]

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11. Given that ( )1( ) 5cos 120= +i t tω

and ( )2 ( ) 5cos 60= +i t tω find i (in Amperes) in the sinusoidal steady state.

load (t)

(1) ( )5cos tω (2) ( )10cos tω

(3) ( )5 3 cos tω (4) ( )5 3 cos 90tω +

(5) ( )10cos 90tω + (6) ( )10cos 180tω +

(7) ( )10 3 cos 90tω +

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12. For the following circuit, which of the following is a valid correct solution for the phasor voltage V1? (Hint: Use superposition.)

1) ( )25j= Ω1 sV I 2) ( )25j= − Ω1 sV I 3) 0=1V

4) ( )50 5025j

Ω= − Ω

Ω1 sV V sI

5) ( )50 5025j

Ω= + Ω

Ω1 sV V sI

6) ( )

( )( )50 25j25j50 25j 50 25j

Ω ΩΩ= +

+ Ω Ω + Ω1 sV V sI

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13. Determine the phasor short-circuit current ISC of the following network.

1) 0 2) -0.2jA 3) 0.25 A 4) 0.2 A1 j−

5) 0.2 A1 j+

6) 0.2 A1 j−

+

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14. Consider the following circuit. Vin and Vout are the phasor representations of the input and output voltages, respectively. The input Vin is a sinusoidal signal of frequency ω in its steady-state.

Vout Vin

1 mF

1 Ω

Which of the following graphs is a correct sketch of out

in

VV

vs. ω ?

out

in

VV

1

0 ω 102 103

out

in

VV

1

0 ω 102 103

out

in

VV

1

0 ω

out

in

VV

1

0 ω 102 103

102 103

(1) (2)

(3) (4)

(5) None of the above

(rad/sec) (rad/sec)

(rad/sec) (rad/sec)

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Useful formulas

( )ot t /ox(t) x( ) x(t ) x( ) e

+− − τ+⎡ ⎤= ∞ + − ∞⎣ ⎦ τ = L/R τ = RC

( ) td dx(t) x( ) A cos t Bsin t e−σ= ∞ + ω + ω

( ) tx(t) x( ) A Bt e−σ= ∞ + +

( )1 2s t s tx(t) x( ) Ae Be= ∞ + +

R / 2L (series)

1 (parallel)2RC

⎧⎪σ = ⎨⎪⎩

2 2

1,2 os = −σ ± σ − ω

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1 2b b 4cs ,s for s bs c 0

2− ± −

= + + =

2

2 2d o

4c b2−

ω = = ω − σ

o 1LC

ω =

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