ECE 20100 Spring 2015 Final Exam - WeeklyJoys · ECE 20100 – Spring 2015 Final Exam May 7, 2015 Section (circle below) ... phasor of the current through the inductor, I L, in Amps?

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ECE 20100 – Spring 2015

Final Exam

May 7, 2015

Section (circle below)

Jung (1:30) – 0001 Qi (12:30) – 0002 Peleato (9:30) – 0004

Allen (10:30) – 0005 Zhu (4:30) – 0006

Name ____________________________ PUID____________

Instructions

1. DO NOT START UNTIL TOLD TO DO SO.

2. Write your name, section, professor, and student ID# on your Scantron sheet. We may check PUIDs.

3. This is a CLOSED BOOKS and CLOSED NOTES exam.

4. The use of a TI-30X IIS calculator is allowed, but not necessary.

5. If extra paper is needed, use the back of test pages.

6. Cheating will not be tolerated. Cheating in this exam will result in, at the minimum, an F grade for the

course. In particular, continuing to write after the exam time is up is regarded as cheating.

7. If you cannot solve a question, be sure to look at the other ones, and come back to it if time permits.

Course

Learning

Objective

Exam Questions Total Points Possible Minimum Points

Required to Satisfy

Learning Objective

i 1–6 54 27

ii 7–10 36 18

iii 11–16 54 27

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Question 1: In the circuit below, the power absorbed by the voltage controlled current source,

Pdep, is:

(1) Pdep = –10 W

(2) Pdep = –6 W

(3) Pdep = –4 W

(4) Pdep = –2 W

(5) Pdep = 0 W

(6) Pdep = 2 W

(7) Pdep = 4 W

(8) Pdep = 6 W

(9) Pdep = 8 W

(10) Pdep = 10 W

3

Question 2: Find the value of the voltage source, Vs, in the circuit below:

(1) Vs = -20 V

(2) Vs = -10 V

(3) Vs = -5 V

(4) Vs = 0 V

(5) Vs = 2 V

(6) Vs = 3 V

(7) Vs = 5 V

(8) Vs = 10V

(9) Vs = 20V

(10) Vs = 35V

Vs

4

Question 3. The loop current I1 in the circuit below is (note the resistors are marked by their

conductance):

GND

-8A

0.2S

0.15S

0.2

5S

25A

0.0

5S

440V

-3A

I2

I1IO I3

I4

(1) I1 = – 3 A

(2) I1 = 0 A

(3) I1 = 4.5 A

(4) I1 = 5.0 A

(5) I1 = 5.5 A

(6) I1 = 6.0 A

(7) I1 = 6.5 A

(8) I1 = 7.0 A

(9) I1 = 7.5 A

(10) I1 = 8.0 A

5

Question 4: A device is connected to a circuit that has only two independent sources. Two

identical measurements are made on the device with different independent sources turned on:

Measurement A Measurement B

Source 1 is turned ON Source 1 is turned OFF

Source 2 is turned OFF Source 2 is turned ON

Voltage drop Voltage drop

across the device: V1 = 2V across the device: V

2 = 1V

Power absorbed Current through

by the device: P1 = –6 W the device: I

2 = 1 A

Assume that the measured voltage drop and current through the device follow the passive sign

convention. What is the power absorbed by the device when both sources are ON?

(1) –6 W (2) –5 W (3) –4 W (4) –3 W (5) –1 W

(6) 1 W (7) 3 W (8) 4 W (9) 5 W (10) 6 W

6

Question 5: In the circuit below, I0 = 1 A, find Is and Vs:

(1) Is = 2 A; Vs = 4 V (2) Is = 3 A; Vs = 5 V

(3) Is = 2 A; Vs = 6 V (4) Is = 1 A; Vs = 4 V

(5) Is = 5 A; Vs = 2 V (6) Is = 3 A; Vs = 6 V

(7) Is = 7 A; Vs = 3 V (8) Is = 1 A; Vs = 3 V

(9) Is = 5 A; Vs = 5 V (10) Is = 2 A; Vs = 7 V

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Question 6: In the circuit below, find RL so that VL =20 V:

(1) RL = 1 Ω

(2) RL = 2 Ω

(3) RL = 4 Ω

(4) RL = 7 Ω

(5) RL = 10 Ω

(6) RL = 13 Ω

(7) RL = 15 Ω

(8) RL = 18 Ω

(9) RL = 20 Ω

(10) RL = 25 Ω

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Question 7: Find the Thevenin equivalent voltage seen by the inductor in the circuit below.

(1) THV 1 V

(2) THV 2 V

(3) THV 3 V

(4) THV 4 V

(5) THV 5 V

(6) THV 6 V

(7) THV 7 V

(8) THV 8 V

(9) THV 9 V

(10) THV 12 V

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Question 8: What is the time constant, , of the following circuit?

(1) 1 ms

(2) 2 ms

(3) 3 ms

(4) 4 ms

(5) 5 ms

(6) 6 ms

(7) 7 ms

(8) 8 ms

(9) 9 ms

(10) 10 ms

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Question 9: Find the zero-input component of the voltage across the capacitor for t > 0 if the

voltage at time t = 0- is Vc(0

-) = 15 V (with the plus sign on top). The zero-input component is

due to only initial conditions.

(1) Vc(t) = 15

(2) Vc(t) = 15 exp(- t / 3)

(3) Vc(t) = 15 exp(- 4*t / 3)

(4) Vc(t) = 15 exp(- 2*t )

(5) Vc(t) = 15 exp(- 2*t / 3)

(6) Vc(t) = 11 + 4 exp(- t / 2)

(7) Vc(t) = 11 - 11 exp(- 3*t / 2)

(8) Vc(t) = 11 + 22 exp(- 3*t)

(9) Vc(t) = 11 - 4 exp(- t / 3)

(10) None of the above

+

-

(t)CV

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Question 10: Find the current through the inductor, IL, at t = ∞

(1) IL = - 1 A

(2) IL = - 0.75 A

(3) IL = - 0.5 A

(4) IL = - 0.25 A

(5) IL = 0 A

(6) IL = 0.25 A

(7) IL = 0.5 A

(8) IL = 0.75 A

(9) IL = 1 A


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Question 11: Find the voltage OUT. Assume the op-amps are ideal.

(1) 1 11cos Vt

(2) 1 12cos Vt

(3) 1 13cos Vt

(4) 1 14cos Vt

(5) 1 15cos Vt

(6) 1 16cos Vt

(7) 1 17cos Vt

(8) 1 18cos Vt

(9) 1 19cos Vt

2 k

1 k

1 k2 k

2 k

1 k

2 k

1 12cos Vt

1 13cos Vt

OUT

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Question 12:

In the circuit above, the switch has been closed for a long time and opens at t = 0+, find

Cdv t

dt

at t = 0+

(1)

0

C

t

dv t

dt

= 2 A/s (2)

0

C

t

dv t

dt

= 4 A/s

(3)

0

C

t

dv t

dt

= 6 A/s (4)

0

C

t

dv t

dt

= 8 A/s

(5)

0

C

t

dv t

dt

= 10 A/s (6)

0

C

t

dv t

dt

= –2 A/s

(7)

0

C

t

dv t

dt

= –4 A/s (8)

0

C

t

dv t

dt

= –6 A/s

(9)

0

C

t

dv t

dt

= –8 A/s (10)

0

C

t

dv t

dt

= –10 A/s

14

Question 13: In the circuit below, ( ) cos(5 )si t t A. In sinusoidal steady state, what is the

phasor of the current through the inductor, LI , in Amps?

(1) 10 -90

(2) 10 90

(3) 5 -90

(4) 5 90

(5) 0.2 -90

(6) 0.2 90

(7) 0.1 -90

(8) 0.1 90

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Question 14: Find Iout in the following circuit:

(1) 0 mA

(2) 5 mA

(3) 15 mA

(4) 30 mA

(5) 45 mA

(6) 60 mA

(7) 90 mA

(8) 150 mA

(9) 300 mA


16

Question 15: Which one of the plots is the most accurate one for the frequency response of the

circuit below? (Vout and Vin are phasor voltages, and observe the numbers at the axis)

(1) (2) (3)

(4) (5) (6)

(7) (8)

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Question 16: Find the Thevenin equivalent voltage, TH, and resistance, THR , of the following

circuit as seen from nodes A–B. Assume the op-amp is ideal.

(1) 3cos V, 450 TH THt R









4 k

INB

A

900 900

900

4 k

4 k

4 k

2cos VIN t

TH

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Question 17: Find the reactive power absorbed by the load if 𝑉𝑙𝑜𝑎𝑑 = 3√2 cos(10𝑡 + 45°) V.

(1) 1.5 W

(2) 1.5 VAR

(3) 3 W

(4) 3 VA

(5) 4.5 VAR

(6) 4.5 VA

(7) 9 W

(8) 9 VAR

(9) 0 VA


Amps

19

Question 18: What is the effective voltage of the periodic signal plotted below?

(1) 0.25 V

(2) 0.5 V

(3) 0.75 V

(4) 1 V

(5) 1.25 V

(6) 1.5 V

(7) 1.75 V

(8) 2 V

(9) 2.25 V

(10) 2.5 V

2

1

1 2

3

4

1

5 64 3

2 1

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Question 19: What is the apparent power generated by the voltage source in the circuit below?

iS , where i = 1,2,3,4,5, are the complex power absorbed by the circuit elements.

(1) 2 2 VA (2) 2 VA

(3) 4 2 VA (4) 4 VA

(5) 6 2 VA (6) 6 VA

(7) 8 2 VA (8) 8 VA

(9) 10 2 VA (10) 10 VA

1 1 VAS j

SV

2 2 2 VAS j 3 1 VAS j

4 2 2 VAS j

5 2 2 VAS j

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Question 20: An inductor of 5 nHL is put in series with a source that has an internal

resistance of 50 and an antenna. In sinusoidal steady state at 636 MHz ( 94 10 rad/s),

what impedance AntZ should the antenna have in order to ensure maximum power delivered to

the antenna? (Note: 1 nH = 91 10 H)

(1) 20 (2) 30 (3) 50 (4) 80

(5) 20 j (6) 50 j (7) 50 20 j (8) 50 20 j

(9) 50 0.05 j (10) 50 0.05 j

22

Question 21: You have built a startup which produces cool plasma TVs that consume 150W

when connected to a wall outlet (Vrms = 120 V, ω = 2× 60 rad/sec). You would like to start

selling in California but their laws require a power factor above 0.9, while your current product

has a power factor of 0.75 (lagging). Which of the following elements, when added in parallel at

the input as shown below, can make the plasma TV to satisfy the California law?

(1) 100 Ω resistor

(2) 210 Ω resistor

(3) 0.9 F capacitor

(4) 11 F capacitor

(5) 137 F capacitor

(6) 1 mF capacitor

(7) 0.9 m inductor

(8) 11 mH inductor

(9) 137 mH inductor

(10) 1 H inductor

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Question 22: In the circuit below, is(t) = 5 2 cos(2000t) A, find R and L so that maximum

power is transferred to R:

(1) R = 1 Ω; L = 1 mH (2) R = 1 Ω; L = 2 mH

(3) R = 6 Ω; L = 1 mH (4) R = 2 Ω; L = 2 mH

(5) R = 4 Ω; L = 2 mH (6) R = 2 Ω; L = 4 mH

(7) R = 2 Ω; L = 1 mH (8) R = 1 Ω; L = 4 mH

(9) R = 1 Ω; L = 6 mH (10) R = 4 Ω; L = 6 mH

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Potentially Useful Formulas (2nd

Midterm)

00

( )/( ) ( ) ( ) ( )

t tx t x x t x e , where THR C or

TH

L

R

00

2 2

0 1 1 0

( )( )

1( ) ( ) ( )

( , ) ( ) ( )2

LL

t

L L Lt

L L L

di tv t L

dt

i t i t v t dtL

LW t t i t i t

00

2 2

0 1 1 0

( )( )

1( ) ( ) ( )

( , ) ( ) ( )2

CC

t

C C Ct

C C C

dv ti t C

dt

v t v t i t dtC

CW t t v t v t

1ln lnx

x

Elapsed time formula: t2 - t1 = ln[(X1 - x(∞))/(X2 - x(∞))]

25

Potentially Useful Formulas (3rd

midterm)

First order circuit: ot t /

ox(t) x( ) x(t ) x( ) e , = L/R or = RC

Series RLC: 2 R 1s s 0

L LC

Parallel RLC: 2 1 1s s 0

RC LC

td dx(t) x( ) Acos t Bsin t e

tx(t) x( ) A Bt e

1 2s t s tx(t) x( ) Ae Be

22

1 2b b 4c

s ,s for s bs c 02

, where

1c LC

R / 2L (series)b

12 (parallel)

2RC

o1

LC

2 21,2 os

22 2

d o4c b

2

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Potentially Useful Formulas (since Exam 3)

0

1( ) cos( ) cos( ) cos( )

2

T

m mave V I eff eff V I rms rms V I

V IP p t dt V I V I

T

* * *1

2m m eff eff rms rms P jQ S = V I V I V I = VA

2 2cos( )V I

P Ppf

P Q

S , ( )V Ipfa

Documents

ECE 20100 Spring 2015 Final Exam - WeeklyJoys · ECE 20100 – Spring 2015 Final Exam May 7, 2015 Section (circle below) ... phasor of the current through the inductor, I L, in Amps?