ECE 20100 – Fall 2014 Final Exam December 19, 2014...ECE 20100 Final Exam, December 19, 2014 9....

ECE 20100 – Fall 2014 Final Exam

December 19, 2014

Section (circle below and enter on your Scantron)

Bermel – 0002 Peleato-Inarrea – 0004 Qi – 0005

Gray – 0006 Allen – 0007 Lin - 0010

Name ______________________________ PUID____________

Instructions

1. DO NOT OPEN THE EXAM BOOKLET UNTIL TOLD TO DO SO. 2. Write your name, section (all 4 digits, as shown above), professor, and PUID on your Scantron sheet. It is

very important that you enter this information accurately. We may check PUIDs. #2 pencils only. 3. This is a CLOSED BOOKS and CLOSED NOTES exam. However, the last four pages of the exam

contain potentially useful formulas – the formula sheets provided on Exams 2 & 3 and an added sheet for material covered since Exam 3.

4. The use of a simple scientific calculator without memory or communication capabilities (such as TI-30Xa or TI-30X IIS) is allowed and some calculation will be 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 – set your pencil down immediately when the exam period ends.

7. If you cannot solve a question, be sure to look at the other ones, and come back to it if time permits. 8. The Final Exam assesses satisfaction of the ECE 20100 Learning Objectives, below. Satisfaction of the

Learning Objectives is independent of the course grade. This information is used as part of the continuous improvement process of the EE and CmpE programs.

i. An ability to analyze linear resistive circuits. [Questions 1-5, 12, 13, 19] ii. An ability to analyze 1st order linear circuits with sources and/or passive elements.

[Questions 6-11, 17, 23] iii. An ability to analyze 2nd order linear circuits with sources and/or passive elements.

[Questions 14-16, 18, 20-22, 24, 25]

Honor Pledge: I have neither given nor received unauthorized assistance on this exam

Signature: ____________________________________________

ECE 20100 Final Exam, December 19, 2014

1. Consider the circuit below. If the independent voltage source 𝑉𝑉𝑠𝑠 = 4 V, what current, I2 (in Amperes) will flow through the 1 Ω resistor as shown below?

(1) 1 A (2) 4.5 A (3) 6 A (4) 2 A (5) 0.5 A (6) 1.0 A (7) 0.5 A (8) 0.25 A (9) 3 A (10) Insufficient information provided


2. Using the Principle of Superposition, what is the nodal equation that can be used to calculate only the contribution of the 12 V source to the nodal voltage, Va,?

(1) ( ) ( )2 12 2 5 0a a aV V V− + − + =

(2) ( ) ( )12 50

2 2a a

a

V VV

− −+ + =

(3) ( )2 2 5 0a a aV V V+ − + =

(4) 2 2 0a a aV V V+ + =

(5) ( )50

2a

a

VV

−+ =

(6) ( )120

2a

a

VV

−+ =

(7) 02 2a a

aV V V+ + =

(8) ( )120

2 2a a

a

V V V−

+ + =

(9) ( )2 12 2 0a a aV V V− + + =

(10) 0aV =

5 V


3. What is the equivalent resistance of this series-parallel network of resistors (in Ω)?

(1) 2 Ω (2) 4 Ω (3) 8 Ω (4) 12 Ω (5) 16 Ω (6) 36 Ω (7) 48 Ω (8) 54 Ω (9) None of the above


4. What is the current I (in Amperes) flowing through the 1 Ω resistor depicted in the figure below?

(1) 1 A (2) 2 A (3) 3 A (4) 4 A (5) 5 A

(6) -1 A (7) -2 A (8) -3 A (9) -4 A (10) 0 A

6 V

+ - 3 Ω

I

2 A

1 Ω

4 A

+ –

5 V


5. If the following relationship between input current and output voltage is observed for a linear circuit, what are its Norton equivalent current source (ISC) and resistance (Rth) values? [Note: when terminals A and B are shorted, IA = -ISC]

(1) Isc = 0 A; Rth = 18 Ω (2) Isc = 1 A; Rth = 15 Ω (3) Isc = 3 A; Rth = 6 Ω (4) Isc = 4 A; Rth = 6 Ω (5) Isc = 2 A; Rth = 3 Ω (6) Isc = 4 A; Rth = 3 Ω (7) Isc = 3 A; Rth = 2 Ω (8) Isc = 4 A; Rth = 2 Ω (9) Isc = 9 A; Rth = 1 Ω

Current 𝑰𝑰𝑨𝑨 Voltage 𝑽𝑽𝑨𝑨𝑨𝑨 -2 A 6 V 1 A 15 V 4 A 24 V

IA A

+ VAB Linear Circuit _ B


6. The switch is closed for a long time and then opened at 1 st = . At 0t −= , the inductor current is known to be (0 ) 0 ALi

− = . For t > 1 s, what is the time constant, τ?

(1) 0 sτ = (2) 1 s4

τ = (3) 1 s2

τ = (4) 1 sτ = (5) 2 sτ =

(6) 4sτ = (7) 8sτ = (8) 16sτ = (9) sτ = ∞


7. Find the voltage 𝑉𝑉𝐶𝐶(𝑡𝑡) (in V), for 𝑡𝑡 > 0.

Recall: 𝑢𝑢(𝑡𝑡) = 1 if 𝑡𝑡 > 0 (and 0 otherwise).

(1) -18e-10t (2) 6e-t - 6 (3) 12 - 12 e-10t (4) 12 - 18e2t (5) -6 +18e-2t (6) 12 - 18e-10t (7) 600 – 600e-2t (8) None of the above

Volts


8. A capacitor is charged to (0 ) 2.5 VCv − = . At 0t = , the switch in the circuit below is closed. At what time 1t in seconds will the capacitor voltage be 1( ) 1.5 VCv t = ?

(1) 1 5ln2 2

(2) 1 ln102

(3) 52ln2

(4) 2 ln10

(5) 5 1ln2 2

(6) 5 ln 22

(7) 110ln2

(8) 10ln 2

(9) None of the above


9. What is the energy stored in the capacitor of Question 8 when its voltage, Cv , has fallen to 2.0 V? The circuit is duplicated below.

(1) 10 Joules (2) 20 Joules (3) 5 Joules (4) 2 Joules (5) 0 Joules (6) -10 Joules (7) -20 Joules (8) -5 Joules (9) -2 Joules (10) None of the above


10. What is the current si through the voltage source if ( ) 3 Vsv t t= ? The initial charge on each capacitor at t = 0 is zero.

(1) 2 A (2) 8 A (3) 9 A (4) 36 A (5) 2 At (6) 8 At (7) 9 At (8) 36 At (9) None of the above


11. What is the voltage across the capacitor, ( )Cv t , for 1 st ≥ ? The switch has been closed for

a very long time and opens at 1 st = .

(1) 1

163 Vt

e−

− (2) 163 V

t

e−

(3) 1

243 Vt

e−

− (4) 243 V

t

e−

(5) 1

164 Vt

e−

− (6) 164 V

t

e−

(7) 1

244 Vt

e−

− (8) 244 V

t

e−



12. Assume an ideal op-amp. The Thevenin equivalent resistance, Rth, of the circuit when looking into terminals A and B is:

(1) 10thR = − Ω (2) 6thR = − Ω (3) 4thR = − Ω (4) 2.4thR = − Ω (5) 2.4thR = Ω (6) 4thR = Ω (7) 6thR = Ω (8) 10thR = Ω


13. Find R so that Vout/Vin = − 6. Assume that the Op-Amp is ideal.

(1) R = 0.1 kΩ

(2) R = 0.2 kΩ

(3) R = 0.3 kΩ

(4) R = 0.4 kΩ

(5) R = 0.5 kΩ

(6) R = 0.6 kΩ

(7) R = 1 kΩ

(8) R = 2 kΩ

(9) R = 3 kΩ



14. In the circuit below, find the phasor voltage, VL, across the 0.5H inductor. Assume sinusoidal steady-state.

(1) 12 V

(2) −12 V

(3) 6 − 𝑗𝑗6 V

(4) 6 + 𝑗𝑗6 V

(5) 185

+ 𝑗𝑗 65 V

(6) 185− 𝑗𝑗 6

5 V

(7) 65

+ 𝑗𝑗 185

V

(8) 65− 𝑗𝑗 18

5 V


15. We wish to build a HPF (high pass filter) with the structure shown below, such that for a sinusoidal signal with ω = 42 10× rad/sec, it gives |Vout (rms)| = 0.8 |Vin (rms)|. What circuit element should the one marked X be?

(1) A resistor, with R = 750 Ohms

(2) A resistor, with R = 12 kOhms

(3) A capacitor, with C = 0.2 F

(4) A capacitor, with C = 0.5 F

(5) A capacitor, with C = 2 F

(6) A capacitor, with C = 12 F

(7) An inductor, with L = 0.2 H

(8) An inductor, with L = 0.5 H

(9) An inductor, with L = 2 H

(10) An inductor, with L = 12 H


16. Find 0

( )(0 ) CC

t

dv tvdt +

+

=

′ = (the derivative of the voltage across the capacitor, vC(t), at t=0+)

(1) (0 )Cv +′ = - 6 V/s, independent of R

(2) (0 )Cv +′ = 6 V/s, independent of R

(3) (0 )Cv +′ = -2 V/s, independent of R

(4) (0 )Cv +′ = 2 V/s, independent of R

(5) (0 )Cv +′ = 3𝑅𝑅 V/s

(6) (0 )Cv +′ = - 3R V/s

(7) (0 )Cv +′ = 2 − 3𝑅𝑅

V/s

(8) (0 )Cv +′ = 2 + 𝑅𝑅3

V/s

(9) (0 )Cv +′ = 5𝑅𝑅

V/s



17. Which of the following is the closest match to the magnitude of the frequency response,

( ) ( )( )

= out

in

VH

Vω

ωω

, of the circuit below:

Vin Vout

1 Ω

1 Ω

0.01 F

ω (rad/s)

1

(1)

102 10

1 104 10

3 ω (rad/s)

1

(2)

102 10

1 104 10

3 ω (rad/s)

1

(3)

102 10

1 104 10

3

ω (rad/s)

1

(4)

102 10

1 104 10

3 ω (rad/s)

1

(5)

102 10

1 104 10

3 ω (rad/s)

1

(6)

102 10

1 104 10

3

ω (rad/s)

1

(7)

102 10

1 104 10

3 ω (rad/s)

1

(8)

102 10

1 104 10

3 ω (rad/s)

1

(9)

102 10

1 104 10

3

0 0 0

0 0 0

0 0 0


18. In sinusoidal steady-state at 318 MHz ( 92 10= ×ω rad/s), an antenna has impedance 50 10 AntZ j= − Ω . What inductor L could you put in series with the antenna to ensure

maximum power is delivered to the antenna for the source in the circuit below?

Note: [1 nH = 10-9 H]

(1) 0 nH (2) 5 nH (3) 10 nH (4) 15 nH

(5) 20 nH (6) 25 nH (7) 50 nH (8) 100 nH (9) ∞ nH


19. When resistor R is connected to a voltage source of ( ) 2 2 cos(10 ) Vsv t t= , the average power it consumes is P W (in sinusoidal steady-state). When it is connected to a DC voltage source, DCV , it consumes the same power, P W. What is the value of DCV ?

(1) 1 V (2) 2 V (3) 4 V (4) 1 2 V (5) 2 2 V (6) 4 2 V (7) 8 V (8) 10 V (9) 0 V


20. Find the power factor seen by the source if load 1 consumes P1 = 200 W with pf = 1 and load

2 has an apparent power of S2 = 1000 VA with pf = 0.6 lagging.

(1) pf = 0

(2) pf = cos(15°) lagging

(3) pf = cos(15°) leading

(4) pf = 1








21. In the circuit below, find the phasor loop current, I1. Assume sinusoidal steady-state.

(1) 20 A

(2) −20 A

(3) 2 + 𝑗𝑗6 A

(4) 2 − 𝑗𝑗6 A

(5) 4 + 𝑗𝑗8 A

(6) 4 − 𝑗𝑗8 A

(7) 10 + 𝑗𝑗10 A

(8) 10 − 𝑗𝑗10 A


22. For the following circuit in sinusoidal steady state, find the load impedance ZL so that the maximum amount of power will be transferred to the load ZL.

(1) ZL = 19.2+j14.4 Ω (2) ZL = 19.2 – j14.4Ω

(3) ZL = 14.4+j19.2 Ω (4) ZL = 14.4 – j19.2 Ω

(5) ZL = 30+j40 Ω (6) ZL = 30 – j40Ω

(7) ZL = 40+j30 Ω (8) ZL = 40– j30 Ω


23. The circuit below is in steady state. The phasor voltage and current at ω = 1 rad/s are as shown graphically. The value of the complex power 𝑺𝑺 absorbed by the resistor and inductor, combined, is given by:

(1) 16 VA

(2) 8 − 8√3𝑗𝑗 VA

(3) 8√3 + 8𝑗𝑗 VA

(4) 8 − 8𝑗𝑗 VA

(5) 16 + 16𝑗𝑗 VA

(6) −8√3 + 8𝑗𝑗 VA

(7) 16𝑗𝑗 VA

(8) 8 + 8√3𝑗𝑗 VA

(9) 8√3 − 8𝑗𝑗 VA

8

2

Vrms

Arms


24. What is the complex power, 𝑺𝑺𝑳𝑳, absorbed by the inductor in the series RLC circuit below?

[round to 3 decimal places]

(1) 0 ∠ 0° VA

(2) 0.080 ∠ -45° VA

(3) 0.080 ∠ 90° VA

(4) 0.128 ∠ -90° VA

(5) 0.128 ∠ 90° VA

(6) 0.800 ∠ -90° VA

(7) 0.800 ∠ 90° VA


𝑺𝑺𝑳𝑳

V (Note: V is not the same as Vrms)


25. In the figure below, Veff = 230 ∠ 0° V at 60 Hz and the complex power absorbed by the motor is 12.0 16.0MS j= + kVA. A capacitor is added as shown below to correct the power factor (pf). What reactive power, QC, must the capacitor have in order to correct the power factor from 0.60pf = lagging to pf = 0.80 lagging? [Assume Veff is unchanged after the addition of the capacitor.]

(1) QC = 0.0 kVAR

(2) QC = 16.0 kVAR

(3) QC = -16.0 kVAR

(4) QC = 9.0 kVAR

(5) QC = -9.0 kVAR

(6) QC = 12.0 kVAR

(7) QC = -12.0 kVAR

(8) QC = 28.0 kVAR

(9) QC = 7.0 kVAR

(10) QC = -7.0 kVAR


Potentially Useful Formulas (from Exam 2)

00

( )/( ) ( ) ( ) ( )+ − − τ = ∞ + − ∞ t tx t x x t x e , where THR Cτ = or

TH

LR

τ =

( ) ( )

00

2 20 1 1 0

( )( )

1( ) ( ) ( )

( , ) ( ) ( )2

LL

t

L L Lt

L L L

di tv t Ldt

i t i t v t dtL

LW t t i t i t

=

′ ′= +

= −

∫

( ) ( )

00

2 20 1 1 0

( )( )

1( ) ( ) ( )

( , ) ( ) ( )2

CC

t

C C Ct

C C C

dv ti t Cdt

v t v t i t dtC

CW t t v t v t

=

′ ′= +

= −

∫

1ln lnxx

− =


Potentially Useful Formulas (from Exam 3)

First order circuit: ( )ot t /ox(t) x( ) x(t ) x( ) e

+− − τ+ = ∞ + − ∞ , τ = L/R for LR circuit, τ = RC for RC circuit

Characteristic equation for series RLC: 2 2 R 1s bs c s s 0L LC

+ + = + + =

Characteristic equation for parallel RLC: 2 2 1 1s bs c s s 0RC LC

+ + = + + =

( )1 2s t s tx(t) x( ) Ae Be= ∞ + + --- Over-damped, 2 4 0b c− >

( ) tx(t) x( ) A Bt e−σ= ∞ + + --- Critically-damped, 2 4 0b c− =

( ) td dx(t) x( ) A cos t Bsin t e−σ= ∞ + ω + ω --- Under-damped, 2 4 0b c− <

2

21 2

b b 4cs ,s for s bs c 02

− ± −= + + = , where ( ) 1c LC −=

R / 2L (series)

b12 (parallel)

2RC

σ = =

o 1LC

ω =

2 21,2 os = −σ ± σ − ω

2

2 2d o

4c b2−

ω = = ω − σ


Potentially Useful Formulas (since Exam 3)

0

1 ( ) cos( ) cos( ) cos( )2

Tm m

ave V I eff eff V I rms rms V IV IP p t dt V I V I

T= = θ − θ = θ − θ = θ − θ∫

* * *12 m m eff eff P jQ= = +S = V I V I VI = VA

2 2cos( )V I

P PpfS P Q

= = = θ − θ+

, ( )V Ipfa = θ − θ

ECE 20100 – Fall 2014 Final Exam December 19, 2014...ECE 20100 Final Exam, December 19, 2014 9....

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