ece 20100 spring 2015 final exam - weeklyjoys · ece 20100 – spring 2015 final exam may 7, 2015...
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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
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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
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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
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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
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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
(10) None of the above
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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
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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
(10) None of the above
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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
(2) 3cos V, 600 TH THt R
(3) 3cos V, 900 TH THt R
(4) 2cos V, 450 TH THt R
(5) 2cos V, 600 TH THt R
(6) 2cos V, 900 TH THt R
(7) 1cos V, 450 TH THt R
(8) 1cos V, 600 TH THt R
(9) 1cos V, 900 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
(10) None of the above
Amps
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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
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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(∞))]
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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