ECE 20100 Fall 2016 Final Exam - WeeklyJoys 20100 – Fall 2016 Final Exam ... Answer (7) Solution:...
Transcript of ECE 20100 Fall 2016 Final Exam - WeeklyJoys 20100 – Fall 2016 Final Exam ... Answer (7) Solution:...
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ECE 20100 – Fall 2016
Final Exam
December 12, 2016
Sections (include on scantron)
Hosseini (9:30) – 0002 Peleato-Inarrea (3:30) – 0004 Michelusi (1:30) – 0005
Qi (10:30) – 0011 Cui (8:30) – 0012
Peroulis (11:30) – 0013 Kildishev (1:30) – 0014
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 and will be dealt with according to the policy in your section. 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.
By signing the scantron sheet, you affirm you have not received or provided assistance
on this exam.
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Question 1
A resistor of unknown resistance has a 4 A current passing through it, as shown. The power
absorbed by the resistor is 12 W. Find the voltage (in V) across the resistor for these conditions.
(1) 1 V
(2) 2 V
(3) 3 V
(4) 4 V
(5) -1 V
(6) -2 V
(7) -3 V
(8) -4 V
(9) None of the above
Answer (7)
Solution:
We can write that I = - 4 A to follow passive sign convention. As P = V I, then for given
signs, V = 12 W/ (-4 A) = - 3V.
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Question 2
For the circuit below, find the voltage, VBC (in V):
(1) 2
(2) 8
(3) 10
(4) 12
(5) 20
(6) -2
(7) -8
(8) -10
(9) -12
(10) None of the above
Answer (8)
Solution
I1 = 5 V /(3 + 2) = 1 A; VB = 1 A 2 = 2 V;
VC = 4 A (2 + 1) = 12 V; VBC = VB – VC = 2 – 12 = – 10 V.
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Question 3
Find the node voltage, VA (in V).
(1) 1.2
(2) 2.4
(3) 3.6
(4) 4.0
(5) 5.0
(6) 6.8
(7) 7.2
(8) 8.0
(9) None of the above
Answer (7)
Solution
Using nodal analysis for node A:
2103 ,
2 1
10where
2
102
10 2Therefore 3 2 10
2 1
10 6 4 20
36So = =7.2
5
A xA
Ax
AA
AA
A A
A
V iV
Vi
VV
VV
V V
V
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Question 4
Using source transformations or any other analysis technique, find the value of the Norton
Equivalent current source (in A) for the circuit below.
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5
(6) 6
(7) 7
(8) 8
(9) 9
(10) None of the above
Answer (4)
SOLUTION
Transform 6A || 4 into 24V connected in series with 4 +4 then add two current sources
(3A+1A) with 8 || 8 to obtain (4A).
Alternatively, one can short the two terminals, then the 8 resistor is shorted, and the 6A
current source is split even (3 A each) for the two 4 resistors. This 3 A plus the other current
source (1 A) is the current flowing through the shorted terminals. So the short circuit current is
4 A.
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Question 5
The linear circuit below consists of resistors and sources only. Experiments were performed to
evaluate circuit parameters. Two current/voltage relationships were found to be:
v = 10 V, iL = 0 A
v = -10 V, iL = 2 A
Find the value of the Thevenin equivalent resistance, RTH, for the linear circuit (in ).
(1) 10
(2) 2
(3) 30
(4) 4
(5) 5
(6) 20
(7) 40
(8) 8
(9) None of the above
Answer (1)
SOLUTION
v = voc - Rth iL. At iL = 0 voc = 10; and for iL = 2 A, -20 = - 2 Rth; Rth = 10 [
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Question 6
The current through an inductor when connected to a resistor, R, in a zero-input (undriven), first-
order RL circuit is:
iL(t) = 2 e-3t A t 0
The instantaneous stored energy in the inductor at t = 0 is known to be 400 mJ. Find the value of
the resistance to which the inductor is connected (in ).
1) 0.1
2) 0.2
3) 0.3
4) 0.4
5) 0.5
6) 0.6
7) 0.7
8) 0.8
9) 0.9
10) none of the above
Answer (6)
SOLUTION
As 400 mJ = 0.5LiL2 and = L/R = 1/3 s-1; iL (0) = 2 A; L = 0.2 H and R = 0.6
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Question 7
The capacitor voltage in a driven, first-order RC circuit with a constant voltage source is:
vC(t) = 5 - 8 e-4t V t 0
Find the value of vC(0-) (in V).
1) 1
2) 2
3) 3
4) 4
5) 5
6) -1
7) -2
8) -3
9) There is insufficient information
10) none of the above
Answer (8)
SOLUTION
vC(t) = 5 - 8 e-4t V
vC(0-) = vC(∞) – (vC(to+) - vC(∞)) V and since vC(∞) = 5 V; vC(0-) = – 3 V.
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Question 8
In the circuit below, the switch has been closed for a long time. At t = 0 sec, the switch opens.
Assuming a solution for the capacitor voltage of the form,
vC(t) = A cos(t) + B sin(t) V
find the closest value for the constant B.
(1) -0.1
(2) -0.2
(3) -0.3
(4) -0.4
(5) -0.5
(6) -0.6
(7) -0.7
(8) -0.8
(9) None of the above
Answer (3)
SOLUTION
Since (the inductor is a short) vC(0+) = A cos(0) = 0 V; then constant A = 0.
So at t = 0, iC = C dvC(t)/dt = C B (1/√LC) = – iL = – 0.5 0.2 A; and finally B = – 0.3 .
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Question 9
The switch in the circuit below has been opened for a long time. It closes at t = 0 s. Find dt
dvc at
t = 0+ (in V/s).
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5
(6) 6
(7) 7
(8) 8
(9) None of the above
Answer (6) Solution:
0
1Since , 0
At 0 , all energy stored in the capacitor and inductor has been dissipated,
So 0 0, 0 0.
Due to the continuity, 0 0 0, 0 0 0
After the switch is cl
C C
C C
t
C L
C C L L
dv t dv tC i t i
dt dt C
t
v i
v v i i
osed, at the moment of 0 , the capacitor can be viewed as
a voltage source of 0 (short circuit), and the inductor can be viewed as a current source
of 0 (open circuit). Therefore the 1 resistor is
t
0
shorted by the capacitor, and
12 V0 3 A
4
1 3and 0 6 V/sec.
0.5
C
C
C
t
i
dv ti
dt C
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Question 10
Find the characteristic equation for the circuit below for t > 0.
(1) s2 + 0.1s + 0.2 = 0
(2) s2 + 4s + 0.2 = 0
(3) s2 + 8.33s + 0.833 = 0
(4) s2 + 2s + 0.2 = 0
(5) s2 + 16.67s + 0.833 = 0
(6) s2 + 0.2s + 0.833 = 0
(7) s2 + 5s + 1 = 0
(8) None of the above
Answer (4) SOLUTION: Introduce node A shown below.
2
2
Choose as the variable, we have
8KCL at node A: 2 3 ; therefore: 2 10 3
0.1
5 10 0 So the characteristic equation i
I IL A
L L
A A A LL L
L LL
di dii V L
dt dt
di did d
d V V dV didt dti i
dt dt dt dt dt
d i dii
dt dt
2 2s: 5 10 1 0, or 2 0.5 0s s s s
A
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Question 11
In the circuit below, find the power delivered by the current source (in mW).
(1) 10
(2) 20
(3) 30
(4) 40
(5) 50
(6) 60
(7) 70
(8) 80
(9) None of the above
Answer (6)
SOLUTION
The power delivered by the 10 mA current source is: P=(VA)(10) mW.
Since (VA – v-)/100 = 10 mA, where v- = v+ = 5 V;
VA = 6 V
P=(VA)(10) mW = 60 mW
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Question 12
Given the phasor currents shown below, determine the phasor current (in A) through the
capacitor.
(1) j
(2) 2j
(3) 3j
(4) 4j
(5) 5j
(6) 3 + 4j
(7) 3 + 2j
(8) 3 + 5j
(9) None of the above
Answer (2)
SOLUTION
Is = IC + IL + IR; IC = 2j;
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Question 13
The input impedance in the circuit below is 1 -60o . Find the instantaneous power generated
by the source (in W).
(1) 100 + 50 cos(10t - 60)
(2) 100 + 25 cos(10t - 120)
(3) 25 + 25 cos(10t - 60)
(4) 25 + 50 cos(10t - 120)
(5) 50 + 25 cos(10t - 60)
(6) 50 + 50 cos(10t - 60)
(7) 75 + 50 cos(10t - 120)
(8) 75 + 25 cos(10t - 60)
(9) 100
(10) None of the above
Answer (4)
SOLUTION
iIN = 10 -90o / 1 -60o = 10 -30o = 10cos(5t - 30)
=> pIN (t) = 10cos(5t - 90)10cos(5t - 30) = 0.5[100cos(60) + 100cos(10t - 120)]
= 25 + 50 cos(10t - 120);
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Question 14
Find the average power (in W) absorbed by the 6 resistor in the circuit shown below.
(1) 1
(2) 2
(3) 3
(4) 0.5
(5) 0.33
(6) 6
(7) 12
(8) 1.5
(9) None of the above
Answer (1)
SOLUTION
is(t) = t , 0 < t < 1 s; 0, is(t) = t, 1 < t < 2 s;
Pave = 1
2∫ 𝑡2𝑅𝑑𝑡 =
1
0
1
6136 =1W
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Question 15
Find the effective (rms) voltage for the waveform shown below (in V).
(1) 1
(2) 2
(3) √3
(4) 4
(5) √2
(6) 6
(7) 2√2
(8) (2/3)√3
(9) None of the above
Answer (5)
SOLUTION
VRMS = √∫ 𝑣(𝑡)23
0
3= √
2+4
3= √2
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Question 16
Find the reactive power (in VAR) for the capacitor (rms) voltage given a source voltage of
v(t) = 50√2 sin(10 t) V.
(1) 25
(2) -25
(3) 50
(4) -50
(5) 100
(6) -100
(7) 50√2
(8) -50√2
(9) None of the above
Answer (2)
SOLUTION
v(t) = 50√2 sin(10 t) V
𝑣(𝑡) = 50√2 cos(10𝑡 − 90°) ; 𝑽𝒆𝒇𝒇 = −𝑗50 V;
𝑰𝒆𝒇𝒇 =(−𝑗50 V)
−𝑗100Ω= 0.5A; 𝐼𝑚[ 𝑽𝒆𝒇𝒇𝑰𝒆𝒇𝒇
∗ ] = -25 VAR.
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Information for next 2 problems
A voltage source supplies power to four machines. The complex power absorbed by each
machine is:
S1 = 75 + 60 j S2 = 110 + 80 j
S3 = 125 + 100 j S4 = 90 + 60 j
Question 17
Find the apparent power (in VA) delivered by the source (i.e. │SS│).
(1) 100
(2) 200
(3) 300
(4) 400
(5) 500
(6) 600
(7) 700
(8) 800
(9) None of the above
Answer (5) SOLUTION
│SS│ = | 75 + 110 + 125 + 90 + 60 j + 60 j + 80 j + 100 j | = | 400 + j300| = 500 VA
Question 18
Find the power factor for the source.
(1) 0.70 leading
(2) 0.75 leading
(3) 0.80 leading
(4) 0.85 leading
(5) 0.70 lagging
(6) 0.75 lagging
(7) 0.80 lagging
(8) 0.85 lagging
(9) None of the above
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Answer (7) SOLUTION: pf = 400/500 = 0.8 lagging, since Q = Im[Ss] > 0.
Question 19
In the circuit below, the complex power absorbed by loads ‘1’ and ‘2’ are,
S1 = 12 kW – j16 kVAR
S2 = 8 kW + j 6 kVAR
If a capacitor or an inductor is added to the circuit (as shown in dashed line), what is proper
selection below that gives a power factor of 0.95 leading:
(1) Capacitor: C = 0.116 F
(2) Capacitor: C = 0.584 F
(3) Capacitor: C = 1.713 F
(4) Capacitor: C = 2.919 F
(5) Inductor: L = 0.116 H
(6) Inductor: L = 0.584 H
(7) Inductor: L = 1.713 H
(8) Inductor: L = 2.919 H
(9) None of the above
Answer (6)
SOLUTION
Ss = 20 kW – j10 kVA; 𝑄 = −𝑃√1
𝑝𝑓2 − 1; 𝑄𝑛𝑒𝑤 = −20√1
0.952 − 1 = −6.5737kVAR → ΔQ =
3.4263 =10−3100100
5L; L = 0.5837.
1002 cos(5t) V
+ _ 1 2 C or
L
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Question 20
In the circuit below, choose the load impedance for maximum power transfer to the load resistor
Rload.
(1) 1 + j
(2) 1 – j
(3) 1 + 2j
(4) 1 – 2j
(5) 2 + 4j
(6) 2 – 4j
(7) 2 + 2j
(8) 2 – 2j
(9) None of the above
Answer (5)
SOLUTION
VOC = -40j V; isc = 10 (1/2)/(1/2 - 1/j4) = 20(2 - j) = 4(2 - j);
Zth = -40j/4(2 - j) = -2j(2 - j) = (2 - 4j) -> ZL = (2 - 4j)* = (2 + 4j)
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Potentially Useful Formulas
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