Download EE-301 Name (Print): Spring AY 2016 6 Week Exam 22 FEB 2016

EE-301
Spring AY 2016
6 Week Exam
22 FEB 2016
Time Limit: 50 Minutes
Name (Print):
Section
This exam contains 9 pages (including this cover page) and 5 problems. Check to see if any pages
are missing. Enter all requested information on the top of this page, and put your initials on the
top of every page, in case the pages become separated. The course equation sheet is attached to
the last page of the examination.
The examination is closed-book/closed-notes. An approved calculator may be used on the exam.
Notes, equations, formulas, example problems, etc., may not be stored in the memory of your your
calculator. You may not use a laptop computer, cell phone or any other device with wireless connectivity in lieu of an approved calculator (TI NSPIRE for example).
You are required to show your work on each problem. The following rules apply:
• Organize your work, in a reasonably neat and
coherent manner, in the space provided. Work
scattered all over the page without a clear ordering is difficult to grade and as a result may receive
very little credit.
• Show your work. A correct answer, unsupported
by calculations, explanation, or algebraic work will
receive no credit; an incorrect answer supported by
substantially correct calculations and explanations
may receive partial credit.
• Express all answers using proper Engineering
prefixes and units. Express all numeric answers
to at least 3 significant digits.
• If you need more space, use the back of the pages;
clearly indicate when you have done this.
• Similar makeup exams will be taken by individuals
at different times. No communication is permitted
concerning the content of this exam with any individual who has not yet taken the examination.
Do not write in the table to the right.
Problem
Points
1
25
2
30
3
20
4
20
5
5
Total:
100
Score
EE-301
6 Week Exam - Page 2 of 9
PAGE
INTENTIONALLY
BLANK
22 FEB 2016
EE-301
6 Week Exam - Page 3 of 9
I0
20V
30Ω
I1
−V2 +
60Ω
22 FEB 2016
+
V1
−
10Ω
Figure 1: Problem 1
1. (25 points) Consider the circuit in Fig. 1,
(a) (5 points) Determine the total resistance, RT otal , seen by the source.
(b) (5 points) Determine the source current, I0 .
60Ω
EE-301
6 Week Exam - Page 4 of 9
22 FEB 2016
(c) (5 points) Determine the current I1 through the 30Ω resistor using the current divider
rule.
(d) (5 points) Determine the voltage, V1 , across the 60Ω resistor.
(e) (5 points) Determine the power dissipated in 10Ω resistor.
EE-301
6 Week Exam - Page 5 of 9
100Ω
22 FEB 2016
50Ω
100Ω
100Ω
RLoad
200V
50Ω
Figure 2: Problem 2.
2. (30 points) Determine the Thevénin equivalent circuit for the circuit displayed in Fig. 2.
(a) (10 points) Determine the Thevénin equivalent voltage, ET h .
(b) (10 points) Determine the Thevénin equivalent resistance, RT h .
(c) (10 points) If RLoad = 300Ω, determine the load power, PLoad .
EE-301
6 Week Exam - Page 6 of 9
10Ω
4Ω
+
20V
22 FEB 2016
V
−
I
5Ω
6Ω
30V
Figure 3: Problem 3.
3. (20 points) Consider the circuit in Fig.3 , use Nodal Analysis to determine:
(a) (15 points) The voltage, V , across the 5Ω resistor.
(b) (5 points) The current, I, through the 4Ω resistor.
EE-301
6 Week Exam - Page 7 of 9
5Ω
22 FEB 2016
a
50V
3A
30Ω
10Ω
15Ω
b
Figure 4: Problem 4.
4. (20 points) For the circuit in Fig. 4,
(a) (10 points) Perform a source conversion of the 50V battery and draw the circuit with the
converted source.
(b) (10 points) Determine the voltage across the 30Ω resistor.
EE-301
6 Week Exam - Page 8 of 9
22 FEB 2016
PIN = 200W
η1 = 0.9
POUT
η2 = 0.8
η3 = 0.5
Figure 5: Problem 5.
5. (5 points) Given the system in Fig. 5, determine the output power, POU T , in Watts.
Voltage, current, resistance
Elementary charge
q = 1.6 × 10−19 C
V=
Capacity (A. hr)
Battery life (hr) =
Discharge rate (A)
W
Q
KVL
�
�
closed loop
ρ𝑙𝑙
A
2
P = VI = I2 R = V �R
Max power transfer
R L = R Th
Pmax =
𝑖𝑖C (𝑡𝑡) = C
W or E =
2
Transient analysis
τ = RC
−𝑡𝑡�
τcharging
𝑣𝑣C (𝑡𝑡) = Vfinal �1 − e
−𝑡𝑡�
τdischarging
�
𝑣𝑣C (𝑡𝑡) = Vfinal + (Vinitial − Vfinal ) e
Electrical
−𝑡𝑡�
τ (general)
Pmech loss=Tmech lossω
Pd = Pin − Pelec loss = Pmech loss + Pout
VDC − Ia R a = Ea
R3
R1 R2
R1 +R2
+⋯�
CDR
−1
(parallel)
(2 parallel)
R2
� (2 parallel)
R1 +R2
d𝑣𝑣C (𝑡𝑡)
d𝑡𝑡
𝑣𝑣L (𝑡𝑡) = L
1 2
CV
2
W or E =
Inductors
τ = L�R
−𝑡𝑡�
τdischarging
�
d𝑖𝑖L (𝑡𝑡)
d𝑡𝑡
1 2
LI
2
(charging)
(discharging)
−𝑡𝑡�
τ (general)
RPM
ω = 2π �
� rad/sec
60
Pin = Pelec loss + Pmech loss + Pout
VDC
1
Linear Motor
Mechanical Loss
Ia
+
Inductors
Pout=Tloadω
1 hp = 746 W
Ra
R2
RT =
𝑖𝑖L (𝑡𝑡) = Ifinal + (Iinitial − Ifinal ) e
DC Motor
1
I1 = IEQ �
𝑖𝑖L (𝑡𝑡) = Iinitial × e
(discharging)
Pd=Tdω
Pelec loss=Ia2R
R1
−𝑡𝑡�
τcharging
Mechanical
Electrical Loss
+
𝑖𝑖L (𝑡𝑡) = Ifinal �1 − e
(charging)
Pd=EaIa=KvIaω
Pin=VDCIa
1
(series)
R EQ
IX = IEQ �
�
RX
Capacitors
Pout
Pin
Capacitors
RT = �
RX
VX = VEQ �
�
R EQ
node
VTh
4 × R Th
R T = R1 + R 2 + R 3 + ⋯
VDR
I=0
η=
Resistances
V=I×R
� Iin = � Iout
Power
𝑣𝑣C (𝑡𝑡) = V𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 × e
R=
Ohm’s law
KCL
V=0
� Erise = � Vdrop
Q
t
I=
DC
Ea
VB
I
R
Fd = ILB (Newton)
Eind = uBL (Volts)
VB − Eind = IR
Pin = VB I
Pout = Eind I = FLoad u
Eind