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EE-301 Name (Print): Spring AY 2021 12-Week Exam Time Limit: 50 Minutes Section
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This exam contains 6 pages (including this cover page) and 5 problems. The Equation Sheet, which is attached to the back of this exam, is not part of the 6 pages. Check to see if any of the 6 pages are missing, or if you are missing the Equation Sheet. 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 examination is closed-book/closed-notes. An approved calculator may be used on the exam. The TI-36X is approved. All other calculators need to be verified with your instructor. Notes, equations, formulas, example problems, etc., may not be stored in the memory of your calculator. You may not use a laptop computer, cell phone, or any other device with wireless connectivity in lieu of an approved calculator.
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 to the right side of the decimal point.
• 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 Score
1 25
2 22
3 13
4 13
5 27
Total: 100
EE-301 12-Week Exam - Page 2 of 6 Mar 2021
1. (25 pts) The waveforms below represent the voltage and current of the load (ZLD) in the circuit
below. Use the waveform information and circuit to answer the following question
a) (5 pts) Determine the frequency.
b) (8 pts) Calculate the RMS phasors of VLD and ILD. Treat ILD as the reference phasor.
c) (3 pts) Name the component(s) represented by 𝐙𝐙LD (Resistor, Capacitor, and/or Inductor).
d) (9 pts) Calculate the source voltage, 𝐄𝐄S (in polar form).
EE-301 12-Week Exam - Page 3 of 6 Mar 2021
2. (22 pts) In the circuit below angular frequency is 800 rad/sec.
a) (12 pts) Perform a source conversion on the current source and redraw the circuit labelingall sources and impedances.
b) (10 pts) Calculate the phasor current going through the inductor (in polar form).
EE-301 12-Week Exam - Page 4 of 6 Mar 2021
3. (13 pts) In the circuit below, calculate the phasor current through the 8kΩ inductor, IL. Express in
polar form.
EE-301 12-Week Exam - Page 5 of 6 Mar 2021
4. (13 pts) Based on the given circuit:
a) (5 pts) Calculate the cutoff frequency, fC, for this filter.
b) (5 pts) Determine the gain factor, AV, at the cutoff frequency.
c) (3 pts) What type of filter is this? (a) High Pass (b) Low Pass (c) Band Pass (d) Band Stop
EE-301 12-Week Exam - Page 6 of 6 Mar 2021
5. (27 pts) In the circuit below the frequency is 60 Hz.
a) (10 pts) Calculate and draw the Total Power Triangle without the Added component. Include
PT, QT, ST, and θS.
b) (7 pts) Calculate the phasor current from the source (IS). Express in polar form.
c) (10 pts) Name and calculate the value of the Added Component on the above circuit needed to correct the power factor to unity. (Hint: either an inductor in Henries or a capacitor in Farads).
DC
Elementary charge q = 1.6 × 10−19C
Battery life (hr) = Capacity (A. hr)
Discharge rate (A)
Voltage, current, resistance Ohm’s law Resistances
V =WQ I =
Qt R =
ρ𝑙𝑙A V = IR
RT = R1 + R2 + R3 + ⋯ (series)
RT = 1R1
+ 1R2
+ 1R3
+ ⋯−1
(parallel)
RT = R1R2R1+R2
(2 parallel)
KVL KCL VDR CDR
Vclosed loop
= 0
Erise = Vdrop
Inode
= 0
Iin = Iout VX = VEQ
RX
REQ
IX = IEQ REQ
RX
I1 = IEQ R2
R1+R2 (2 parallel)
Power Capacitors Inductors
P = VI = I2R =𝑉𝑉2
𝑅𝑅 η =PoutPin
𝑖𝑖C(𝑡𝑡) = Cd𝑣𝑣C(𝑡𝑡)
d𝑡𝑡 𝑣𝑣L(𝑡𝑡) = Ld𝑖𝑖L(𝑡𝑡)
d𝑡𝑡
Max power transfer
W or E = 12 CV2 W or E =
12 LI2
RL = RTh
Pmax =VTh2
4RTh
Transient analysis Capacitors Inductors
τ = RC τ =𝐿𝐿𝑅𝑅
𝑣𝑣C(𝑡𝑡) = Vfinal + (Vinitial − Vfinal) e−𝑡𝑡 τ (general) 𝑖𝑖L(𝑡𝑡) = Ifinal + (Iinitial − Ifinal) e−𝑡𝑡 τ (general) DC Motor Linear Motor
Electrical Mechanical
Pin=VDCIa
Electrical Loss
Pd=EaIa=Tdω=KvIaω Pout=Tloadω1 hp = 746 W Td=KvIa
Pelec loss=Ia2R
Mechanical Loss
Pmech loss=Tmech lossω
Ea=Kvω
VB Eind
RI
Fd = ILB (Newton)
Eind = uBL (Volts)
VB − Eind = IR
Pin = VBI
Pout = EindI = FLoadu
Pin = Pelec loss + Pmech loss + Pout
Pd = Pin − Pelec loss = Pmech loss + Pout
VDC Ea
RaIa
VDC − IaRa = Ea
ω = 2π RPM
60 rad/sec
AC Ohm’s law Impedances Sinusoids
𝐕𝐕 = 𝐈𝐈 𝐙𝐙
𝐙𝐙 = R + jXL − jXC = R + jX 𝐙𝐙T = 𝐙𝐙1 + 𝐙𝐙2 + 𝐙𝐙3 + ⋯ (series) 𝑥𝑥(𝑡𝑡) = XPeaksin (ω𝑡𝑡 ± θ)
XL = ωL = 2π𝑓𝑓L 𝐙𝐙T = 1𝐙𝐙1
+ 1𝐙𝐙2
+ 1𝐙𝐙3
+ ⋯−1
(parallel)
𝐗𝐗 = XRMS∠± θ XRMS =𝑋𝑋𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃√2
XC =1𝜔𝜔𝜔𝜔
=1
2𝜋𝜋𝑓𝑓𝜔𝜔 𝐙𝐙T = 𝐙𝐙1𝐙𝐙2
𝐙𝐙1+𝐙𝐙2 (2 parallel) ω = 2π𝑓𝑓 T =
1𝑓𝑓
θ =ΔtT
× 360o
KVL KCL VDR CDR
𝐕𝐕closed loop
= 0
𝐄𝐄rise = 𝐄𝐄drop
𝐈node
= 0
𝐈in = 𝐈out 𝐕𝐕X = 𝐕𝐕EQ
𝐙𝐙X𝐙𝐙EQ
𝐈X = 𝐈EQ
𝐙𝐙EQ𝐙𝐙X
𝐈1 = 𝐈EQ 𝐙𝐙2
𝐙𝐙1+𝐙𝐙2 (2 parallel)
Single-phase power Max power transfer Transformers
|S| = P2 + Q2 = |V||I| 𝐒𝐒 = P + jQL − jQC = P + jQ 𝐙𝐙L = 𝐙𝐙Th∗ a =
Npri
Nsec=
EpriEsec
=IsecIpri
𝐒𝐒 = 𝐕𝐕 𝐈∗
𝐒𝐒 = |I|𝟐𝟐𝐙𝐙
𝐒𝐒 =|V|𝟐𝟐
𝐙𝐙∗
P = |V||I| cos(θ)
P = |I|𝟐𝟐R
P =|V|2
R
Q = |V||I| sin(θ)
Q = |I|𝟐𝟐X
Q =|V|2
X
Pmax =|VTh|2
4RTh 𝐙𝐙pri−reflected = a2 𝐙𝐙sec
Fp = cos(θ𝐒𝐒) =P
|S| θ𝐒𝐒 = θ𝐙𝐙 = θ𝐕𝐕 − θ𝐈𝐈 Three-phase connections
Three-phase power Y Δ
ST = √3|VLine||ILine| = 3SΦ 𝐈Line = 𝐈Phase 𝐕𝐕Line = 𝐕𝐕Phase
PT = √3|VLine||ILine| cos(θ) = 3PΦ 𝐕𝐕Line = 𝐕𝐕Phase(√3∠30o) 𝐈Line = 𝐈Phase(√3∠−30o)
QT = √3|VLine||ILine| sin(θ) = 3QΦ Δ to Y transformation: 𝐙𝐙Y = 𝐙𝐙Δ3
Three-phase AC generator
ωrotor = 2Poles
2π𝑓𝑓 (rad/sec)
Np = 120𝑓𝑓Poles
(RPM)
Mechanical Electrical
Pin = Tinω 1 hp = 746 W
Mechanical LossElectrical Loss
Pelec loss=3ILine2 Rarmature
Pin = Pmech loss + Pelec loss + Pout Series Resonant circuits (Q ≥ 10) Filters
Low pass (RC) High pass (RC)
ωs =1
√𝐿𝐿𝜔𝜔 𝑓𝑓s =
12𝜋𝜋√𝐿𝐿𝜔𝜔
=𝜔𝜔𝑠𝑠2𝜋𝜋 𝑓𝑓c =
12𝜋𝜋𝑅𝑅𝜔𝜔
Qs =𝑋𝑋𝑅𝑅
=ωsL𝑅𝑅
=1
ωsRC θ = − tan−1
𝑅𝑅Xc θ = tan−1
XcR
BW =𝑓𝑓sQs
= 𝑓𝑓2 − 𝑓𝑓1 Av =VoVi
=1
RXc2
+ 1
Av =VoVi
=1
XcR
2+ 1
𝑓𝑓1,2 = 𝑓𝑓s ±BW
2
Pout = √3VLineILinecos(θ)
ω = 2π RPM
60 rad/sec