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.

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• 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.

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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