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ENGS2613 Intro Electrical Science Week 15 Dr. George Scheets
Read: 10.1 – 10.4, 10.6 Problems: 10.2, 4, 5, 6, 11, 12, 18, 41, & 43 Quiz 8 Results Hi = 10, Lo = 1.5, Ave = 7.29, Deviation = 2.04 Versions A & B +1 point Take Home Project Results Hi = 30, Lo = 0, Ave = 28.67, Deviation = 2.34
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Final Exam Closed Book & Neighbor
Up to 4 unattached sheets of notes OK Keep on your desk Have them on hand before you start No Textbooks Calculator OK Smart Phones NOT OK Reaching down for something in backpack? Get permission first
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Final Exam Friday, 12 May, 1000 – 1150 4 pages, comprehensive, 1 hour & 50 minutes R circuit, Opamp, Phasor, Something Else 10 pt Extra Credit Thevenin Equivalent of Circuit with Dependent Source No TA Office Hours during Finals Week Though they will respond to ed questions Final Exam Review Session Wednesday, 10 May, 7:00 – 9:00 pm ES 317 Instructor Available Thursday, 11 May, 10:00 am – 5:00 pm
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What to Study? Electrical Science Notes & Homework Power Point
Problems Readings Anything in the circles is fair game. Overlaps more likely.
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Calculus You Might Need on Final
Integrals or Derivatives (with respect to time t) of… time raised to some power (i.e. tα) sines or cosines; e.g. sin(2πft) et raised to some power (i.e. eαt) integral of form 1/x = ln(x)
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Quiz 7A Transfer Function H(f)
Vout = VinH(360) = (8∟328º)∙ (1.299∟-1.52º) = 10.39∟326.5 º
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Quiz 7B Transfer Function H(f)
Vout = VinH(300) = (4∟21º)∙ (0.181∟74.21º) = 0.724∟95.21 º
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Quiz 7C Transfer Function H(f)
Vout = VinH(400) = (6∟101º)∙ (0.786∟38.18º) = 4.716∟139.2 º Null occurs when Zeq denominator = 0 at f = Hz Here jωL || 1/(jωC) = ∞
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Increasing R
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RLC Bandstop Filter Vout/Vin ≡ H(f) H(f) = R/(L||C + R) EKG or EEG
Weak Signals contaminated by 60 Hz Run thru Narrow Bandstop to eliminate 60 Hz
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Eyeballing Filter Type
Examine at ω = 0 & ω = ∞ rads/sec Output at ω = 0 > Output at ω = ∞ Probably Low Pass Output at ω = 0 < Output at ω = ∞ Probably High Pass Output at ω = 0 close or equal to zero Output at ω = ∞ close or equal to zero Probably Band Pass Output at ω = 0 ≈ Output at ω = ∞ both > 0 Probably Band Reject
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2R & 2C Circuit Vout = VinH(350) = (5∟239º)∙ (0.15∟-51.65º) = 0.75∟187.4 º
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What kind of volts? Have a Time Varying Signal? Specify type of volts.
Vrms Take square root of time average of waveform squared Delivers same average power as DC Voltage of same value Vpeak Max + or – swing from average voltage Average Voltage = Volts DC Vpeak-to-peak Max + swing minus Max - swing
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Power P = VI (for DC) p = vi (for a time varying signal)
p = Vpcos(ωt + θv)Ipcos(ωt + θi) = [VpIp/2]{cos(θv - θi) + cos(2ωt + θv + θi)} for a sinusoid waveform Power Factor Average Power Instantaneous Power
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Power in a Resistor p = vi (for a time varying signal)
p = Vpcos(ωt + θv)Ipcos(ωt + θi) = [VpIp/2]{cos(0) + cos(2ωt + 2θv)} for a sinusoid waveform Power Factor Average Power Instantaneous Power
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100 Hz Cosine thru Capacitor
v = C (dv/dt) Current leads Voltage by 90º =1/4 wavelength = (1/4)(1/100) = 1/400 second
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ELI the ICE man. E = IR Voltage leads current Inductor
E = voltage here Voltage leads current Inductor Current leads voltage Capacitor
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Capacitor Current, Voltage, & Power
+ - On 1st Power 1/2 cycle, v and i are positive, and capacitor is absorbing power. On the 2nd Power 1/2 cycle, i goes negative and the current direction flips. The capacitor is now releasing the absorbed power.
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Power in a Capacitor p = vi (for a time varying signal)
p = Vpcos(ωt + θv)Ipcos(ωt + θi) = [VpIp/2]{cos(-90º) + cos(2ωt + 2θv + 90º)} for a sinusoid waveform Power Factor Average Power Instantaneous Power
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Power p = vi (for a time varying signal)
p = Vpcos(ωt + θv)Ipcos(ωt + θi) = [VpIp/2]{cos(-90º) + cos(2ωt + 2θv + 90º)} for a sinusoid waveform 0 < Power Factor < 1 Current can't lead or lag voltage by more than 90º Power Company wants this = 1 Power Factor Average Power Instantaneous Power
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Maximizing Power Delivered to Load
Both Adjustable Decrease Rsource & Rload as much as possible Cancel Reactive Component Neither Adjustable Use Capacitance or Inductance & Cancel Reactive Component Source Impedance Adjustable, Load Not? Decrease Rsource as much as possible Load Impedance Adjustable, Source Not? Set Zload = Z*source
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Reducing Line Impedance to 0
250/0ᵒ Load 39 + j26 46.87/33.69ᵒ Power to load was watts When Zline = 1 + j4 Average load power now = watts Power to load was watts When Zline+capacitor = 1 – j26
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Reducing Line Resistance to 0 & Canceling Reactance
0 - j26 250/0ᵒ Load 39 + j26 46.87/33.69ᵒ Power to load = watts when Zline = 1 + j4 Power to load = watts when Zline = 0 Power to load was watts Zline+capacitor = 1 – j26 Power to load = watts when Zline = -j26
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