lecture 12 electromagnetic oscillations and alternating current chp. 33 cartoon -. opening demo -...
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![Page 1: Lecture 12 Electromagnetic Oscillations and Alternating Current Chp. 33 Cartoon -. Opening Demo - Warm-up problem Physlet Topics –LC Circuit Qualitatively](https://reader036.vdocument.in/reader036/viewer/2022062309/56649f175503460f94c2e49b/html5/thumbnails/1.jpg)
Lecture 12 Electromagnetic Oscillations and Alternating Current Chp. 33
• Cartoon -. Opening Demo - Warm-up problem
• Physlet
• Topics
– LC Circuit Qualitatively – Electrical and Magnetic energy oscillations– Alternating current– \Pure R and L, circuti– Series RLC circuit– Power and Transfomers
• Demos– LR circuit
– Series LRC circuit
![Page 2: Lecture 12 Electromagnetic Oscillations and Alternating Current Chp. 33 Cartoon -. Opening Demo - Warm-up problem Physlet Topics –LC Circuit Qualitatively](https://reader036.vdocument.in/reader036/viewer/2022062309/56649f175503460f94c2e49b/html5/thumbnails/2.jpg)
dAnBm ˆ⋅=•r
φ
AdBrr
⋅=dAB θcos=
θn̂
B
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ε =−dΦ
dt= −d(BAcosθ)
dt= −BA
dcosθ
dt= BAsinθ
dθ
dt
= BAω sinθ but θ =ωt so dθ
dt=ω
€
ε =BAω sinωt
€
ε =εm sinωt
Where is the rotational angular frequency of the generator
f and f= 60 Hz
Axis of rotationCoil of wire
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ε =εm sin(ωt −φ)
Instantaneous voltage
Amplitude
Angular frequency
time Phase constat
phase
€
εm
€
ε
€
t −φ
Phasor diagram
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ε =εm sinωt
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i = I sin(ωt)
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vL = Ldi
dt= LωIcos(ωt)
€
di
dt=ωIcos(ωt)
€
vL = LωIcos(ωt)
€
vR=Ri
€
R=VRI
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XL=VLI
=Lω
VR=RI VL= XLIL or VL= (LI since I=IL
L = 4.22mH
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=2πf
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f =1000Hz
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ε =vR + vL
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Impedance Z: New quantity for AC circuits. This is analogous to resistance in DC circuits
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Z = R2 + (ωL)2
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I =εmZ
€
I =εm
R2 + (ωL)2
€
XL =ωL
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RL Circuit Example
Suppose εm = 100 volts, f=1000 Hz, R=10 Ohms, L=4.22 mH,Find XL, Z, I, VR, and Vl.
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XL =ωL = 6.28 ×1000 × 0.00422H = 26.5Ω
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Z = 102 + (26.5)2 = 28.3Ω
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Z=R2
+(ωL)2
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I =εmZ
=100
28.3= 3.53A
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VR = RI =10 × 3.53 = 35.3v
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VL = XLI = 26.5 × 3.53 = 93.5v
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Power in AC circuits
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P = i2R = (I sin(ωt))2R
Instantaneous power doesn’t mean anything
Need to average over time or one period of the sine wave
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Pavg = 12π Rdθ
0
2π
∫ (I sin(θ))2 = 12π RI
2 sin2
0
2π
∫ θdθ = RI2 1
2= (
I
2)2R
Note
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Irms =I
2
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Pavg = Irms2 R
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Averaging over a sine curve
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Calculate Power lost in resistor from example
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Pavg = Irms2 R
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Irms =I
2=
3.53A
1.414= 2.50A
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Pavg = (2.50A)210 = 62.5Watts
To calculate power produced by the generator you need to take account of the phase difference between the voltage and the current. In general you can write:
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Pavg = εrmsIrms cosφ
For an inductor P = 0 because the phase difference between current through the inductor and voltage across the inductor is 90 degrees
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Series LRC circuit
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ε =vR + vC + vL
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ε =εm sinωt
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i = I sin(ωt −φ)
VL
VC
VR
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I =εm
R2 + (XL − XC )2=εmZ
XL=L
XC=1/(C)
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Z = R2 + (ωL −1
ωC)2
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R2 + (XL − XC )2
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tanφ =ωL −
1
ωCR
ELI the ICE man
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Resonance
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XL = XC
ωL =1
ωC
ω =1
LC
Series LRC demo
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10 uF 4.25 mH
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f =1
6.28 LC=
1
6.28 4.25 ×10−3H ×10−6Ff = 2442Hz
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Series LCR circuit