CH 31  Electromagnetic Oscillations and Alternating Current  I. Review

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March 10, 2016 CH 31 Electromagnetic Oscillations and Alternating Current I.
Review
A.
LastSemesteryoulearnedinMechanics(CH‐15)
1.
LinearSimpleHarmonicOscillator(fig15‐15)
2.
UsingF==()Newton’sLaw
3.
Also…F=Hooke’sLaw
4.
Thus:=()k=
5.
Sow=(angularfrequency)
6.
Rememberthat:
a)
B.
velocity
b)
PotentialEnergy
c)
KineticEnergy
d)
period
AboutamonthagoinChapter25youlearned:
1.
EnergystoredinanElectricField(CH‐25)
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March 10, 2016 2.
orPotentialEnergyinanElectricField
C.
LastweekinChapter30youlearned:
1.
EnergystoredinaMagneticField(CH‐30)
2.
orPotentialEnergyinaMagnetic
Field
II.
ElectroMagneticOscillations
A.
Inductor/Capacitor(LC)combinedCKTOverview(fig31‐1)
Fig 31‐ 1 Page2
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March 10, 2016 1.
=0.
Lookat(a):AssumeCapacitorischarged,nocurrent.UE=MaxandUB
2.
In(b),Capacitorstarts_______________________________,andcurrentis
___________________…thusInductor’smagneticfieldstartsto_________________.
3.
In(c),Capacitorisfully__________________;Inductor’scurrentisat
_______________.UE=0andUB=Max.
4.
In(d),Capacitorstarts_____________________again,butthistimein
__________________polarity.Inductor’scurrentis_______________________________.
5.
In(e),Capacitorreaches________________________charge.Againnotice
polarity;nocurrent.
6.
In(f),Capacitorstarts___________________________withcurrentflowin
_______________________directionfrom(b),andcurrentis________________________…thus
Inductor’smagneticfieldstartsto___________________________________________
direction.
7.
In(g),Capacitoris____________________________________;Inductor’scurrentis
atmax.UE=0andUB=Max.
8.
B.
Finallyin(h),Capacitorstartschargingagainascurrentdecreases.
UsingMechanicalAnalog
1.
LookbackatA6inyouroutline;Ifwereplacexwithqthen:
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March 10, 2016 1. 2.
ThespringwhichhasPotentialEnergyisreplacedwithaCapacitor

3.
Themassbecomestheinductor
 L=
4.
Thuswherew=
5.
a)
thenw=
b)
  c)
Let’sPROVEwithCalculus
Block‐SpringExample
a)
(1)
(2)
U=UB+US=
= ()=0Totalenergyremains
constantwithtime.
(a)
=0,butrememberthat
(b)
…So
(c)
=0(block‐springOscillations)
;thus=0
(3)
SolutiontothisDiffEQis:
(a)
x=(displacement)
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March 10, 2016 C.
ForOurLCOscillator
1.
U=UB+UE=or=
a)
= ()=0Totalenergy
(1)
b)
Thus
0
But,rememberthat:
and =
(1)
(a)
Therefore:=0
(LCOscillations)
c)
NoticethisequationisofthesametypeofDIFFEQasbefore
(1)
Thereforesolutionis:q=
(charge)
(2)
=
Thus
(current)
(3)
SO
ori=
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March 10, 2016 d)
AngularFrequency(w)
(1)
Since
then
(2)
Usingthestarequationandsubstitutingfromabove:
(a)
=0
(b)
=1
(c)
w=sameaswesaid
before.NOTE:w=
,wherefis
frequencyinHz.
2.
ElectricalandMagneticEnergyOscillations
a)
b)
andsubstitutingw= /√
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March 10, 2016 D.
ExampleLCProblems
1.
Asshownbelow,aseriesLCcircuitcontainsa100mHInductoranda
36mFCapacitorthatwaschargedbya12Vbattery.Whatisthefrequencyin
Hz.oftheelectromagneticoscillationsinthecircuit(neglectresistanceofthe
circuitwires)?
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March 10, 2016 2.
Exampletwo:AnLCcircuitproduceselectromagneticoscillationswitha
frequencyof 1.50 x10 6 Hz. Ifthecapacitanceis2.00nF,theinductanceinthecircuit
isclosestto:
A. 1.98 x106 H B.
4.35 x106 H C. 3.47 x104 H D. 5.63 x106 H E.
2.22 x104 H Show all work: a)
Answeris.Ifyouchosethenyouforgot
______________________________________________________________________________.
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March 10, 2016 3. Example Three: In the circuit below, R = 2.0 Ω, C = 1.4 F, L = 0.50 H and
E = 6.0 V. The switch has been in position 1 for a long time (t = ∞) and the
capacitor has no charge. The switch is then quickly thrown to position 2.
After the switch is thrown to position 2, the maximum charge on the
capacitor is closest to: A. 1.7 C.
B. 2.5 C.
C. 2.9 C.
D. 3.4 C.
E. 3.8 C. Show all work: Page9
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March 10, 2016 E.
RealWorldLCOscillatorsareDAMPED
1.
RememberdampingisFrictionintheBlock‐Springexample.Its
electricalanalogisResistance.
a)
Note:EveniftheCircuitonlyhasanInductorandaCapacitorin
theCircuit,thewirehasresistance.ResistanceremovesEnergyinthe
formofheat.
2.
IfwestartwithU=UB+UE=andrememberfrom
CH‐26thattheRATEofheattransferis:
a)
=
b)
Then
=but
(1)
Therefore:=0(LRCOscillations)
c)
Thereforesolutionis:q=Q
(Eq)
(1)
/
and =
cos(w’t+f)(charge)
w’=andw= /√
F.
ExampleLRCProblem
1.
a)
Answeris
Page
10
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