induction1 time varying circuits 2008 induction2 a look into the future we have one more week after...
TRANSCRIPT
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Induction 1
Time Varying CircuitsTime Varying Circuits
20082008
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Induction 2
A look into the futureA look into the future
We have one more week after today (+ one We have one more week after today (+ one day)day) Time Varying Circuits Including ACTime Varying Circuits Including AC Some additional topics leading to wavesSome additional topics leading to waves A bit of review if there is time.A bit of review if there is time.
There will be one more Friday morning quiz.There will be one more Friday morning quiz. I hope to be able to return the exams on I hope to be able to return the exams on
Monday at which time we will briefly review Monday at which time we will briefly review the solutions.the solutions.
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Induction Induction 33
The Final ExamThe Final Exam
8-10 Problems 8-10 Problems similar to (or similar to (or exactly) Web-exactly) Web-AssignmentsAssignments
Covers the entire Covers the entire semester’s worksemester’s work
May contain some May contain some short answer short answer questions.questions.
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Induction 4
Max Current Rate ofincrease = max emfVR=iR
~current
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Induction 5
constant) (time
)1( /
R
L
eR
Ei LRt
We Solved the lo
op equation.
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Induction 6
We also showed that
2
0
2
0
2
1
2
1
E
B
capacitor
inductor
u
u
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Induction 7
At t=0, the charged capacitor is now connected to the inductor. What would you expect to happen??
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Induction 8
The math …For an RLC circuit with no driving potential (AC or DC source):
2/12
2max
2
2
2
1
)cos(
:
0
0
L
R
LC
where
teQQ
Solutiondt
QdL
C
Q
dt
dQR
dt
diL
C
QiR
d
dL
Rt
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Induction 9
The Graph of that LR (no emf) circuit ..
L
Rt
e 2
L
Rt
e 2
I
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Induction 10
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Induction 11
Mass on a Spring Result
Energy will swap back and forth. Add friction
Oscillation will slow down Not a perfect analogy
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Induction 12
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Induction 13
LC Circuit
High
Q/CLow
Low
High
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Induction 14
The Math Solution (R=0):
LC
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Induction 15
New Feature of Circuits with L and C
These circuits produce oscillations in the currents and voltages
Without a resistance, the oscillations would continue in an un-driven circuit.
With resistance, the current would eventually die out.
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Induction 16
-1.5
-1
-0.5
0
0.5
1
1.5
0 1 2 3 4 5 6 7 8 9 10
Time
Vo
lts
Variable Emf Applied
emf
Sinusoidal
DC
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Induction 17
Sinusoidal Stuff
)sin( tAemf
“Angle”
Phase Angle
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Induction 18
Same Frequencywith
PHASE SHIFT
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Induction 19
Different Frequencies
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Induction 20
Note – Power is delivered to our homes as an oscillating source (AC)
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Induction 21
Producing AC Generator
x x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x xx x x x x x x x x x x x x x x x x x x x x x x
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Induction 22
The Real World
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Induction 23
A
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Induction 24
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Induction 25
The Flux:
tAR
emfi
tBAemf
t
BA
bulb
sin
sin
cos
AB
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Induction 26
problems …
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Induction 27
14. Calculate the resistance in an RL circuit in which L = 2.50 H and the current increases to 90.0% of its final value in 3.00 s.
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Induction 28
18. In the circuit shown in Figure P32.17, let L = 7.00 H, R = 9.00 Ω, and ε = 120 V. What is the self-induced emf 0.200 s after the switch is closed?
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Induction 29
32. At t = 0, an emf of 500 V is applied to a coil that has an inductance of 0.800 H and a resistance of 30.0 Ω. (a) Find the energy stored in the magnetic field when the current reaches half its maximum value. (b) After the emf is connected, how long does it take the current to reach this value?
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Induction 30
16. Show that I = I0 e – t/τ is a solution of the differential equation where τ = L/R and I0 is the current at t = 0.
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Induction 31
17. Consider the circuit in Figure P32.17, taking ε = 6.00 V, L = 8.00 mH, and R = 4.00 Ω. (a) What is the inductive time constant of the circuit? (b) Calculate the current in the circuit 250 μs after the switch is closed. (c) What is the value of the final steady-state current? (d) How long does it take the current to reach 80.0% of its maximum value?
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Induction 32
27. A 140-mH inductor and a 4.90-Ω resistor are connected with a switch to a 6.00-V battery as shown in Figure P32.27. (a) If the switch is thrown to the left (connecting the battery), how much time elapses before the current reaches 220 mA? (b) What is the current in the inductor 10.0 s after the switch is closed? (c) Now the switch is quickly thrown from a to b. How much time elapses before the current falls to 160 mA?
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Induction 33
52. The switch in Figure P32.52 is connected to point a for a long time. After the switch is thrown to point b, what are (a) the frequency of oscillation of the LC circuit, (b) the maximum charge that appears on the capacitor, (c) the maximum current in the inductor, and (d) the total energy the circuit possesses at t = 3.00 s?
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Induction 34
Source Voltage:
)sin(0 tVVemf
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Induction 35
Average value of anything:
Area under the curve = area under in the average box
T
T
dttfT
h
dttfTh
0
0
)(1
)(
T
h
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Induction 36
Average Value
T
dttVT
V0
)(1
0sin1
0
0 T
dttVT
V
For AC:
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Induction 37
So …
Average value of current will be zero. Power is proportional to i2R and is ONLY
dissipated in the resistor, The average value of i2 is NOT zero because
it is always POSITIVE
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Induction 38
Average Value
0)(1
0
T
dttVT
V
2VVrms
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Induction 39
RMS
2
2)(
2
2)
2(
2
1
)2
(1
0
02
0
20
0
20
0
20
220
VV
VdSin
VV
tT
dtT
SinT
TVV
dttT
SinT
VtSinVV
rms
rms
T
rms
T
rms
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Induction 40
Usually Written as:
2
2
rmspeak
peakrms
VV
VV
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Induction 41
Example: What Is the RMS AVERAGE of the power delivered to the resistor in the circuit:
E
R
~
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Induction 42
Power
tR
VRt
R
VRitP
tR
V
R
Vi
tVV
22
0
2
02
0
0
sin)sin()(
)sin(
)sin(
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Induction 43
More Power - Details
R
VVV
RR
VP
R
VdSin
R
VP
tdtSinR
VP
dttSinTR
VP
tSinR
VtSin
R
VP
rms
T
T
200
20
20
2
0
22
0
0
22
0
0
22
0
22
022
0
22
1
2
1
2
1)(
2
1
)(1
2
)(1
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Induction 44
Resistive Circuit
We apply an AC voltage to the circuit. Ohm’s Law Applies
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Induction 45
Con
sid
er
this
cir
cuit
CURRENT ANDVOLTAGE IN PHASE
R
emfi
iRe
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Induction 46
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Induction 47
Alternating Current Circuits
is the angular frequency (angular speed) [radians per second].
Sometimes instead of we use the frequency f [cycles per second]
Frequency f [cycles per second, or Hertz (Hz)] f
V = VP sin (t -v ) I = IP sin (t -I )
An “AC” circuit is one in which the driving voltage andhence the current are sinusoidal in time.
v
V(t)
t
Vp
-Vp
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Induction 48
v
V(t)
t
Vp
-Vp
V = VP sin (t - v )Phase Term
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Induction 49
Vp and Ip are the peak current and voltage. We also use the
“root-mean-square” values: Vrms = Vp / and Irms=Ip /
v and I are called phase differences (these determine whenV and I are zero). Usually we’re free to set v=0 (but not I).
2 2
Alternating Current Circuits
V = VP sin (t -v ) I = IP sin (t -I )
v
V(t)
t
Vp
-Vp
Vrms
I/
I(t)
t
Ip
-Ip
Irms
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Induction 50
Example: household voltage
In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.
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Induction 51
Example: household voltage
In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.
This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.2
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Induction 52
Example: household voltage
In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.
This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.This 60 Hz is the frequency f: so =2f=377 s -1.
2
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Induction 53
Example: household voltage
In the U.S., standard wiring supplies 120 V at 60 Hz. Write this in sinusoidal form, assuming V(t)=0 at t=0.
This 120 V is the RMS amplitude: so Vp=Vrms = 170 V.This 60 Hz is the frequency f: so =2f=377 s -1.
So V(t) = 170 sin(377t + v).Choose v=0 so that V(t)=0 at t=0: V(t) = 170 sin(377t).
2
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Induction 54
Review: Resistors in AC Circuits
ER
~EMF (and also voltage across resistor): V = VP sin (t)Hence by Ohm’s law, I=V/R:
I = (VP /R) sin(t) = IP sin(t) (with IP=VP/R)
V and I“In-phase”
V
t
I
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Induction 55
This looks like IP=VP/R for a resistor (except for the phase change). So we call Xc = 1/(C) the Capacitive Reactance
Capacitors in AC Circuits
E
~C Start from: q = C V [V=Vpsin(t)]
Take derivative: dq/dt = C dV/dtSo I = C dV/dt = C VP cos (t)
I = C VP sin (t + /2)
The reactance is sort of like resistance in that IP=VP/Xc. Also, the current leads the voltage by 90o (phase difference).
V
t
I
V and I “out of phase” by 90º. I leads V by 90º.
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Induction 56
I Leads V???What the **(&@ does that mean??
I
V
Current reaches it’s maximum at an earlier time than the voltage!
1
2
I = C VP sin (t +/2)
Phase=
-(/2)
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Induction 57
Capacitor Example
E
~
CA 100 nF capacitor isconnected to an AC supply of peak voltage 170V and frequency 60 Hz.
What is the peak current?What is the phase of the current?
MX
f
C 65.2C
1
1077.3C
rad/sec 77.360227
Also, the current leads the voltage by 90o (phase difference).
I=V/XC
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Induction 58
Again this looks like IP=VP/R for aresistor (except for the phase change).
So we call XL = L the Inductive Reactance
Inductors in AC Circuits
LV = VP sin (t)Loop law: V +VL= 0 where VL = -L dI/dtHence: dI/dt = (VP/L) sin(t).Integrate: I = - (VP / L cos (t)
or I = [VP /(L)] sin (t - /2)
~
Here the current lags the voltage by 90o.
V
t
I
V and I “out of phase” by 90º. I lags V by 90º.
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Induction 59
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Induction 60
Phasor Diagrams
Vp
Ipt
Resistor
A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.
The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.
Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.
A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.
The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.
Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.
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Induction 61
Phasor Diagrams
Vp
Ipt
Vp
Ip
t
Resistor Capacitor
A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.
A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.
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Induction 62
Phasor Diagrams
Vp
Ipt
Vp
IpVp Ip
Resistor Capacitor Inductor
A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.
A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.The “y component” is the actual voltage or current.
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Induction 63
Steady State Solution for AC Current (2)
• Expand sin & cos expressions
• Collect sindt & cosdt terms separately
• These equations can be solved for Im and (next slide)
1/ cos sin 0
1/ sin cos
d d
m d d m m
L C R
I L C I R
sin sin cos cos sin
cos cos cos sin sin
d d d
d d d
t t t
t t t
High school trig!
cosdt terms
sindt terms
cos sin cos sinmm d d m d d m d
d
II L I R t t t
C
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Induction 64
Steady State Solution for AC Current (2)
• Expand sin & cos expressions
• Collect sindt & cosdt terms separately
• These equations can be solved for Im and (next slide)
1/ cos sin 0
1/ sin cos
d d
m d d m m
L C R
I L C I R
sin sin cos cos sin
cos cos cos sin sin
d d d
d d d
t t t
t t t
High school trig!
cosdt terms
sindt terms
cos sin cos sinmm d d m d d m d
d
II L I R t t t
C
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Induction 65
• Solve for and Im in terms of
• R, XL, XC and Z have dimensions of resistance
• Let’s try to understand this solution using “phasors”
Steady State Solution for AC Current (3)
1/tan d d L CL C X X
R R
m
mIZ
22L CZ R X X
L dX L
1/C dX CInductive “reactance”
Capacitive “reactance”
Total “impedance”
1/ cos sin 0
1/ sin cos
d d
m d d m m
L C R
I L C I R
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Induction 66
REMEMBER Phasor Diagrams?
Vp
Ipt
Vp
Ip
t
Vp Ip
t
Resistor Capacitor Inductor
A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.
A phasor is an arrow whose length represents the amplitude ofan AC voltage or current.The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity.Phasor diagrams are useful in solving complex AC circuits.
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Induction 67
Reactance - Phasor Diagrams
Vp
Ipt
Vp
Ip
t
Vp Ip
t
Resistor Capacitor Inductor
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Induction 68
“Impedance” of an AC Circuit
R
L
C~
The impedance, Z, of a circuit relates peakcurrent to peak voltage:
IV
Zpp (Units: OHMS)
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Induction 69
“Impedance” of an AC Circuit
R
L
C~
The impedance, Z, of a circuit relates peakcurrent to peak voltage:
IV
Zpp (Units: OHMS)
(This is the AC equivalent of Ohm’s law.)
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Induction 70
Impedance of an RLC Circuit
R
L
C~
As in DC circuits, we can use the loop method:
- VR - VC - VL = 0 I is same through all components.
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Induction 71
Impedance of an RLC Circuit
R
L
C~
As in DC circuits, we can use the loop method:
- VR - VC - VL = 0 I is same through all components.
BUT: Voltages have different PHASES
they add as PHASORS.
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Induction 72
Phasors for a Series RLC Circuit
Ip
VRp
(VCp- VLp)
VP
VCp
VLp
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Induction 73
Phasors for a Series RLC Circuit
By Pythagoras’ theorem:
(VP )2 = [ (VRp )2 + (VCp - VLp)2 ]
Ip
VRp
(VCp- VLp)
VP
VCp
VLp
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Induction 74
Phasors for a Series RLC Circuit
By Pythagoras’ theorem:
(VP )2 = [ (VRp )2 + (VCp - VLp)2 ]
= Ip2 R2 + (Ip XC - Ip XL)
2
Ip
VRp
(VCp- VLp)
VP
VCp
VLp
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Induction 75
Impedance of an RLC Circuit
Solve for the current:
Ip Vp
R2 (Xc XL )2
Vp
Z
R
L
C~
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Induction 76
Impedance of an RLC Circuit
Solve for the current:
Impedance:
Ip Vp
R2 (Xc XL )2
Vp
Z
Z R2 1
C L
2
R
L
C~
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Induction 77
The circuit hits resonance when 1/C-L=0: r=1/When this happens the capacitor and inductor cancel each otherand the circuit behaves purely resistively: IP=VP/R.
Impedance of an RLC Circuit
Ip Vp
Z
Z R2 1
C L
2
The current’s magnitude depends onthe driving frequency. When Z is aminimum, the current is a maximum.This happens at a resonance frequency:
LC
The current dies awayat both low and highfrequencies.
IP
01 0
21 0
31 0
41 0
5
R = 1 0 0
R = 1 0
r
L=1mHC=10F
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Induction 78
Phase in an RLC CircuitIp
VRp
(VCp- VLp)
VP
VCp
VLp
We can also find the phase:
tan = (VCp - VLp)/ VRp
or; tan = (XC-XL)/R.
or tan = (1/C - L) / R
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Induction 79
Phase in an RLC Circuit
At resonance the phase goes to zero (when the circuit becomespurely resistive, the current and voltage are in phase).
IpVRp
(VCp- VLp)
VP
VCp
VLp
We can also find the phase:
tan = (VCp - VLp)/ VRp
or; tan = (XC-XL)/R.
or tan = (1/C - L) / R
More generally, in terms of impedance:
cos R/Z
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Induction 80
Power in an AC Circuit
V(t) = VP sin (t)
I(t) = IP sin (t)
P(t) = IV = IP VP sin 2(t) Note this oscillates
twice as fast.
V
t
I
t
P
= 0
(This is for a purely resistive circuit.)
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Induction 81
The power is P=IV. Since both I and V vary in time, sodoes the power: P is a function of time.
Power in an AC Circuit
Use, V = VP sin (t) and I = IP sin (t+) :
P(t) = IpVpsin(t) sin (t+)
This wiggles in time, usually very fast. What we usually care about is the time average of this:
PT
P t dtT
10
( ) (T=1/f )
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Induction 82
Power in an AC Circuit
Now: sin( ) sin( )cos cos( )sin t t t
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Induction 83
Power in an AC Circuit
P t I V t tI V t t t
P P
P P
( ) sin( )sin( )sin ( )cos sin( )cos( )sin
2
Now: sin( ) sin( )cos cos( )sin t t t
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Induction 84
Power in an AC Circuit
P t I V t tI V t t t
P P
P P
( ) sin( )sin( )sin ( )cos sin( )cos( )sin
2
sin ( )
sin( ) cos( )
2 1
2
0
t
t t
Use:
and:
So P I VP P1
2cos
Now: sin( ) sin( )cos cos( )sin t t t
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Induction 85
Power in an AC Circuit
P t I V t tI V t t t
P P
P P
( ) sin( )sin( )sin ( )cos sin( )cos( )sin
2
sin ( )
sin( ) cos( )
2 1
2
0
t
t t
Use:
and:
So P I VP P1
2cos
Now:
which we usually write as P I Vrms rms cos
sin( ) sin( )cos cos( )sin t t t
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Induction 86
Power in an AC Circuit
P I Vrms rms cos goes from -900 to 900, so the average power is positive)
cos( is called the power factor.
For a purely resistive circuit the power factor is 1.When R=0, cos()=0 (energy is traded but not dissipated).Usually the power factor depends on frequency.
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Induction 87
16. Show that I = I0 e – t/τ is a solution of the differential equation where τ = L/R and I0 is the current at t = 0.
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Induction 88
17. Consider the circuit in Figure P32.17, taking ε = 6.00 V, L = 8.00 mH, and R = 4.00 Ω. (a) What is the inductive time constant of the circuit? (b) Calculate the current in the circuit 250 μs after the switch is closed. (c) What is the value of the final steady-state current? (d) How long does it take the current to reach 80.0% of its maximum value?
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Induction 89
27. A 140-mH inductor and a 4.90-Ω resistor are connected with a switch to a 6.00-V battery as shown in Figure P32.27. (a) If the switch is thrown to the left (connecting the battery), how much time elapses before the current reaches 220 mA? (b) What is the current in the inductor 10.0 s after the switch is closed? (c) Now the switch is quickly thrown from a to b. How much time elapses before the current falls to 160 mA?
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Induction 90
52. The switch in Figure P32.52 is connected to point a for a long time. After the switch is thrown to point b, what are (a) the frequency of oscillation of the LC circuit, (b) the maximum charge that appears on the capacitor, (c) the maximum current in the inductor, and (d) the total energy the circuit possesses at t = 3.00 s?