Chapter 30 Inductance - UTK Department of Physics and ... · PDF fileChapter 30 Inductance....
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Fall 2008Physics 231 Lecture 10-1
Chapter 30Inductance
Fall 2008Physics 231 Lecture 10-2
Magnetic Effects
As we have seen previously, changes in the magnetic flux due to one circuit can effect what goes on in other circuits
The changing magnetic flux induces an emf in the second circuit
Fall 2008Physics 231 Lecture 10-3
Mutual InductanceSuppose that we have two coils,Coil 1 with N1 turns and Coil 2 with N2 turns
Coil 1 has a current i1 which produces a magnetic flux, ΦΒ2 , going through one turn of Coil 2
If i1 changes, then the flux changes and an emf is induced in Coil 2 which is given by
dtd
N B222
Φ−=ε
Fall 2008Physics 231 Lecture 10-4
Mutual InductanceThe flux through the second coil is proportional to the current in the first coil
12122 iMN B =Φ
where M21 is called the mutual inductance
Taking the time derivative of this we get
dtdiM
dtdN B 1
212
2 =Φ
dtdiM 1
212 −=εor
Fall 2008Physics 231 Lecture 10-5
Mutual InductanceIf we were to start with the second coil having a varying current, we would end up with a similar equation with an M12
MMM == 1221We would find that
The two mutual inductances are the same because the mutual inductance is a geometrical property of the arrangement of the two coils
To measure the value of the mutual inductance you can use either
dtdIM 1
2 −=ε ordt
dIM 21 −=ε
Fall 2008Physics 231 Lecture 10-6
Units of Inductance
2AmpJ1
AmpsecV1 Henry 1 =
⋅=
Fall 2008Physics 231 Lecture 10-7
Self InductanceSuppose that we have a coil having N turns carrying a current I
That means that there is a magnetic flux through the coil
This flux can also be written as being proportional to the current
ILN B =Φ
with L being the self inductance having the same units as the mutual inductance
Fall 2008Physics 231 Lecture 10-8
Self InductanceIf the current changes, then the magnetic flux through the coil will also change, giving rise to an induced emf in the coil
This induced emf will be such as to oppose the change in the current with its value given by
dtdIL−=ε
If the current I is increasing, then
If the current I is decreasing, then
Fall 2008Physics 231 Lecture 10-9
Self InductanceThere are circuit elements that behave in this manner and they are called inductors and they are used to oppose any change in the current in the circuit
As to how they actually affect a circuit’s behavior will be discussed shortly
Fall 2008Physics 231 Lecture 10-10
What Haven’t We Talked AboutThere is one topic that we have not mentioned with
respect to magnetic fields
Just as with the electric field, the magnetic field has energy stored in it
We will derive the general relation from a special case
Fall 2008Physics 231 Lecture 10-11
Magnetic Field EnergyWhen a current is being established in a circuit, work has to bedone
If the current is i at a given instant and its rate of change is given by di/dt then the power being supplied by theexternal source is given by
dtdiiLiVP L ==
The energy supplied is given by PdtdU =
The total energy stored in the inductor is then
2
0 21 ILdiiLU
I== ∫
Fall 2008Physics 231 Lecture 10-12
Magnetic Field EnergyThis energy that is stored in the magnetic field is available to act as source of emf in case the current starts to decrease
We will just present the result for the energy density of the magnetic field
0
2
21
µBuB =
This can then be compared to the energy density of an electric field
202
1 EuE ε=
Fall 2008Physics 231 Lecture 10-13
R-L CircuitWe are given the following circuit
and we then close S1 andleave S2 open
It will take some finite amount of time for the circuit to reachits maximum current which is given by RI ε=
Kirchoff’s Law for potential drops still holds
Fall 2008Physics 231 Lecture 10-14
R-L CircuitSuppose that at some time t the current is i
The voltage drop across the resistor is given by RiVab =The magnitude of the voltage drop across the inductor is given by
dtdiLVbc =
The sense of this voltage drop is that point b is at a higher potential than point c so that it adds in as a negative quantity
0=−−dtdiLRiε
Fall 2008Physics 231 Lecture 10-15
R-L Circuiti
LR
Ldtdi
−=εWe take this last equation and
solve for di/dt
Ldtdi
initial
ε=⎟
⎠⎞
⎜⎝⎛Notice that at t = 0 when I = 0 we
have that
Also that when the current is no longer changing, di/dt = 0, that the current is given by R
I ε=
as expected
But what about the behavior between t = 0 and t = ∞
Fall 2008Physics 231 Lecture 10-16
R-L CircuitWe rearrange the original equation and then integrate
( ) ∫∫ −=−
tidt
LR
Ridi
00
'/''
ε( ) dtLR
Ridi
−=− /ε
( )( )tLReR
i /1 −−=εThe solution for this is
Which looks like
Fall 2008Physics 231 Lecture 10-17
R-L CircuitAs we had with the R-C Circuit, there is a time constantassociated with R-L Circuits
RL
=τ
Initially the power supplied by the emf goes into dissipative heating in the resistor and energy stored in the magnetic field
dtdiiLRii += 2ε
After a long time has elapsed, the energy supplied by the emf goes strictly into dissipative heating in the resistor
Fall 2008Physics 231 Lecture 10-18
R-L Circuit
We now quickly open S1and close S2
The current does not immediatelygo to zero
The inductor will try to keep the current, in the same direction, at its initial value to maintain the magnetic flux through it
Fall 2008Physics 231 Lecture 10-19
R-L CircuitApplying Kirchoff’s Law to the bottom loop we get
0=−−dtdiLiR
dtLR
idi
−=Rearranging this we have
and then integrating this
∫∫ −=t
I
dtLR
idi
0
0'
''
0
( )tLReII /0
−=
where I0 is the current at t = 0
Fall 2008Physics 231 Lecture 10-20
R-L Circuit
This is a decaying exponentialwhich looks like
The energy that was stored in the inductor will be dissipated in the resistor
Fall 2008Physics 231 Lecture 10-21
L-C CircuitSuppose that we are now given a fully charged capacitor and an inductor that are hooked together in a circuit
Since the capacitor is fully charged there is a potential difference across it given by Vc = Q / C
The capacitor will begin to discharge as soon as the switch is closed
Fall 2008Physics 231 Lecture 10-22
L-C CircuitWe apply Kirchoff’s Law to this circuit
0=−−Cq
dtdiL
dtdqi =Remembering that
2
2
dtqd
dtdi
=We then have that
012
2=+ q
LCdtqd
The circuit equation then becomes
Fall 2008Physics 231 Lecture 10-23
L-C CircuitThis equation is the same as that for the Simple Harmonic Oscillator and the solution will be similar
)cos(0 ϕω += tQq
The system oscillates with angular frequency
LC1
=ω
ϕ is a phase angle determined from initial conditions
)sin(0 ϕωω +−== tQdtdqiThe current is given by
Fall 2008Physics 231 Lecture 10-24
L-C Circuit
Both the charge on the capacitor and the current in the circuit are oscillatory
The maximum charge and the maximum current occur π / (2ω) seconds apart
For an ideal situation, this circuit will oscillate forever
Fall 2008Physics 231 Lecture 10-25
L-C Circuit
Fall 2008Physics 231 Lecture 10-26
L-C CircuitJust as both the charge on the capacitor and the current through the inductor oscillate with time, so does the energy that is contained in the electric field of the capacitor and themagnetic field of the inductor
Even though the energy content of the electric and magnetic fields are varying with time, the sum of the two at any given time is a constant
BETotal UUU +=
Fall 2008Physics 231 Lecture 10-27
L-R-C Circuit
Instead of just having an L-C circuit with no resistance, what happens when there is a resistance R in the circuit
Again let us start with the capacitor fully charged with a charge Q0 on it
The switch is now closed
Fall 2008Physics 231 Lecture 10-28
L-R-C CircuitThe circuit now looks like
The capacitor will start to discharge and a current will start to flow
We apply Kirchoff’s Law to this circuit and get
0=−−−Cq
dtdiLiR
dtdqi =And remembering that we get
012
2=++ q
LCdtdq
LR
dtqd
Fall 2008Physics 231 Lecture 10-29
L-R-C CircuitThe solution to this second order differential equation is similar to that of the damped harmonic oscillator
The are three different solutions
Underdamped
Critically Damped
Overdamped
Which solution we have is dependent upon the relative values of R2 and 4L/C
Fall 2008Physics 231 Lecture 10-30
L-R-C CircuitUnderdamped:
CLR 42 <
The solution to the second differential equation is then
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛+
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
⎟⎠⎞
⎜⎝⎛−
φtL
RLC
eQqt
LR
2
22
0 41cos
The system still oscillates but with decreasing amplitude, which is represented by the decaying exponential
This solution looks like
This decaying amplitude is often referred to as the envelope
Fall 2008Physics 231 Lecture 10-31
L-R-C Circuit
CLR 42 =Critically Damped:
Here the solution is given by
tL
R
etL
RQq⎟⎠⎞
⎜⎝⎛−
⎟⎠⎞
⎜⎝⎛ += 2
0 21
This solution looks like
This is the situation when the system most quickly reachesq = 0
Fall 2008Physics 231 Lecture 10-32
L-R-C Circuit
CLR 42 >Overdamped:
Here the solution has the form
⎥⎥⎦
⎤
⎢⎢⎣
⎡
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−+⎟
⎟
⎠
⎞
⎜⎜
⎝
⎛+= −⎟
⎠⎞
⎜⎝⎛−
ttt
LR
eLR
eLR
eQq ''20'
21'
212
ωω
ωω
This solution looks like
LCLR 14
' 2
2−=ωwith
Fall 2008Physics 231 Lecture 10-33
L-R-C Circuit
The solutions that have been developed for this L-R-C circuit are only good for the initial conditions at t = 0 that q = Q0 and that i = 0