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InductorsA coil of wire can create a magnetic field if a
current is run through it. If that current
changes (as in the AC case), the magnetic field
created by the coil will change. Will this
changing magnetic field through the coil causea voltage to be created across the coil? YES!
This is called self-inductance and is the basis
behind the circuit element called the inductor.
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Inductors
Vinductor = -L dI /dt
&ere the minus sign means that when the
current is increasing, the voltage across the
inductor will tend to o!!ose the increase,
and it also means when the current is
decreasing, the voltage across the inductorwill tend to o!!ose the decrease.
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Units: Henry
'rom Vinductor = -L dI /dt
# has units of olt *Am!sec+ which is
called a &enry"
1 Henry = Volt-sec / Am .
This is the same unit we had for mutual
inductance before (recall the transformer).
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Solenoid tye inductor
'or a ca!acitor, we started with a !arallel
!late configuration since the !arallel !lates
!rovided a uniform electric field betweenthe !lates.
In the same way, we will start with a device
that !rovides a uniform magnetic fieldinside the device" a solenoid.
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olenoid inductor
solenoid = µnI where n -#ength .
ecall 'araday/s #aw" V = d/dt "∫∫ •dA # .
ince the I in the 0 is a constant with res!ect to the variable of
integration (dA), we have"
V = L dI/dt where # ∫∫ (µn) dA , and since 0 is uniformover the area for the solenoid (and in the same direction asarea)"
# (µn)A. 0ut the area is for each of - loo!s, so" # µ -1A#ength where Aeach loo! π 1.
L =µπ
$%& % / Len't( )
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i2e of a &enry
'rom the inductance of a solenoid"L =
µπ
$%& % / Len't(
we see that with vacuum inside the solenoid, µ becomes µo which has a value of 3π 4 5678 T7mA, a rather smallnumber. will normally be less than a meter, and so 1
will also ma%e # small.
&owever, - can be large, and µ can be a lot larger than µo if we use a magnetic material.
Altogether, a (enry is a rat(er lar'e inductance.
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Inductors
Lsolenoid = µπ$%& % / Len't( )
As we indicated before, the value of the
inductance deends only on the sha!e (,
#ength, -) and materials (µ).The inductance relates the voltage across it to
the changing current through it"
V = - L dI/dt . We again use #en2/s #aw to
give us the direction (sign) of the voltage.
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Ener'y Stored in an Inductor
We start from the definition of voltage" V =
*E/+ (or 9: ;). 0ut since the voltage
across an inductor is related to the currentchange, we might e4!ress ; in terms of I"
I = d+/dt, or d; I dt. Therefore, we have"
:stored Σ ;i i ∫ d; ∫ I dt and now weuse VL = L dI/dt to get"Estored ∫ (# dIdt) I dt ∫ # I dI ,1/%LI%.
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&e.ie of Ener'y in 0ircuits
There is ener'y stored in a ca!acitor (that has
:lectric 'ield)" Estored = ,1/%0V% .
ecall the is related to :.
There is ener'y stored in an inductor (that has
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eview of Circuit :lements
esistor" V& = & I where I = ∆+/∆t
Ca!acitor" V0 = ,1/0+ (from 0 = +/V)
Inductor" VL = -L ∆I/∆t
We can ma%e an analogy with mechanics"
+ is li%e = t is li%e t,
I =∆
+/∆
t is li%e . =∆
/∆
t=∆
I/∆
t is li%e a =∆
./∆
t
V is li%e 2= 0 is li%e 1/3 (s!ring)=
& is li%e air resistance, L is li%e m.
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&L 0ircuit
What ha!!ens when we have a resistor in series with aninductor in a circuit?
2rom t(e mec(anical analo'y, this should be li%e having
a mass with air resistance. If we have a constant force
(li%e gravity), the ob$ect will accelerate u! to a terminals!eed (due to force of air resistance increasing u! to the
!oint where it balances the gravity).
Σ2=ma → − bv − mg ma, or
m d./dt 4 5. = -m'
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# Circuit (cont.)
If we connect the resistor and the inductor to a
battery and then turn the switch on, from
the mechanical analogy we would e4!ectthe current (which is li%e velocity) to begin
to increase until it reaches a constant
amount.
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# Circuit 7 ;ualitative loo%
'rom the circuit !oint of view, initially we have 2erocurrent so there is no (voltage dro! across the
resistor). Thus the full voltage of the battery is trying
to change the current, hence # battery, and so
dIdt battery #.
&owever, as the current increases, there is more
voltage dro! across the resistor, , which reduces
the voltage across the inductor (# battery 7 ),and hence reduces the rate of change of the current>
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# Circuit"
6ifferential E+uationTo see this behavior ;uantitatively, we need to
get an e;uation.
We can get a differential e;uation by using the0onser.ation of Ener'y (Σi 6)"
V5attery - Vresistor - Vinductor = 7.
*This loo%s li%e an ordinary algebraic e;uation.0ut all the /s are not constants" we have
relations for resistor and inductor .+
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# Circuit"
ifferential :;uation
V5attery - Vresistor - Vinductor = 7)
With V5attery = constant, Vresistor = I& , and
Vinductor = L dI/dt, we have the differentiale;uation (for I(t))"
V5attery - I& - L dI/dt = 7. This can be rewritten as"I& 4 L dI/dt = V5attery
which is an inhomogeneous first order differentiale;uation, $ust li%e we had for the mechanical analogy"
m d./dt 4 5. = -m' .
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# Circuit (cont.)
I& 4 L dI/dt = V5attery
The (omo'eneous e;uation is" LdI/dt 4 &I = 7.
This has a dying e4!onential solution"
IH,t = Io e-,&/L t .
The in(omo'eneous e;uation is"L dI/dt 4 &I = V5attery .
This has the sim!le solution" II,t = V5attery / & .
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# Circuit (cont.)
I,t I&(t) @ II(t) Io e-,&/L t @ V5attery / & .
To find the com!lete solution (that is, find Io ,
we a!!ly the initial conditions"
I(t6) 6 Io @ battery so Io 7 battery .
Therefore, we have"
I,t = "V5attery/"1 - e-,&/L t # .
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# Circuit
I,t = "V5attery/"1 - e-,&/L t # .
As time goes on, the current does increase andfinally reaches the value battery which is
what it would be without the inductor !resent.The gra!h on the ne4t slide shows this function
when V5attery = %9 volts= & = %9 Ω, and L = 1
H. -ote that the ma4 current is 5 Am!, andthe time scale is in milliseconds.
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# Circuit" I(t) versus t
I ( t ) v s t f o r R L C i r c u i t
0
0 . 2
0 . 4
0 . 6
0 . 8
1
1 . 2
1 6
1
1
1
6
2
1
2
6
3
1
3
6
4
1
4
6
t in m i l l i s e c o n d s
I i n
a m
p s
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& L and 0 in a circuit
We have already considered an C circuit (in
9art 1) and an # circuit ($ust now). We
loo%ed at these cases for a circuit in which
we had a battery and then threw the switch.
&owever, the main reason these circuit
elements are im!ortant is in AC circuits. We
loo% ne4t at the case of the three elements in
a series circuit with an AC voltage a!!lied.
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#C Circuit" scillations
$eton;s Second La" Σ ' ma can be writtenas" Σ ' 7 ma 6 . With a s!ring and resistance,this becomes" -3 -5. -ma = 7
This is li%e the e;uation we get from0onser.ation of Ener'y when we have aca!acitor, resistor and inductor" Σ 6 , or,1/0< 4 &I 4 L dI/dt = 7 ,
where V0 = ,1/0
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#C Circ it"
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#C Circuit"
6ifferential E+uation
tarting with Conservation of :nergy
(ΣVi = 7) and using I , V0 =
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#C Circuit"
ifferential :;uation,1/0< 4 &d
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#C Circuit
When we substitute this e4!ression
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#C Circuit
0y !lacing a resistor in the circuit, the
differential e;uation becomes a little harder.
We need to consider either both sines andcosines, or we need to consider e4!onentials
with imaginary numbers in the e4!onent.
This can be done, and reasonable solutionscan be found, but we will not !ursue that
here. We will !ursue an alternative way.
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#C Circuit"
ImedanceAn alternative way of considering the #C
Circuit is to use the conce!t of imedance.
The idea of im!edance is that all three of thema$or circuit elements im!ede the flow of
current.
A resistor obviously limits the current in acircuit. 0ut a ca!acitor and an inductor also
limit the current in an AC circuit.
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#C Circuit
The basic idea we will !ursue is that in a
series circuit,
a) the same current flows through all of theelements" I = Io sin,ωt = and
b) the voltages at any instant add u! to 2ero
around the circuit"
V& ,t 4 V0,t 4 VL,t = VA0,t .
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0aaciti.e &eactance
'or a ca!acitor" V0 = ,1/0
-ote that the constant (5ωC) acts $ust li%e . We call
this the ca!acitive reactance,>0 = ,1/ω0 .
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oltage across the ca!acitor
V0 = -V0o cos,ωt
I = Io sin,ωt
V0o = Io>0 where >0 = 1/ω0
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Inducti.e &eactance
'or an inductor, VL = L dI/dt, if I = Iosin,ω
tthen dI/dt = Ioω cos,ωt) -ote that cosine isE6o different than sine 7 we say it is E6o out of !hase. This means that VL is 7o out of (ase
it( t(e current)
ince VL = L dI/dt, VL = ωLIo cos,ωt , and wesee that the constant ω# acts $ust li%e . We
call this the inductive reactance,>L = ωL .
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oltage across the inductor
VL = VLo cos,ωt
I = Io sin,ωt
VLo = Io>L where >L = ωL
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L&0 in series
-ote that in a series combination, the current
must be the same in all the elements, while
the voltage adds. &owever, the voltagemust add to 2ero across a com!lete circuit at
every instant of time. 0ut since the
voltages are out of !hase with each other,the am!litudes of the voltages (and hence
the rms voltages) will not add u! to 2ero>
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#C in series
'or" I = Io sin,ωt
V& = I& = &Io sin,ωt
V0 = ,1/08< = -,1/ω0 Io cos,ωt
VL = L8dI/dt = ,ωL Io cos,ωt
V& 4 V0 4 VL = VA0 =
Io "& sin,ωt ? ,1/ω0 cos,ωt 4 ωL cos,ωt#
= Io @ sin,ωt4 = Vo sin,ωt4
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Imedance
In 17 s!ace, 4 and y are E6o a!art. We combine thetotal s!ace se!aration by the 9ythagorean
Theorem" r *41@y1+51 .
If we add u! the /s, this is e;uivalent to adding u!
the reactances. 0ut we must ta%e the !hases into
account. 'or the total im!edance, F, we get " Vrms
= Irms @ where @ = "& % 4 ,ωL - 1/ω0%#1/% .
(-ote that we had to subtract G# from GC because of thesigns involved.)
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&esonance
Vrms = Irms@ where @ = "& % 4 ,ω
L-1/ω
0%#1/%
-ote that when ,ωL - 1/ω0 = 7, F is smallest
and so I is biggest> This is the condition for
resonance. Thus whenω = [1
L0#1/%, we haveresonance. This is the same result we would
get using the differential e;uation route.
-ote that this is e;uivalent to the resonance of as!ring when ω = [3 m#1/%.
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V,t .ersus Vrms
2rom 0onser.ation of Ener'yΣ
Vi = 7) B(is (oldstrue at e.ery time. Thus"
V0,t 4 V& ,t 4 VL,t = Vosin,ωt4θο . 0ut due to the
!hase differences, C will be ma4imum at a different
time than , etc.
&owever, when we deal with rms voltages, we ta%e
into account the !hase differences by using the
9ythagorean Theorem. Thus,V0-rms 4 V&-rms 4 VL-rms C VA0-rms .
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$et *oer 6eli.ered
What about energy delivered?'or a resistor, electrical energy is changed into heat, so the
resistor will remove !ower from the circuit.
2or an inductor and a caacitor t(e ener'y is
merely stored for use later) Hence t(e a.era'e
oer used is Dero for both of these. -ote that
9ower IH, but I and are out of !hase by E6o
for both the inductor and ca!acitor, so on average
there is 2ero !ower delivered.
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-et 9ower elivered
Thus, even though we have V=I@ as a
generali2ed hm/s #aw, !ower is still"
*a.' = I%& (not 9I1F).
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Fa'netism in Fatter
ust as materials affect the electric fields in s!ace,
so do materials affect the magnetic fields in
s!ace.ecall that we described the effect of materials on
the electric fields with the dielectric constant, J.
This measured the stretchabilityD of the electric
charges in the materials. This stretching due toa!!lied electric fields caused electric fields itself.
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Two main effects
Atoms have electrons that orbitD the !ositive
nuclei. These orbitingD electrons act li%e
little current loo!s and can create smallmagnets.
:ffect 5" A diamagnetic effect similar to the
dielectric effect.:ffect 1" An aligning effect.
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iamagnetic :ffect
:ffect 5" 0y #en2/s law, when the a!!lied
magnetic field changes, there is a tendency in
the circuit to resist the change. This effecttends to create a magnetic field o!!osing the
change. In this effect, the material acts to
reduce any a!!lied e4ternal magnetic field.
This is similar to the dielectric effect that leads
to the dielectric constant for electric fields.
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Aligning :ffect
:ffect 1" ince the normal currentsD due to
the orbitingD electrons act li%e tiny
magnets, these magnets will tend to alignwith an e4ternal magnetic field. This
tendency to align will tend to add to any
e4ternal a!!lied magnetic field.&eat tends to destroy this ordering tendency.
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-et esult
When both of these effects (#en2/s law and aligning)are combined, we find three different ty!es of
results"
5. iamagnetic"
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'urther elations
0 µ& µo(& @
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