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Unformatted text preview: 12. Lecture 12
12.1 LR circuit, comparison with RC
An inductor stores energy in the form of a magnetic ﬁeld similarly as a capacitor stores
energy in the form of an electric ﬁeld. In both cases we can ”charge” the device by
loading energy into it taking it it for example from a battery. In the case of a capacitor
we did that through a resistor in an RC circuit. The charging time is given by τ = RC .
See the ﬁg.68 to remind yourself how the current and voltage behave as a function of
time in such case. Figure 68: AS seen before, energy from a battery can be stored in a capacitor using an RC
circuit. The time constant of the circuit is given by τ = RC . Compare with the RL circuit
below. We can do the same for an inductor by using an RL circuit, a resistor in series
with an inductor. See ﬁg.69. Initially, when we turn on the switch, the battery tries
to increase the current in the circuit and that will occur almost instantaneously if the
inductor is replaced by a cable. However such large changes in current are opposed by
the inductor which creates between its terminals a voltage that opposes the increase in
current. For that reason the current increases at a slow rate until it reaches its ﬁnal
value given by:
1
If = ∆Vbattery
(12.1)
R
at that time the current does no change anymore and the inductor behaves as a cable,
namely no voltage cross it terminals. However the inductor is generating a magnetic
ﬁeld so it has energy stored in it. To see how long it takes to reach such a stationary
state, consider the voltage across the inductor
Va − Vb = L – 80 – ∆I
∆t (12.2) Initially, as we said, the inductor opposes the battery and does not allow any current
to circulate. This means that Va − Vb = ∆Vbattery and the initial slope with which
the current start increasing is then:
∆I
= 1 ∆V
(12.3)
∆t L battery
As the current increases, this slope decreases as can be see in the plot in ﬁg.69. However,
we get a good estimate of the time it takes to reach the steady state if we assume the
slope to be constant and ask how long it would take with that slope to reach the current
If . It is
∆Vbattery
If
1
L
∆ t = ∆I =
=
(12.4)
∆V
R
R
battery
∆t
L So we ﬁnd the characteristic time for an RL circuit is
τ= L
R (12.5) The larger the inductance the longer it takes to charge. If the resistance is small it also
takes longer since we want to reach a larger current in the end. Figure 69: Energy from a battery can be loaded in an inductor using an RL circuit. The
time constant of the circuit is given by τ = R . Pay attention to the diﬀerent voltage and
L
current curves compared to the capacitor case. 12.2 Using the oscilloscope
A very useful tool to check what we discussed in the previous section is the oscilloscope
(sometimes called simply ”scope”). It measures the voltage across its terminals and
plots it as a function of time. One terminal is usually grounded and the other, which is
called a probe, has a conducting pin that you can touch to diﬀerent parts of the circuit
to measure the voltage. Digital oscilloscopes have integrated circuits called analog to – 81 – digital converters which take the voltage and covert it into a number which they store
in a memory. The instrument then displays the voltage as a function of time using
an LCD display. On the screen the horizontal scale is time and the vertical scale is
voltage. The scale can be set manually or can be left automatic (easier). In the past,
and also for high frequencies oscilloscopes work with a vacuum tube where an electron
beam was directed toward a screen. The path of the beam was changed by a potential
applied to two parallel plates. In the horizontal direction is was moved back and forth
at a constant rate (backward much faster than forward). In the vertical direction it
was moved according to the input voltage. The resulting point on the screen traces the
voltage as a function of time. Both type of oscilloscopes are perfectly good for looking
at the RL circuit. In the demo we do that and check the voltage and current as a
function of time. notice that to measure the current we can use Ohm’s law:
∆V = IR (12.6) So we can just measure the voltage cross the resistor to know the current. The humble
resistor can be though as a current to voltage converter!. Figure 70: Demo. An RL circuit is analyzed with the aid of an oscilloscope. 12.3 Energy contained in a solenoid
We mentioned several times that a solenoid or inductor contain energy. We are going
to derive that such energy is given by
1
U = LI 2
2 (12.7) where L is the inductance and I the current circulating through the solenoid. To see
how this works recall that for a solenoid of area A, length and number of turns N , – 82 – Figure 71: Demo. A vacuum tube where an electron beam is directed toward a screen
producing a bright spot. The beam can be deﬂected by electric and magnetic ﬁelds. This
allows to build an oscilloscope by using external input to deﬂect the beam. the inductance is given by
L = µ0 N 2A
(12.8) The energy therefore should be given by
1 N 2A 2
U = µ0
I
2
(12.9) To verify this result we are going to is a procedure analogous to what we did for a
capacitor. In that case we computed how much energy was required to separate the
parallel plates by a small distance ∆d. This allowed us to compute the energy in the
capacitor. Here we can see how much energy is required to expand the solenoid so that
the area is increased by ∆A. The reason we do that is that, from eq.(12.9) we expect
that the energy is linear in A so it should be easy to compute the energy necessary to
change A. The ﬁrst consideration is that if we increase A, the ﬂux through the solenoid
changes and therefore a voltage is induced:
Va − Vb = 2
∆Φ
∆ A = − µ0 N I ∆ A
= −N B 
∆t
∆t
∆t – 83 – (12.10) This means that as we do this, we need to push the current against such potential
diﬀerence. The power necessary to do that is
P = I ( V a − V b ) = µ0 N 2 2 ∆A
I
∆t (12.11) Since power is energy consumed in unit time, to get the total energy necessary to do
this we multiply the power by ∆t obtaining
∆U1 = P ∆t = µ0 N2 2
I ∆A
(12.12) This energy we loose, so it is incorporated into the solenoid. If we compare with
eq.(12.9) we see that there is a factor of two. So we must be gaining energy somewhere
else because we overestimated the energy we needed to do this. The reason is that the
magnetic ﬁeld exerts a force on the solenoid that tends to expand it. This forces does
work when we expand the solenoid and therefore we get energy form there. Let us see
how much this is. The force on a small portion of the cable is radial and per unit length
equal to
F 
= B I
(12.13)
λ
The magnetic ﬁeld is indicated as B because we should not consider the magnetic ﬁeld
created by the small portion of cable itself. To see what this is we can use Ampere’s law
to compute the magnetic ﬁeld generated by a small piece of the solenoid (see ﬁg.72).
It gives:
N
2  B  ∆ = µ0 ∆ I
(12.14)
implying that
µ0 N
B  =
I
(12.15)
2
which is precisely half the magnetic ﬁeld of the solenoid!. What that from the magnetic
ﬁeld acting on a small portion of the cable, half is created by the rest of the solenoid
and half is created by itself and should be excluded (because the small portion of cable
cannot make a force on itself). We conclude that
µ0 N
I
2 (12.16)
F 
µ0 N 2
=
I
l
2 (12.17)
B  =
We can now compute the force – 84 – The length of each turn is 2π R where R is the radius of the coil. The displacement if
we increase the radius by ∆R is precisely ∆R. We can the compute the work as
W=
F 
µ0 N 2
2π R ∆ R =
I ∆A
l
2 (12.18) where we used that, if we change the radius by ∆R, the area changes precisely by
∆A = 2π R∆R (assuming ∆R R). This is work done by the solenoid, so we gain
energy. Therefore the total change in the energy of the solenoid is
∆ U = ∆ U1 − W = µ0 N 2
I ∆A
2 (12.19) If we now assume that U = 0 when A = 0 because when A = 0 there is no magnetic
ﬁeld to speak of, we obtain that
U= µ0 N 2
1
I A = LI 2
2
2 (12.20) as we wanted to show. Admittedly this derivation is a bit lengthy but is a very useful
exercise if you want to understand several properties of magnetic ﬁeld, work, energy,
etc. If you go through it carefully, the basic idea should appear quite simple and then
one has to do the calculations with great care. – 85 – Figure 72: Energy contained in a coil. There is a force trying to expand the coil. It is
created by the magnetic ﬁeld acting on the current. However the magnetic ﬁeld created by
the small piece of coil considered should be excluded. If we expand the coil this force does
work and the energy of the coil is reduced. On the other hand, a voltage is induced and we
need to do work to keep the current circulating. In total the energy of the coil increases. It
is given by U = 1 LI 2 .
2 – 86 – ...
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This note was uploaded on 12/07/2011 for the course PHY 219 taught by Professor Na during the Fall '11 term at Purdue UniversityWest Lafayette.
 Fall '11
 NA
 Charge, Energy

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