Chapter 32 Inductance

32.1 Self-Inductance and Inductance

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A time-varying current created by an emf source in a circuit induces an emf in the same circuit. That emf is called the induced emf and
the current related to it called the induced current.
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Convention: when only emf or current term (without an adjective) is used, it means the quantities are related to a physical source (like a
battery). When the induced emf or the induced current term is used, it means the quantities are related to a change in magnetic field.

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Self-Induction: Consider the circuit shown in Figure 32.1 There is a resistor, an emf source and a switch are
present in the circuit.
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When the switch is closed, a current is created in the circuit. The current in turn produces a magnetic field
around the wire.
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The magnetic field causes a a nonzero magnetic flux through the circuit.
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According to Faraday’s law, an emf (a back emf) is induced in the circuit. That induced emf is called self-
induced emf ƐL . The effect is called self-induction.
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The self-induced emf creates an induced current in the circuit. Lenz’s law say that the current should be in the
direction opposite that of the current created by the emf source.
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Because of the induced current, the current in the circuit does not immediately rise to the maximum value Ɛ/R.
It gradually increases to that maximum value.
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The self-induced emf is a result of the changing magnetic flux, or changing the magnetic field or the changing
current which creates the field. So we may write for ƐL
ƐL = -L dI/dt
where L is a proportionality constant It is called the inductance of the loop.
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Faraday’s law says that ƐL = -N dΦB/dt for a loop with N turns.
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Combine it with the previous equation: -N dΦ B/dt = -L dI/dt or d(NΦB)/dt = d(LI)/dt. It follows that
L = NΦB / I
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The first equation also becomes
L = - ƐL / (dI/dt)
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Compare it with R = V/I. R in here is the measure of opposition to current. So L is a measure of opposition to dI/dt, a change in current.
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SI unit of inductance is the henry (H), after Joseph Henry. 1 H = 1 V/(1 A/s) or 1 H = 1 V · s / A.
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The inductance of a coil is depends on its geometry like the capacitance of a capacitor.
32.2 RL Circuits
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Consider a solenoid (or a any other coil). Take and connect it together with a resistor to a circuit which has an emf source.
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The coil creates an induced emf when a current starts in the circuits. The inductance of the coil thus prevents a sudden increase, or
decrease, in the current.
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Such a coil is called inductor and is shown by in a circuits.
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Such inductors have large inductance values compared to that of the circuit. The inductance of a circuit will be omitted in the rest of the
chapter though it exists in reality.
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Consider the circuit shown in Figure 32.2. A source of emf, two switches S 1 and S2 , a resistor and an inductor
are connected to the circuit.
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The circuit is called an RL circuit because of the resistor and the inductor.
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The switch S2 has a curved part so that it is always set to either a or b. That is, it is always closed.
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Let’s initially, that is when t > 0, the switch S 1 is open and S2 is set to a.
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Switch S1 is closed at t = 0. A current begins in the circuit. But an induced current is created by the back emf
due to the inductor at the same instant. The induced current opposes the current.
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Let’s find the current in the circuit. If we apply Kirchhoff’s loop rule to the circuit, we get
Ɛ - IR - LdI/dt = 0
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That is a differential equation for I. We have to solve it to get the current as a function time.
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Organize the differential equation: Ɛ/R - I - (L/R)dI/dt = 0 – > I - Ɛ/R = - (L/R)dI/dt – > dI / ( I – Ɛ/R)
= -R dt/L.
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The left hand side of the last equation is in terms I (current) only, while its right hand side is in terms of t (time)
only. Integrate both sides.
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The results is ln(I – Ɛ/R) + ln(A) = -R t/ L where A is just a constant to be determined.
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Organize the result: ln[(I – Ɛ/R)/ A] = -R t/ L – > (I – Ɛ/R)/ A = e-Rt/L – > I = A e-Rt/L + Ɛ/R.
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Find A. We know that at t = 0 the current I = 0, so let t = 0 in the equation: 0 = A (1) + Ɛ/R – > A = -Ɛ/R.
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The final form of the equation is I = Ɛ(1 - e-Rt/L)/R.
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What does the final equation say?
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The current function has an exponentially decreasing part.
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So, the current does not immediately increase to its final value. It takes time.
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The exponential part goes to zero as time goes to infinity, then the current approaches its final value, Ɛ/R.
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We can recover the current in the case of a R only by letting L goes to zero in the exponential part: as L - >
zero R/L - > infinity and therefore e -Rt/L – > zero.
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Note that L/R has a dimension of time: L/R = (V s / A)(A/V) = s – > time. It is said that L/R is the time
constant τ of the RL circuit: τ = L/R.
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In terms of time constant, the current function becomes
I = Ɛ(1 - e-t/τ)/R
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The current reaches 63.2% of its final value Ɛ/R at t = τ.
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Figure 32.3 shows a graph of the current versus time in the RL circuit. The current
approaches to its final value as time goes to infinity. That is, the curve becomes
tangent to the dashed line which intersects I axis at I = Ɛ/R as time goes infinity.
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What is the time rate of change of the current? Take the time derivative of the
current function:
dI/dt = Ɛ e-Rt/L/L
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The rate is an exponential function of time, it decays exponentially with time as
shown in Figure 32.4.
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The rate is maximum at t = 0.
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Go back to the differential equation: Ɛ - IR -LdI/dt = 0. Rewrite it: Ɛ = IR +
LdI/dt.
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The term on the left is the emf of the emf source, which is constant in time.
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The first term on the right is the potential difference across the resistor and the
second term is the potential difference across the inductor. Both terms change
with time. They are not constants.
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But the sum of the terms on the right should be constant as their sum is equal to
the emf, Ɛ.
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That means one of the terms should decrease while the other one increases, or
vice versa. In fact, that is what we get from the solution. See Figure
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When the current has reached its final value, set the switch S2 to b as shown in Figure 32.2.
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Now the loop consists of the resistor and the inductor. The source of emf is not in the loop.
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Rewrite the equation derived from the Kirchhoff’s loop rule by setting Ɛ = 0 :
- IR - LdI/dt = 0 or IR + LdI/dt = 0
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Now we have another differential equation for current. Let’s solve it:
IR + LdI/dt = 0 → LdI/dt = -IR → (1/I)dI/dt = -Rdt/L → ln(I) + ln(A) = -Rt/L →ln(I/A) = -Rt/L →
I = A e-Rt/L
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Find the constant A. We know that at t = 0 the current is I = Ɛ/R. So Ɛ/R= A e0 →A = Ɛ/R
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Insert it into the result and write the solution in the final form: I= Ɛe-t/τ /R
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The equation tells that the current decreases exponentially and it goes zero as time goes to infinity.

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The current goes to zero immediately if the inductor is not in the circuit. The inductor
opposes the decrease in the circuit. As shown in Figure 32.5, the current decreases
exponentially in the presence of the inductor.
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The differential equation is equal to zero. That means the changes in the two terms
should cancel each other. That is, while one of them is increasing, the other one should
be decreasing.
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Since the we expect a decreasing current in the circuit, RI (the potential difference
across R) term should decrease and LdI/dt (the potential difference across L) term
should increase at the same time.
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What is dI/dt? It is -Ɛe-Rt/L /L. It is negative. As time goes to infinity, the LdI/dt term goes
to zero. So the term increases.
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As you guess RI term is equal to -Ɛe-Rt/L which is decreasing.
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Next Figure plots both terms.
32.3 Energy in a Magnetic Field
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Multiply the differential equation for the current by I:
ƐI - I2R - LIdI/dt = 0 or ƐI = I2R + LIdI/dt
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The term on the left hand (ƐI) is the power delivered by the source of emf to the circuit at an instant of
time.
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The first term on the right hand side (I2R) is the power delivered to the resistor at the same instant.
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The conservation of energy requires that the second term on the right hand side (LIdI/dt ) should be the
power delivered to the inductor at the same instant. See Figure
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The resistor converts the energy provided it to internal energy (e.g. heat).
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The inductor stores the energy provided it.
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Let’s calculate the amount of the stored energy (U) in the inductor:
dU/dt = LIdI/dt → dU/dt = d(LI2/2)/dt →
U = LI2/2
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The result says that at a given instant of time t, the stored energy, U, in the inductor is equal to LI 2/2.
Do not forget the current I in the expression is the current at that instant.
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If we substitute the current expression in the equation, we get U = Ɛ2(1 - e-t/τ)2L/2R2. See Figure
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The stored energy in the inductor has a from similar to the one stored in a capacitor: CV 2/2.
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Consider the solenoid asked in Example 32.1. We have found that its inductance is L = μ0n2V.
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The magnetic field inside the solenoid is given by B = μ0nI.
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Substitute them into the expression for the potential energy of the inductor:
U = μ0n2V I2 / 2 = μ0n2 V B2 / 2n2μ20 = B2V/2μ0 or U = B2V/2μ0.
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The potential energy stored has been written in terms of the magnetic field inside the solenoid. That
means the energy provided by the source of emf is used to establish a magnetic field inside the
solenoid. This is similar to the case where the source of emf provides energy to create an electric field
between the plates of a capacitor.
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We can define an energy density u B by dividing the stored energy by volume:
U/V = B2/2μ0 → uB = B2V/2μ0
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The energy density uB is valid for any region of space in which a magnetic field exists. Remember we
derived a similar expression for electric field (u E = Ɛ0 E2/2).
Example: Find when does the power delivered to an
inductor in LR circuit makes maximum?
Solution
The power makes a maximum as shown in Figure where
PR, PL, and PT. To find its position write the power
expression for the inductor:P L = LIdI/dt.
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Insert I and dI/dt there: P L = L(Ɛe-t/τ /L)Ɛ(1 - e-t/τ)/R =
Ɛ2e-t/τ(1-e-t/τ)/R.

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Takes its time derivative:
dPL/dt = (Ɛ2/R)[(-1/τ)e-t/τ(1-e-t/τ)+e-t/τ(-1/τ)(-e-t/τ)]
= (Ɛ2/R) (1/τ)e-t/τ(-1+2e-t/τ)
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Set it to zero:
(Ɛ2/R) (1/τ)e-t/τ(-1+2e-t/τ) = 0 → e-t/τ = 0 or -1+2e-t/τ = 0
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The first one says t = ∞ and the second one says e-t/τ
= ½ → t = τ ln(2).
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As t approaches zero, the power goes to zero. So the
answer is t = τ ln(2) = L ln(2)/R.
32.4 Mutual Inductance
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Consider the two coils shown in Figure 32.8. Coil 1 has N 1 turns and carries the current I 1. Coil 2
has N2 turns and carries the current I 2.
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The magnetic field due to coil 1 creates a magnetic flux Φ12 through coil 2.
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We define the mutual inductance M 12 of coil 2 with respect to coil 1 as
M12 = N2 Φ12 / I1
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The mutual inductance depends on the geometry of both coils and their orientation with respect to
each other.
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If the distance between coils increases, the mutual inductance decrease because of the decreasing
flux.
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Now assume that the current I1 in coil 1 changes with time. Then Faraday’s law says that an emf
should be induced in coil 2. The induced emf Ɛ2 is given by
Ɛ2 = -N2d(Φ12)/dt = -N2d(M12I1/N2)/dt = -M12dI1/dt
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We could consider the mutual inductance M 21 of coil 1 with respect to coil 2. If the current I2
changes with time, the induced emf Ɛ1 in coil 1 is
Ɛ1 = -M21dI2/dt
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As you see, the induced emf in one coil depends on the time rate of change of the current in the
other coil.
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The induced emf is proportional to the mutual inductance of each coil.
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It can be shown that the mutual inductance of each coil is equal to each other:
M12 = M21 = M
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Therefore, each induced emf becomes
Ɛ1 = -MdI2/dt and Ɛ2 = -MdI1/dt
32.4 Oscillations in an LC Circuit
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Consider the circuit shown in Figure 32.10. A capacitor is connected to an inductor. There is an open switch
in the circuit.
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Assume that all the wires have negligible resistance and there is no energy lost via radiation.
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Imagin that the capacitor has a charge given by Q max on it. So , the capacitor has an energy given by
Q2max/2C stored in the electric field between its plates.
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Let’s say the switch is closed at t = 0. As soon as the switch is closed, the charges leave or enter the plates
of the capacitor.Therefore, a current is created in the circuit. The current is equal to the time rate of the
change in the amount of charge on the plates of the capacitor.
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At the same time, the energy stored in the capacitor decreases. The decrease appears as an increase in the
energy stored in the magnetic field of the inductor. That is, the energy of the electric field is transformed to
the magnetic field.
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Later, the capacitor is full discharged; it stores no energy; the current reaches its maximum value. The
inductor stores all the energy.
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Then, the current decreases in the same direction but the capacitor is being charged. Finally, the capacitor is
fully charged again. But its plates carry opposite signs.
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The capacitor is discharged again. The current reaches its maximum. The inductor transfer all the energy.
Then the charging of the capacitor starts again. So the previous steps repeat themselves. That is, the energy
is transformed between the capacitor and the inductor repeatedly. The energy, the charge and the current
oscillate.
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The situation is similar to the energy oscillations in a spring-mass system. The potential energy stored in the
spring is transformed to the mass as a kinetic energy. Then the kinetic energy is transformed back to the
spring as the potential energy.
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The similarity is shown in Figure 32.11.
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The energy stored in the capacitor is the analog of the energy stored in the spring, kx 2/2.
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At t = 0, the energy stored in the capacitor starts to decrease and the energy stored in the inductor starts to
increase. The analog of the energy stored in the inductor is the kinetic energy of the mass, mv 2/2.
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When the current reaches its maximum value I max, all the energy given by LI 2max/2 is stored in the inductor.
The capacitor has zero energy.
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Between zero current and the maximum current, the total energy is partially stored in the capacitor and
partially stored in the inductor.
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The total energy of the circuit is conserved. The total energy of the system is the sum of the potential energy stored in the capacitor and
the potential stored in the inductor at an instant of time.
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So if the total energy is U, then we write U = UC+ UL = Q 2/2C + LI2/2.
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Since the total energy is constant in time, we get dU/dt = 0, so it is possible to write dU/dt = (Q/C)dQ/dt + LIdI/dt = 0.
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The resulting equation has two variables Q and I. It contains their first derivatives, too.
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If we eliminate one of the variables and its derivatives, we can solve the resulting differential equation.
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Let’s use I = dQ/dt and dI/dt = d 2Q/dt2 in the equation: (Q/C)dQ/dt + LdQ/dt d2Q/dt2 = 0.
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Everything is in terms of Q and its derivatives: (Q/C + Ld2Q/dt2)dQ/dt = 0.
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Here we have two equations: dQ/dt = 0 and Q/C + Ld2Q/dt2 = 0.
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The first one gives Q = constant. But this is not a solution we want. We know that Q is not constant in time.
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The second one is a differential equation to be solved: Q/C + Ld2Q/dt2 = 0.
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The solution is given by Q = Q max cos(ωt + φ) where ω =1/ √ LC and φ is a constant. We can find the constant using the initial conditon Q
= Qmax when t = 0. So we get Q max = Qmax cos(φ). Form there we have φ = 0. Therefore, the solution becomes Q = Q max cos(ωt)
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Using I = dQ/dt, we find the current I = -ωQmax sin(ωt) = -Imax sin(ωt).
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The solutions are oscillating functions of time. Both Q and I oscillate between a maximum and a minimum value as time changes. The
frequency of the oscillations is given by ω =1/ √ LC . The period of the oscillations becomes T=2 π √ LC .
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Figure 32.12 shows Q and I as functions of time. As you see, Q oscillates between Q max and -Qmax while
the I oscillates between I max and -Imax.
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An important point is that the current is 90° out of phase with the charge. That is, the current is zero
when the charge is maximum or the charge is zero when the current is maximum. Note: cos(ωt+90°)= -
sin(ωt).
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Let’s write the total energy in terms of the solutions:
U = UC+ UL = Q2/2C + LI2/2 = Q2max cos2(ωt)/2C + LI2max sin2(ωt)/2
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The total energy oscillates between the energy stored in the capacitor and the energy stored in the
inductor.
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When t = 0, the total energy Q2max /2C is stored in the capacitor.
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When t = T/4 = π/2ω, the total energy is stored in the inductor. It is equal to LI2max
/2. Since the energy is conserved, Q 2max /2C should be equal to LI 2max /2. That is,
Q2max /2C = LI2max /2 holds.
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Therefore, we have U = Q2max cos2(ωt)/2C + Q2max sin2(ωt)/2C = Q2max [cos2(ωt) +
sin2(ωt)]/2C = Q2max /2C (or = LI2max /2)
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Figure 32.13 shows plots of time variations of U C and UL .
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Note that sum of the two curves gives the constant Q2max /2C.
How do we obtain the solution? One of the ways is given
here.
32.6 The RLC Circuit
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Consider the circuit shown in Figure 32.15. The circuit consists of an emf source, a switch, an inductor, a
resistor and a capacitor.
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Switch is initially set to a. The capacitor is charged to a maximum value Q max.
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Now throw the switch to position b. The circuit has the inductor, the resistor and the capacitor, which is fully
charged.
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A current is created in the circuit. The capacitor discharges. Energy stored in the capacitor is transferred to
the resistor and the inductor.
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The resistor does not store any energy but it transforms the energy it receives to internal energy. The time
rate of the energy transformation is given by I 2R.
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If U is the stored energy in the circuit, we write dU/dt = -I 2R. That is the energy decreases at a rate given by
I2R.
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Substitute U = Q2/2C+ LI2/2 in the equation: LIdI/dt + (Q/C)dQ/dt = -I 2R.
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This is an another differentia equation in terms of Q, I and their derivatives. Replace dQ/dt with I and dI/dt
with d2Q/dt2: LId2Q/dt2 + (Q/C)I + I2R = 0.
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Divide the result by I: Ld2Q/dt2 + Q/C + IR = 0.
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Replace I with dQ/dt:
Ld2Q/dt2 + RdQ/dt + Q/C = 0

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This is a differential equation in terms Q and its derivatives only.
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Its solution is
Q = Qmax e-Rt/2L cos(ωdt)
where ωd, the angular frequency of the oscillation, is given by
ωd = [1/LC – (R/2L)2]1/2
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This equation says that the value of the charge on the capacitor undergoes a damped harmonic oscillation.
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Figure 32.16 a shows a plot of the charge against the time.
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Since the current is dQ/dt, it undergoes a damped harmonic oscillation, too.
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The damping is due to the presence of the resistor. The exponential term leads to the damping. While the cosine term oscillates, the
exponential term makes it smaller as time increases.
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The degree of the damping depends on the resistance. In fact, there is no oscillation for R values above R c= (4L/C)1/2. The system in
that case is called overdamped. The value Rc is called the critical resistance.
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When R =Rc, the system is called critically damped.
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As you guess when R = 0, the solution becomes that of the LC circuit.
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The current is given by I = -Q max e-Rt/2L[ R cos(ωdt) / 2L + ωdsin(ωdt)] if
dQ/dt is evaluated.
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Next Figure shows three different current curves. The curves show underdamped
, critically damped and overdamped cases.

Figure: The current in an RLC circuit as a function of time. R1

shows an underdamped, while R2 and R3 show critically and
overdamped cases.