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<p><b>Introduction to LC and RLC Circuits</b></p><p> What happens when we combine an inductor and a capacitor in a circuit? One has an EMF that depends on \(Q\) and one has an EMF that depends on the second time derivative \(\frac{d^2Q}{dt^2}\). We will see that this leads to oscillatory motion, so we will start with a review of Simple Harmonic Motion for the system of a mass on a spring and then explore the analogy between this mechanical system and an LC circuit.</p>
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<li><a href="https://openlearninglibrary.mit.edu/courses/course-v1:MITx+8.02.3x+1T2019/pdfbook/0/#viewer-frame" target="[object Object]">Chapter 11.7-11.8: LC and RLC Circuits</a></li>
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LC circuit
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Consider the circuit shown in the figure. After a long time in position (a), the switch is moved to position (b). After that, the current through the newly formed lower loop rises to some maximum value. At the instant when the current has its maximum value, which of the following statements is (are) true? </p>
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<text>1. The charge on the capacitor has its maximum value.</text>
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<text>2. The magnetic field in the inductor has its maximum magnitude.</text>
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<text>3. The electric field in the capacitor has its maximum magnitude.</text>
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<text>4. The charge on the capacitor is zero.</text>
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<text>5. The induced emf in the inductor has its maximum value.</text>
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<h2 class="hd hd-2 unit-title">L29Q2: Chapters in LC Circuit Oscillation</h2>
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Chapters in LC Circuit Oscillation
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<p>
Consider the circuit shown in the leftmost figure above. Switch S has been in position (a) for a long time. At time [mathjaxinline]t=0[/mathjaxinline], S is moved to position (b). In Figures 1 and 2, the polarity of the charge in the capacitor and the direction of the current through the inductor are shown at two different instants of time after S is moved to position (b). </p>
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Which one of the following statements is true? </p>
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<text> In Figure 1, the E field increases and the B field decreases. In Figure 2, the E field decreases and the B field increases</text>
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<text> In Figures 1 and 2, the E field increases and the B field decreases</text>
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<text> In Figures 1 and 2, the E field decreases and the B field increases</text>
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<text> In Figure 1, the E field decreases and the B field increases. In Figure 2, the E field increases and the B field decreases</text>
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<h2 class="hd hd-2 unit-title">L29Q3: LC Circuit Period of Oscillation</h2>
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LC circuit and Period of Oscillation
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Consider the circuit shown in the leftmost figure above. Switch S has been in position (a) for a long time. At time [mathjaxinline]t=0[/mathjaxinline], S is moved to position (b). In Figures 1 and 2, the polarity of the charge on the capacitor and the direction of the current through the inductor are shown at two different instants of time after S is moved to position (b). Let [mathjaxinline]T[/mathjaxinline] be the period of the charge and current oscillation. </p>
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Which of the following statements about the instant of time [mathjaxinline]t[/mathjaxinline] corresponding to the situation shown in the two figures is true? </p>
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<text> In Figure 1, [mathjaxinline]t[/mathjaxinline] is in the range [mathjaxinline]T/2&lt;t&lt;3T/4[/mathjaxinline]; and in Figure 2, [mathjaxinline]t[/mathjaxinline] is in the range [mathjaxinline]0&lt;t&lt;T/4[/mathjaxinline].</text>
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<text> In Figure 1, [mathjaxinline]t[/mathjaxinline] is in [mathjaxinline]T/2&lt;t&lt;3T/4[/mathjaxinline]; and in Figure 2, [mathjaxinline]t[/mathjaxinline] is in the range [mathjaxinline]3T/4&lt;t&lt;T[/mathjaxinline].</text>
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<text> In Figure 1, [mathjaxinline]t[/mathjaxinline] is in the range [mathjaxinline]T/4&lt;t&lt;T/2[/mathjaxinline]; and in Figure 2, [mathjaxinline]t[/mathjaxinline] is in the range [mathjaxinline]3T/4&lt;t&lt;T[/mathjaxinline].</text>
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<text> In Figure 1, [mathjaxinline]t[/mathjaxinline] is in the range [mathjaxinline]T/4&lt;t&lt;T/2[/mathjaxinline]; and in Figure 2, [mathjaxinline]t[/mathjaxinline] is in the range [mathjaxinline]0&lt;t&lt;T/4[/mathjaxinline].</text>
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LC Oscillator
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Consider the LC circuit above. At time [mathjaxinline]t=0[/mathjaxinline] (left figure), the capacitor has a charge [mathjaxinline]Q_0[/mathjaxinline] and there is no current through the inductor. At a later time [mathjaxinline]t[/mathjaxinline] (right figure), the magnitude of the charge is decreasing, creating the clockwise current shown. </p>
<p><b class="bfseries">(Part a)</b> If we use the convention of clockwise current positive, the current [mathjaxinline]I[/mathjaxinline] is given by: </p>
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<text> [mathjaxinline]I=+\dfrac {dq}{dt}[/mathjaxinline]</text>
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<text> [mathjaxinline]I=-\dfrac {dq}{dt}[/mathjaxinline]</text>
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The differential equation for the charge in the capacitor is given by: </p>
<table id="a0000000002" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000003">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \dfrac {d^2q}{dt^2} = -\dfrac {1}{LC} q[/mathjaxinline]
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<span>(<span>1</span>)</span>
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<p>
and the general solution of this equation is: </p>
<table id="a0000000004" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000005">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle q(t)=A \cos (\omega _0 t+\phi )[/mathjaxinline]
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<span>(<span>2</span>)</span>
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where [mathjaxinline]A[/mathjaxinline] and [mathjaxinline]\phi[/mathjaxinline] are constants to be determined by the initial conditions of the problem. The constant [mathjaxinline]\omega _0[/mathjaxinline] is the angular frequency of the oscillations of units rad/s. </p>
<p><b class="bfseries">(Part b)</b> Calculate [mathjaxinline]\dfrac {dq}{dt}[/mathjaxinline], the time derivative of the charge. Express your answer in terms of [mathjaxinline]A[/mathjaxinline], [mathjaxinline]t[/mathjaxinline], omega_0 for [mathjaxinline]\omega _0[/mathjaxinline], and phi for [mathjaxinline]\phi[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\dfrac {dq}{dt}=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part c)</b> Calculate [mathjaxinline]\dfrac {d^2q}{dt^2}[/mathjaxinline], the second time derivative of the charge. Express your answer in terms of [mathjaxinline]A[/mathjaxinline], [mathjaxinline]t[/mathjaxinline], omega_0 for [mathjaxinline]\omega _0[/mathjaxinline], and phi for [mathjaxinline]\phi[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\dfrac {d^2q}{dt^2}=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part d)</b> Substitute the expressions for [mathjaxinline]q(t)[/mathjaxinline] given in Eq. (2) and the expression for [mathjaxinline]\dfrac {d^2q}{dt^2}[/mathjaxinline] derived in part (c) into Eq. (1). What is the value of [mathjaxinline]\omega _0[/mathjaxinline] needed for Eq. (2) to be a solution of Eq. (1)? Express your answer in terms of [mathjaxinline]A[/mathjaxinline], phi for [mathjaxinline]\phi[/mathjaxinline], [mathjaxinline]L[/mathjaxinline] and [mathjaxinline]C[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]\omega _0=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part e)</b> Use the initial conditions of the problem to determine the constants [mathjaxinline]A[/mathjaxinline] and [mathjaxinline]\phi[/mathjaxinline]. Express your answer in terms of [mathjaxinline]L[/mathjaxinline], [mathjaxinline]C[/mathjaxinline], and Q_0 for [mathjaxinline]Q_0[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]A=[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\phi =[/mathjaxinline] </p>
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The charge and current in an LC oscillator
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<p><b class="bfseries">(Part a)</b> At time [mathjaxinline]t=0[/mathjaxinline], the magnitude of the current through the inductor in an ideal LC circuit is at its maximum value. The figures shown below are possible plots of the current through the inductor [mathjaxinline]I[/mathjaxinline] and charge on the capacitor [mathjaxinline]q[/mathjaxinline] against time [mathjaxinline]t[/mathjaxinline] on the [mathjaxinline]x[/mathjaxinline]-axis. The convention is that the current is positive if it is clockwise as shown above and the charge in the plots is positive if the charge on the <em>left</em> plate is positive. </p>
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Which of the following statements about the depictions in the figures above is true? </p>
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<text> In Figure 3, the solid line is [mathjaxinline]I[/mathjaxinline] and the dashed line is [mathjaxinline]q[/mathjaxinline].</text>
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<p><b class="bfseries">(Part b)</b> Assume that the units of the vertical axis is mA for the current and mC for the charge. What is the value of [mathjaxinline]\omega _0[/mathjaxinline], the angular frequency of the oscillator? </p>
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<p style="display:inline">[mathjaxinline]\omega _0=[/mathjaxinline]</p>
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<span class="trailing_text" id="trailing_text_checkpoint_w12_16_3_1">[mathjaxinline]\mathrm{(rad\cdot s^{-1} ) }[/mathjaxinline]</span>
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<h2 class="hd hd-2 unit-title">L29v6: Energy in the LC Oscillator</h2>
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<h2 class="hd hd-2 unit-title">L29Q6: Energy in an LC Oscillator </h2>
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LC Energy Oscillations
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<p>Consider an LC circuit in which at \(t=0\), the capacitor has its maximum charge and there is no current in the inductor. For these initial conditions, the red dotted and the black dashed graphs show the time dependence of the energy stored in the capacitor and inductor, respectively. Which of the following statements is (are) true about these curves and the underlying physics?</p>
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<text>The frequency at which the stored energy in the capacitor oscillates is twice the frequency with which the voltage drop across the capacitor (the potential difference from one plate to the other) repeats</text>
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<text>The sum of the two curves in the graph must be a constant because energy is conserved</text>
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<text>The sum of the two curves in the graph must be a constant because one curve is proportional to \(\sin^2(\theta)\) and the other is proportional to \(\cos^2(\theta)\)</text>
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<text>The storage of energy in the magnetic field is out of phase with the storage of energy in the electric field</text>
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<text>The scale on the y axis would be correct in joules, and you would get these curves if \(C = 1\, F\), \(L=1\,H\), and \(Q_0 = 1\, C\)</text>
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<h2 class="hd hd-2 unit-title">L29v7: RLC Circuit</h2>
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RLC Circuit
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<p>
Consider a capacitor, an inductor, and a resistor connected in series as shown above. In the left figure, the switch is opened and the capacitor has an initial charge [mathjaxinline]Q_0[/mathjaxinline]. There is no current in the circuit. In the right figure, the switch is closed and charges leaving the upper plate of the capacitor set up a current [mathjaxinline]I[/mathjaxinline] as shown. </p>
<p><b class="bfseries">(Part a)</b> Consistent with the direction of the current shown in the figure, which statement is true? <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_checkpoint_w12_22_2_1">
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</div><b class="bfseries">(Part b)</b> Circulating the loop clockwise (in the same direction as the current) starting at the switch, we obtain: </p>
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<text> [mathjaxinline]+\dfrac {Q}{C}-IR+L\dfrac {dI}{dt}=0[/mathjaxinline]</text>
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<text> [mathjaxinline]+\dfrac {Q}{C}-IR-L\dfrac {dI}{dt}=0[/mathjaxinline]</text>
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<text> [mathjaxinline]+\dfrac {Q}{C}+IR+L\dfrac {dI}{dt}=0[/mathjaxinline]</text>
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<text> [mathjaxinline]+\dfrac {Q}{C}+IR-L\dfrac {dI}{dt}=0[/mathjaxinline]</text>
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<p><b class="bfseries">(Part c)</b> Replace your expression of the current you picked in part (a) in the equation you chose in part (b). The second order differential equation for the charge in the upper plate of the capacitor as a function of time is: </p>
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<text> [mathjaxinline]\dfrac {Q}{C}-\dfrac {dQ}{dt}R-L\dfrac {d^2Q}{dt^2}=0[/mathjaxinline]</text>
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<text> [mathjaxinline]\dfrac {Q}{C}+\dfrac {dQ}{dt}R-L\dfrac {d^2Q}{dt^2}=0[/mathjaxinline]</text>
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<text> [mathjaxinline]\dfrac {Q}{C}-\dfrac {dQ}{dt}R+L\dfrac {d^2Q}{dt^2}=0[/mathjaxinline]</text>
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<text> [mathjaxinline]\dfrac {Q}{C}+\dfrac {dQ}{dt}R+L\dfrac {d^2Q}{dt^2}=0[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">Optional: RLC Circuit</h2>
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Damped oscillator
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In the left figure above, a capacitor and an inductor are connected in series with a resistor. At the instant shown, a current [mathjaxinline]I[/mathjaxinline] is flowing in the circuit and a charge [mathjaxinline]Q[/mathjaxinline] is on the capacitor, with the upper plate positive. The differential equation obtained for the charge on the capacitor is: </p>
<table id="a0000000002" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000003">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle L\dfrac {d^2Q}{dt^2} +R\dfrac {dQ}{dt}+\dfrac {1}{C}Q = 0[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">
<span>(<span>1</span>)</span>
</td>
</tr>
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<p>
In the right figure, a block of mass [mathjaxinline]m[/mathjaxinline] is connected to the end of an ideal spring of spring constant [mathjaxinline]k[/mathjaxinline]. The other end of the spring is fixed to a wall. In addition, a drag force [mathjaxinline]\mathbf{\vec{F}_ D}=-b\mathbf{\vec{v}}[/mathjaxinline], with [mathjaxinline]b&gt;0[/mathjaxinline] and constant, is acting on the box. At the instant shown in the figure, the spring is stretched by [mathjaxinline]x(t)[/mathjaxinline] from its equilibrium position (at [mathjaxinline]x=0[/mathjaxinline]) and the box is moving with velocity [mathjaxinline]\mathbf{\vec{v}}=|v|\mathbf{\hat{i}}[/mathjaxinline]. </p>
<p><b class="bfseries">(Part a)</b> Applying Newton's 2nd law to the box and using the coordinate system shown in the figure, the resulting second order differential equation for the box's position is </p>
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<text> 1. [mathjaxinline]m\dfrac {d^2x}{dt^2} +kx -b\dfrac {dx}{dt} = 0[/mathjaxinline]</text>
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<text> 2. [mathjaxinline]m\dfrac {d^2x}{dt^2} +kx +b\dfrac {dx}{dt} = 0[/mathjaxinline]</text>
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<text> 3. [mathjaxinline]m\dfrac {d^2x}{dt^2} -kx -b\dfrac {dx}{dt} = 0[/mathjaxinline]</text>
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<text> 4. [mathjaxinline]m\dfrac {d^2x}{dt^2} -kx +b\dfrac {dx}{dt} = 0[/mathjaxinline]</text>
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<p><b class="bfseries">(Part b)</b> We can find similarities between the block-spring system and the RLC circuit. Select all that apply: </p>
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<text>1. The resistor plays the role of the drag force</text>
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<text>2. The current plays the role of the box's velocity.</text>
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<text>3. The inductance [mathjaxinline]L[/mathjaxinline] is equivalent to [mathjaxinline]1/k[/mathjaxinline].</text>
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<text>4. The electric energy in the capacitor is equivalent to the potential energy of the spring.</text>
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