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<h2 class="hd hd-2 unit-title">Introduction to RC and LR Circuits</h2>
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<h2>Introduction to RC and LR Circuits</h2><p> Now that we understand inductors, we are ready to put them into circuits. We start this lesson by reviewing the RC circuit that we discussed in a previous part of the course. It turns out that the mathematical description of the RC circuit will be mirrored by the discussion of the LR circuit.</p>
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<p>Textbook Links</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 11.4-11.6: LR Circuit</a></li>
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<h2 class="hd hd-2 unit-title">Review of RC Circuits</h2>
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<h2>RC circuits, charging and discharging a capacitor - Review</h2><h2>Charging a capacitor.</h2><p>An RC circuit is a circuit that consists of only capacitors, resistors, and fixed-voltage power sources. Let us look at the simplest of these.</p><p><center><img src="/assets/courseware/v1/7064a6aa6741979456378f426feca026/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_html_lesson28_01.svg" width="400"/></center></p><p>
</p><p>We will start with the switch in position 1, which includes the battery in the circuit. The battery is going to drive a current \(I\) in the clockwise direction as shown. Because current will be flowing into the top plate of the capacitor, it will add \(+Q\) on the top plate, therefore \(I = \dfrac{dQ}{dt}>0\). At the same time, positive charge is flowing <em>out</em> of the bottom plate, leaving behind a net negative charge that is equal in magnitude to the charge on the top plate. The magnitudes of the charges on the two plates are increasing and so the capacitor is charging. Applying Kirchoff's loop rule:</p>
\[\displaystyle + \mathcal{E} - IR - \frac{Q}{C} = 0\]
<p>This describes, in sequence, the battery, the resistor, and the capacitor. After substituting \(I=\dfrac{dQ}{dt}\), we have a first order, linear, inhomogeneous, ordinary differential equation. The initial condition is the initial charge on the capacitor (the charge at time \(t = 0\)). Assuming that the capacitor starts with zero charge, the solution to this equation is:</p>
\[\displaystyle Q(t) = C \mathcal{E} (1 - e^{-\dfrac{ t}{\tau}}) \Longrightarrow I =\dfrac{dQ}{dt}= \dfrac{\mathcal{E}}{R}e^{-\dfrac{t}{ \tau}} \]
<p>where \(\tau = RC\), is known as the <i>time constant</i> of the circuit. It is the amount of time required for the current to drop by a factor of \(e.\) As time progresses, the current steadily drops, while the charge on the capacitor gets closer and closer to its maximum value, \(C \mathcal{E}\). Plots of the charge and the current as functions of time are shown below.</p>
<p><center><img src="/assets/courseware/v1/32b777129f9cfad229eb5e5f843e7303/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_html_lesson28_01c.svg" width="600"/></center></p>
<p>The current drops as the charge in the capacitor increases. After a long time (many time constants), there is no current flowing, and the capacitor is fully charged.</p>
<h2>Discharging the capacitor. </h2>
<p><center><img src="/assets/courseware/v1/15e1daf9ad2666063603bf1eb52a4bad/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_html_lesson28_01b.svg" width="400"/></center></p>
<p>We now look at what happens if we flip the switch to position 2, such that only the capacitor and the resistor are in the loop. Note that the capacitor starts with a charge \(+Q_0=C \mathcal{E}\) on the top plate. Therefore, the current in the loop flows clockwise again, but now taking charges <em>out</em> from the positive plate. Because the capacitor is discharging (i.e. the time derivative of the charge is negative), the current is given by \(I = -\dfrac{dQ}{dt}\). Applying the loop rule clockwise, we obtain</p>
\[ \frac{Q}{C} - IR = 0 \]
<p>Replacing \(I = -\dfrac{dQ}{dt}\), the equation becomes \(\dfrac{Q}{C} +\dfrac{dQ}{dt}R = 0\). The solution of this equation is:</p>
\[ Q(t) = C\mathcal{E} e^{-\dfrac{t}{RC}} \Longrightarrow I= -\dfrac{dQ}{dt}=\dfrac{\mathcal{E}}{R}e^{-\dfrac{t}{RC}}\]
<p>Plots of the charge and current as functions of time are shown below.</p>
<p><center><img src="/assets/courseware/v1/80bec6ea4164a7462951e9f52ed9f257/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_html_lesson28_01d.svg" width="600"/></center></p>
<p>Note that both the current and the charge are falling with the same time constant \(\tau = RC\) as previously. An RC circuit is useful because you can make the time constant really quite small. This allows for the energy in the capacitor to be dumped very quickly, much more so than, for example, from a battery.</p>
<p>From these simple examples of capacitors, we can form two rules of thumb: When a capacitor has no charge on it, it behaves like a <b>wire</b> (zero potential difference). When a capacitor is fully charged, it behaves like a <b>break</b> (no current flows through it).</p>
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<h2 class="hd hd-2 unit-title">L28v1: Charge on a Capacitor in an RC Circuit</h2>
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<p> This is an optional review of RC Circuits. </p>
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<h2> Math Interlude </h2><p> The video below is a review of the derivation of the solution to the first order differential equation representing the charge (\(Q\)) in the case of a charging capacitor in a circuit with a fixed-voltage battery (\(\mathcal{E}\)), a capacitor (\(C\)), and a resistor (\(R\)), all in series:</p>
\[\displaystyle {\cal E} - \frac{Q}{C} - R\dfrac{dQ}{dt} = 0\]
<p> Note that the differential equation for the charge \(Q(t)\) shown above is similar to the equation for the increasing current \(I(t)\) in an inductor presented later in this lesson and shown below:</p>
\[\displaystyle {\cal E} - IR - L\dfrac{dI}{dt} = 0\]
<p> You can skip this video if you already know how to obtain the solution to these sorts of equations.</p>
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<h3 class="hd hd-2">L28v1: Charge on a Capacitor in an RC Circuit</h3>
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<h2 class="hd hd-2 unit-title">L28Q1: Circuit with Resistors and Capacitor</h2>
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Circuit with Resistors and Capacitor: Part I
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<p>
In the circuit shown in the left figure below, the switch has been open for a long time and so there is no current in any branch and the capacitor is uncharged. The switch is closed at [mathjaxinline]t = 0[/mathjaxinline] (figure below right). </p>
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<img src="/assets/courseware/v1/9d824e76c47b60f4beef36b4b009ec12/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_checkpoint_w12_9.svg" width="605"/>
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The currents through the capacitor and through the bottom resistor at [mathjaxinline]t=0^{+}[/mathjaxinline] (just after the switch is closed) are given by: </p>
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<text> [mathjaxinline]I_1 = 0[/mathjaxinline], and [mathjaxinline]I_2=[/mathjaxinline]E[mathjaxinline]/4R[/mathjaxinline]</text>
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<text> [mathjaxinline]I_1 =[/mathjaxinline]\mathcal{E}[mathjaxinline]/4R[/mathjaxinline], and [mathjaxinline]I_2 = 0[/mathjaxinline]</text>
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<text> [mathjaxinline]I_1 = 0[/mathjaxinline], and [mathjaxinline]I_2=[/mathjaxinline]\mathcal{E}[mathjaxinline]/2R[/mathjaxinline]</text>
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<text> [mathjaxinline]I_1 =[/mathjaxinline]\mathcal{E}[mathjaxinline]/2R[/mathjaxinline], and [mathjaxinline]I_2 = 0[/mathjaxinline]</text>
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Circuit with Resistors and Capacitor: Part II
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In the circuit shown in the left figure below, the switch has been open for a long time and so there is no current in any branch and the capacitor is uncharged. The switch is closed at [mathjaxinline]t = 0[/mathjaxinline] (figure below right). </p>
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<img src="/assets/courseware/v1/9d824e76c47b60f4beef36b4b009ec12/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_checkpoint_w12_9.svg" width="605"/>
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The currents through the capacitor and through the bottom resistor at [mathjaxinline]t=\infty[/mathjaxinline] (long time after the switch is closed) are given by: </p>
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<text> [mathjaxinline]I_1 = 0[/mathjaxinline], and [mathjaxinline]I_2=\mathcal{E}/4R[/mathjaxinline]</text>
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<text> [mathjaxinline]I_1 = 0[/mathjaxinline], and [mathjaxinline]I_2=\mathcal{E}/2R[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">L28Q2: Circuit with Inductor and Resistors</h2>
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Circuit with Inductor and Resistors: Part I
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In the circuit shown in the left figure below, the switch has been open for a long time and so there is no current in any branch. The switch is closed at [mathjaxinline]t = 0[/mathjaxinline] (figure below right). </p>
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<img src="/assets/courseware/v1/ded18770a2a835501106313bd18b42ba/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_checkpoint_w12_5_1.svg" width="605"/>
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The currents through the inductor and through the bottom resistor at [mathjaxinline]t=0^+[/mathjaxinline] (just after the switch is closed) are given by: </p>
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Circuit with Inductor and Resistors: Part II
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In the circuit shown in the left figure below, the switch has been open for a long time and so there is no current in any branch. The switch is closed at [mathjaxinline]t = 0[/mathjaxinline] (figure below right). </p>
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The currents through the inductor and the bottom resistor at [mathjaxinline]t=\infty[/mathjaxinline] (long time after the switch is closed) are given by: </p>
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<text> [mathjaxinline]I_1 = 0[/mathjaxinline], and [mathjaxinline]I_2=\mathcal{E}/4R[/mathjaxinline]</text>
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<text> [mathjaxinline]I_1 = \mathcal{E}/4R[/mathjaxinline], and [mathjaxinline]I_2 = 0[/mathjaxinline]</text>
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<text> [mathjaxinline]I_1 = 0[/mathjaxinline], and [mathjaxinline]I_2=\mathcal{E}/2R[/mathjaxinline]</text>
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<text> [mathjaxinline]I_1 = \mathcal{E}/2R[/mathjaxinline], and [mathjaxinline]I_2 = 0[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">L28Q3: Decaying Current</h2>
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Decaying Current
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<p><b class="bfseries">(Part a)</b> An ideal inductor of inductance [mathjaxinline]L[/mathjaxinline] is connected to a resistor of resistance [mathjaxinline]R[/mathjaxinline] as shown. The equation to determine [mathjaxinline]I[/mathjaxinline], the current in the inductor at a given instant of time is given by: </p>
<p>
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<text> [mathjaxinline]\dfrac {dI}{dt}=\dfrac {I}{LR}[/mathjaxinline]</text>
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<text> [mathjaxinline]\dfrac {dI}{dt}=-\dfrac {I}{LR}[/mathjaxinline]</text>
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<text> [mathjaxinline]\dfrac {dI}{dt}=\dfrac {IR}{L}[/mathjaxinline]</text>
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<text> [mathjaxinline]\dfrac {dI}{dt}=-\dfrac {IR}{L}[/mathjaxinline]</text>
</label>
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<text> [mathjaxinline]\dfrac {dI}{dt}=\dfrac {IL}{R}[/mathjaxinline]</text>
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<text> [mathjaxinline]\dfrac {dI}{dt}=-\dfrac {IL}{R}[/mathjaxinline]</text>
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<p><b class="bfseries">(Part b)</b> Find [mathjaxinline]I(t)[/mathjaxinline], the solution of the equation obtained in part (a) knowing that at time [mathjaxinline]t=0[/mathjaxinline] the current in the inductor is [mathjaxinline]I_0[/mathjaxinline]. Express your answers in terms of [mathjaxinline]L[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]t[/mathjaxinline], and I_0 for [mathjaxinline]I_0[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]I(t)=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part c)</b> The current in the inductor at time [mathjaxinline]t = L/R[/mathjaxinline] is approximately equal to: . <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 3" role="group"><div class="choicegroup capa_inputtype" id="inputtype_checkpoint_w12_7_4_1">
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<input type="radio" name="input_checkpoint_w12_7_4_1" id="input_checkpoint_w12_7_4_1_choice_1" class="field-input input-radio" value="choice_1"/><label id="checkpoint_w12_7_4_1-choice_1-label" for="input_checkpoint_w12_7_4_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w12_7_4_1"> <text> [mathjaxinline]0.6I_0[/mathjaxinline]</text>
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<input type="radio" name="input_checkpoint_w12_7_4_1" id="input_checkpoint_w12_7_4_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="checkpoint_w12_7_4_1-choice_2-label" for="input_checkpoint_w12_7_4_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w12_7_4_1"> <text> [mathjaxinline]0.5I_0[/mathjaxinline]</text>
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<input type="radio" name="input_checkpoint_w12_7_4_1" id="input_checkpoint_w12_7_4_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="checkpoint_w12_7_4_1-choice_3-label" for="input_checkpoint_w12_7_4_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w12_7_4_1"> <text> [mathjaxinline]0.4I_0[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">L28Q4: Energy in a RL Circuit</h2>
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Energy in an LR circuit
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In the circuit shown above, an ideal inductor of inductance [mathjaxinline]L[/mathjaxinline] is connected to a resistor of resistance [mathjaxinline]R[/mathjaxinline]. At the instant shown, the current through the inductor is [mathjaxinline]I[/mathjaxinline]. </p>
<p><b class="bfseries">(Part a)</b> What is [mathjaxinline]U_ B[/mathjaxinline], the magnetic energy in the inductor at the instant when the current through the inductor is [mathjaxinline]I[/mathjaxinline]. Express your answer in terms of [mathjaxinline]I[/mathjaxinline] and [mathjaxinline]L[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]U_ B=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> As a function of time, the magnetic energy in the inductor </p>
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<p><b class="bfseries">(Part b)</b> Which of the following statements about [mathjaxinline]\dfrac {dU_ B}{dt}[/mathjaxinline], the rate of change of magnetic energy, and [mathjaxinline]P_ R = I^2R[/mathjaxinline], the rate at which energy is converted into Joule heat in the resistor, is true? </p>
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<p><b class="bfseries">(Part c)</b> Use the results of parts (a) and (b) to obtain an expression for [mathjaxinline]\dfrac {dI}{dt}[/mathjaxinline] in terms of [mathjaxinline]R[/mathjaxinline], [mathjaxinline]L[/mathjaxinline], and [mathjaxinline]I[/mathjaxinline]. <p style="display:inline">[mathjaxinline]\dfrac {dI}{dt}=[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 4" role="group"><div id="inputtype_checkpoint_w12_8_5_1" class="text-input-dynamath capa_inputtype inline textline">
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<h2 class="hd hd-2 unit-title">L28v5: LR Circuit Demo</h2>
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