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<p>Here we add capacitors into our simple circuits, exploring both how to simpify multiple capactitors in series or parallel as well as looking at the time-dependence of current and charge for charging and discharging capacitors.</p>
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<li><a href="https://prev.openlearninglibrary.mit.edu/courses/course-v1:MITx+8.02.2x+2T2018/pdfbook/0/#viewer-frame" target="[object Object]">Chapter 7.7-7.8: Capacitors in Circuits</a></li>
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<h2 class="hd hd-2 unit-title">L16v1: Circuit with a Capacitor</h2>
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Two capacitors are connected as shown in the left figure above. The upper capacitor has a capacitance [mathjaxinline]C[/mathjaxinline] and an initial charge [mathjaxinline]Q_0[/mathjaxinline], the lower capacitor has a capacitance [mathjaxinline]2C[/mathjaxinline] and no charge. The switch is closed. After a long time that the switch is closed, the charge on the lower capacitor is </p>
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<h2 class="hd hd-2 unit-title">Summary of Equivalent Capacitance</h2>
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<p><b class="bfseries">Summary: Capacitors in Parallel and in Series</b></p><p><b class="bfseries">Parallel Connection</b></p><p>
Suppose we have two capacitors [mathjaxinline]C_1[/mathjaxinline] with charge [mathjaxinline]Q_1[/mathjaxinline] and [mathjaxinline]C_2[/mathjaxinline] with charge [mathjaxinline]Q_2[/mathjaxinline] that are connected in parallel, as shown in Figure 1. </p><table cellspacing="0" class="tabular" style="table-layout:auto"><tr><td style="border-top-style:solid; border-bottom-style:solid; border-bottom-width:1px; border-left:1px solid black; border-right:1px solid black; border-top-color:black; border-top-width:1px; border-bottom-color:black; text-align:center; border:none"><img src="/assets/courseware/v1/0f9e4a93be93db1e8c0d1c95377bbceb/asset-v1:MITx+8.02.2x+2T2018+type@asset+block/images_fridayw5_intro_fig1-1.png" width="220"/></td><td style="border-top-style:solid; border-bottom-style:solid; border-bottom-width:1px; border-bottom-color:black; border-right:1px solid black; border-top-color:black; border-top-width:1px; text-align:center; border:none"><img src="/assets/courseware/v1/b52fa4e178acd081c1e091c04c328428/asset-v1:MITx+8.02.2x+2T2018+type@asset+block/images_fridayw5_intro_fig1-2.png" width="165"/></td><td style="border-top-style:solid; border-bottom-style:solid; border-bottom-width:1px; border-bottom-color:black; border-right:1px solid black; border-top-color:black; border-top-width:1px; text-align:center; border:none"><img src="/assets/courseware/v1/149377df9fb79b9a249eb10930e935ce/asset-v1:MITx+8.02.2x+2T2018+type@asset+block/images_fridayw5_intro_fig1-3.png" width="275"/></td></tr></table><p><center><b class="bfseries">Figure 1 </b>Capacitors in parallel and an equivalent capacitor</center><br/></p><p>
The left plates of both capacitors [mathjaxinline]C_1[/mathjaxinline] and [mathjaxinline]C_2[/mathjaxinline] are connected to the positive terminal of the battery and have the same electric potential as the positive terminal. Similarly, both right plates are negatively charged and have the same potential as the negative terminal. Thus, the potential difference [mathjaxinline]\left| \Delta V \right|[/mathjaxinline] is the same across each capacitor. This gives </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"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle C_1 = \frac{Q_1}{\left| \Delta V \right|},[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle C_2 = \frac{Q_2}{\left| \Delta V \right|}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"><span>(<span>1</span>)</span></td></tr></table><p>
These two capacitors can be replaced by a single equivalent capacitor [mathjaxinline]C_\text {eq}[/mathjaxinline] with a total charge [mathjaxinline]Q[/mathjaxinline] supplied by the battery. However, since [mathjaxinline]Q[/mathjaxinline] is shared by the two capacitors, we must have </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"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle Q = Q_1 + Q_2 = C_1 \left| \Delta V \right| + C_2 \left| \Delta V \right| = \left( C_1 + C_2 \right) \left| \Delta V \right|[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"><span>(<span>2</span>)</span></td></tr></table><p>
By definition the equivalent capacitance is given by </p><table id="a0000000006" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000007"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle C_\text {eq} \equiv \frac{Q}{\left| \Delta V \right|}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"><span>(<span>3</span>)</span></td></tr></table><p>
We can now substitute Eq. (2) into Eq. (3) and find that </p><table id="a0000000008" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000009"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle C_\text {eq} = \frac{Q}{\left| \Delta V \right|} = C_1 + C_2[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"><span>(<span>4</span>)</span></td></tr></table><p>
Thus, capacitors that are connected in parallel add. The generalization to any number of capacitors is </p><table id="a0000000010" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000011"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle C_\text {eq} = C_1 + C_2 + C_3 + \cdots + C_ N = \sum \limits _{i=1}^ N C_ i[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \text {(parallel)}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"><span>(<span>5</span>)</span></td></tr></table><p><b class="bfseries">Series Connection</b></p><p>
Suppose two initially uncharged capacitors and are connected in series, as shown in Figure 2. A potential difference [mathjaxinline]\left| \Delta V \right|[/mathjaxinline] is then applied across both capacitors. The left plate of capacitor 1 is connected to the positive terminal of the battery and becomes positively charged with a charge [mathjaxinline]+Q[/mathjaxinline], while the right plate of capacitor 2 is connected to the negative terminal and becomes negatively charged with charge [mathjaxinline]-Q[/mathjaxinline] as electrons flow in. What about the inner plates? They were initially uncharged; now the outside plates each attract an equal and opposite charge. So the right plate of capacitor 1 will acquire a charge [mathjaxinline]-Q[/mathjaxinline] and the left plate of capacitor [mathjaxinline]+Q[/mathjaxinline]. </p><table cellspacing="0" class="tabular" style="table-layout:auto"><tr><td style="border-top-style:solid; border-bottom-style:solid; border-bottom-width:1px; border-left:1px solid black; border-right:1px solid black; border-top-color:black; border-top-width:1px; border-bottom-color:black; text-align:center; border:none"><img src="/assets/courseware/v1/13a78d4f4e1b72613245b5cbf9728945/asset-v1:MITx+8.02.2x+2T2018+type@asset+block/images_fridayw5_intro_fig2-1.png" width="220"/></td><td style="border-top-style:solid; border-bottom-style:solid; border-bottom-width:1px; border-bottom-color:black; border-right:1px solid black; border-top-color:black; border-top-width:1px; text-align:center; border:none"><img src="/assets/courseware/v1/262419ab0c15905e47939b1b973bb4ae/asset-v1:MITx+8.02.2x+2T2018+type@asset+block/images_fridayw5_intro_fig2-2.png" width="220"/></td><td style="border-top-style:solid; border-bottom-style:solid; border-bottom-width:1px; border-bottom-color:black; border-right:1px solid black; border-top-color:black; border-top-width:1px; text-align:center; border:none"><img src="/assets/courseware/v1/ece1f7c0078922d9c5d4dc7ecc58a200/asset-v1:MITx+8.02.2x+2T2018+type@asset+block/images_fridayw5_intro_fig2-3.png" width="220"/></td></tr></table><p><center><b class="bfseries">Figure 2 </b>Capacitors in series and an equivalent capacitor</center><br/></p><p>
The potential differences across capacitors [mathjaxinline]C_1[/mathjaxinline] and [mathjaxinline]C_2[/mathjaxinline] are </p><table id="a0000000012" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000013"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \left| \Delta V_1 \right| = \frac{Q}{C_1},[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \left| \Delta V_2 \right| = \frac{Q}{C_2}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"><span>(<span>6</span>)</span></td></tr></table><p>
respectively. From Figure 2, we see that the total potential difference is simply the sum of the two individual potential differences: </p><table id="a0000000014" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000015"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \left| \Delta V \right| = \left| \Delta V_1 \right| + \left| \Delta V_2 \right|[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"><span>(<span>7</span>)</span></td></tr></table><p>
In fact, the total potential difference across any number of capacitors in series connection is equal to the sum of potential differences across the individual capacitors. These two capacitors can be replaced by a single equivalent capacitor [mathjaxinline]C_\text {eq} = Q / \left| \Delta V \right|[/mathjaxinline]. Using the fact that the potentials add in series, </p><table id="a0000000016" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000017"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \frac{Q}{C_\text {eq}} = \frac{Q}{C_1} + \frac{Q}{C_2}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"> </td></tr></table><p>
and so the equivalent capacitance for two capacitors in series becomes </p><table id="a0000000018" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000019"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \frac{1}{C_\text {eq}} = \frac{1}{C_1} + \frac{1}{C_2}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"><span>(<span>8</span>)</span></td></tr></table><p>
The generalization to any number of capacitors connected in series is </p><table id="a0000000020" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto"><tr id="a0000000021"><td style="width:40%; border:none"> </td><td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \frac{1}{C_\text {eq}} = \frac{1}{C_1} + \frac{1}{C_2}+ \cdots + \frac{1}{C_ N} = \sum \limits _{i=1}^ N \frac{1}{C_ i}[/mathjaxinline]
</td><td style="vertical-align:middle; text-align:center; border:none">
</td><td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle \text {(series)}[/mathjaxinline]
</td><td style="width:40%; border:none"> </td><td style="width:20%; border:none" class="eqnnum"><span>(<span>9</span>)</span></td></tr></table>
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<h2 class="hd hd-2 unit-title">L16Q2: Equivalent Capacitance</h2>
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Equivalent Capacitance
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Consider the circuit below. What is the equivalent capacitance? </p>
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<text> [mathjaxinline]\frac{1}{C_{eq}} = \frac{1}{C_2+C_3} + \frac{1}{C_1}[/mathjaxinline]</text>
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<text> [mathjaxinline]C_{eq} = \frac{C_2 C_3}{C_2 + C_3} + C_1[/mathjaxinline]</text>
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<text> [mathjaxinline]C_{eq} = C_1 + C_2 + C_3[/mathjaxinline]</text>
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<text> [mathjaxinline]\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}[/mathjaxinline]</text>
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Capacitors in Series and in Parallel. Part a.
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Three capacitors are connected to a battery as shown in the figure. Their capacitances are [mathjaxinline]C_1=3C[/mathjaxinline], [mathjaxinline]C_2=C[/mathjaxinline], and [mathjaxinline]C_3=5C[/mathjaxinline]. </p>
<p><b class="bfseries">(Part a)</b> What is the equivalent capacitance of this set of capacitors? Express your answer in terms of [mathjaxinline]C[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]C_{eq} =[/mathjaxinline] </p>
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Capacitors in series and in parallel, Part b.
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<p><b class="bfseries">(Part b)</b> Assume [mathjaxinline]Q_1[/mathjaxinline], [mathjaxinline]Q_2[/mathjaxinline] and [mathjaxinline]Q_3[/mathjaxinline] are the charges stored in capacitors 1, 2 and 3, respectively. Which of the following statements is true? </p>
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<text> [mathjaxinline]Q_1 &gt;Q_2&gt;Q_3[/mathjaxinline]</text>
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<text> [mathjaxinline]Q_1 &gt;Q_2=Q_3[/mathjaxinline]</text>
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<input type="radio" name="input_pset_802x_pset6_1b_2_1" id="input_pset_802x_pset6_1b_2_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="pset_802x_pset6_1b_2_1-choice_4-label" for="input_pset_802x_pset6_1b_2_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_pset_802x_pset6_1b_2_1">
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Capacitors in series and in parallel, Part c.
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<p><b class="bfseries">(Part c)</b> Calculate the magnitude of the potential difference across each of the capacitors and the magnitudes of their charges. Express your answers in terms of V_0 for [mathjaxinline]V_0[/mathjaxinline] and [mathjaxinline]C[/mathjaxinline] as needed. </p>
<p>
<p style="display:inline">[mathjaxinline]\Delta V_1 =[/mathjaxinline] </p>
<div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_pset_802x_pset6_1c_2_1" class="text-input-dynamath capa_inputtype inline textline">
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<div id="display_pset_802x_pset6_1c_2_1" class="equation">`{::}`</div>
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<p>
<p style="display:inline">[mathjaxinline]Q_1 =[/mathjaxinline] </p>
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<div id="display_pset_802x_pset6_1c_3_1" class="equation">`{::}`</div>
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<p>
<p style="display:inline">[mathjaxinline]\Delta V_2 =[/mathjaxinline] </p>
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<div id="display_pset_802x_pset6_1c_4_1" class="equation">`{::}`</div>
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<p>
<p style="display:inline">[mathjaxinline]Q_2 =[/mathjaxinline] </p>
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<div id="display_pset_802x_pset6_1c_5_1" class="equation">`{::}`</div>
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<p>
<p style="display:inline">[mathjaxinline]\Delta V_3 =[/mathjaxinline] </p>
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<span class="trailing_text" id="trailing_text_pset_802x_pset6_1c_6_1"/>
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<div id="display_pset_802x_pset6_1c_6_1" class="equation">`{::}`</div>
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<p>
<p style="display:inline">[mathjaxinline]Q_3 =[/mathjaxinline] </p>
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<span class="trailing_text" id="trailing_text_pset_802x_pset6_1c_7_1"/>
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<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
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<div id="display_pset_802x_pset6_1c_7_1" class="equation">`{::}`</div>
<textarea style="display:none" id="input_pset_802x_pset6_1c_7_1_dynamath" name="input_pset_802x_pset6_1c_7_1_dynamath"/>
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Capacitors in series and in parallel, Part d.
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<p><b class="bfseries">(Part d)</b> If the capacitance [mathjaxinline]C_3[/mathjaxinline] is increased, what happens to the charge stored by each of the capacitors? </p>
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<h2 class="hd hd-2 unit-title">L16Q4: Single Loop RC-Circuit</h2>
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Single Loop RC Circuit
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An uncharged capacitor is connected to a battery, resistor and switch. The switch is initially open but at [mathjaxinline]t = 0[/mathjaxinline] it is closed. A very long time after the switch is closed, the current in the circuit is </p>
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<h2 class="hd hd-2 unit-title">L16Q5: Current Through Resistor</h2>
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Current Through Resistor
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In the circuit below, the switch is closed at [mathjaxinline]t = 0[/mathjaxinline]. At [mathjaxinline]t = \infty[/mathjaxinline] (a long time later) the current through the lower resistor will be: </p>
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