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    <name>
      <text>20.4.5.1 Engineering Example - Capacitor</text>
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      <text><![CDATA[<h4 class="HELM_example">Example</h4>
<h3>Charge on a capacitor</h3>
<p>In the circuit shown in the figure below, the switch \(S\) is closed at \(t=0\) with a capacitor charge \(q(0)=q_0=\) constant and \(dq/dt(0)=0\).</p>
<p>
<center><img src="@@PLUGINFILE@@/20-4-fig-19-LRC_circuit.svg" alt="A circuit with a resistor an inductor, and a capacitor in series." width="184" height="104" class="img-responsive atto_image_button_text-bottom"></center>
</p>
<p>Show that \( q(t)=q_0(t)\text{e}^{-\alpha t}\left[ \cos\omega t+\dfrac{\alpha}{w}\sin \omega t\right]\) where \(\alpha=\dfrac{R}{2L}\) and \(\omega^2=\dfrac{1}{LC}-\alpha^2\)
</p>
<h4>Laplace transform properties required</h4>
<p>The following properties are needed to solve this problem.</p>
<table style="width:100%;">
<tbody><tr>
  <td>\[F(s+a)={\cal L}\{e^{-at}f(t)\}\]</td>
  <td>(P1)</td>
</tr>
<tr>
  <td>\[{\cal L}\left\{\dfrac{df(t)}{dt}\right\}=s\{f(t)\}-f(0)\]</td>
  <td>(P2)</td>
</tr>
<tr>
  <td>\[{\cal L}\left\{\dfrac{d^2f(t)}{dt^2}\right\}=s^2{\cal L}\{f(t)\}-\dfrac{df}{dt}(0)-s\ f(0)\]</td>
  <td>(P3)</td>
</tr>
<tr>
  <td>\[ {\cal L}\{\sin kt\}=\dfrac{k}{s^2+k^2} \quad \text{with} \quad s&gt;0\]</td>
  <td>(P4)</td>
</tr>
<tr>
  <td>\[{\cal L}\{\cos kt\}=\dfrac{s}{s^2+k^2} \quad \text{with} \quad s&gt;0\]</td>
  <td>(P5)</td>
</tr>
<tr>
  <td>\[{\cal L}^{-1}\{{\cal L}\{f(t)\}\}=f(t)\]</td>
  <td>(P6)</td>
</tr>
</tbody></table>

<h4 class="HELM_solution">Solution</h4>


<p><strong>STEP 1</strong> Establish the differential equation for \(q(t)\) using, for example, Kirchhoff's law.</p>
<p>When the switch \(S\) is closed, the inductance \(L\), capacitance \(C\) and resistance \(R\) give rise to a.c. voltages related by \[ V_A-V_B=L\dfrac{di}{dt},\ V_B-V_D=R\ i,\ V_D-V_F=q/C \text{ respectively}\]  So since \(V_A-V_F=(V_A-V_B)+(V_B-V_D)+(V_D-V_F)=0\) and \(i=\dfrac{dq}{dt}\) we have
</p>\[ L\dfrac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{q}{C}=0 \tag{1}\]  <p></p>


<p><strong>STEP 2</strong> Write the Laplace transform of the differential equation substituting for the initial conditions:</p>
<p>Since the Laplace transform is linear, the transform of differential Equation (1) is
</p><p>\[ {\cal L}\left\{L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{q}{C}\right\}=L{\cal
L}\left\{\frac{d^2q}{dt^2}\right\}+R{\cal L}\left\{\frac{dq}{dt}\right\}+{\cal L}\{\dfrac{q}{C}\}=0. \tag{2}\]</p>
<p> We deal with each derivative term in turn: Using property (P3), \[ {\cal L}\left\{\frac{d^2q}{dt^2}\right\}=s^2{\cal L}\{q(t)\}-\frac{dq}{dt}(0)-s\ q(0).\]  So, using the initial conditions \(q(0)=q_0\) and \(\dfrac{dq}{dt}(0)=0\) \[ {\cal L}\left\{\frac{d^2q}{dt^2}\right\}=s^2{\cal L}\{q(t)\}-s\ q_0. \tag{3}\]</p> 
<p>  By means of property (P2) \[ {\cal L}\left\{\frac{dq}{dt}\right\}=s{\cal L}\{q(t)\}-q_0 \tag{4} \] </p>

<p><strong>STEP 3</strong> Solve for the function \({\cal L}\{q(t)\}\) by substituting from  (3) and (4) into Equation (2):</p>
<p>\[ L[s^2{\cal L}\{q(t)\}-sq_0]+R[s{\cal L}\{q(t)\}-q_0]+\dfrac{1}{C}{\cal L}\{q(t)\}=0\]\[\qquad\quad\Rightarrow\qquad {\cal L}\{q(t)\}[Ls^2+Rs+\dfrac{1}{C}]=Lsq_0+Rq_0\]</p>\[\qquad\quad\Rightarrow\qquad {\cal L}\{q(t)\}=\dfrac{(Ls+R)}{(Ls^2+Rs+\dfrac{1}{C})}q_0 \tag{5} \]<p></p>
<p>Using the definitions \(\alpha=\dfrac{R}{2L}\) and \(\omega^2=\dfrac{1}{LC}-\alpha^2\) enables the denominator in Equation (5) to be expressed as the sum of two squares,
\[ \begin{align} L\ s^2+R \ s+\dfrac{1}{C}&amp;=L[s^2+\dfrac{Rs}{L}+\dfrac{1} {LC}]=L[s^2+2\alpha s+\dfrac{1}{LC}] \\
 &amp;=L[s^2+2\alpha s+\alpha^2+\omega^2]=L[\{s+\alpha\}^2+\omega^2].\end{align}\] Consequently, with the new expression for the denominator, Equation (5) becomes
\[ {\cal L}\{q(t)\}=q_0\left[\frac{s}{(s+\alpha)^2+\omega^2}+\frac{R}{L}\
\frac{1}{(s+\alpha)^2+\omega^2}\right]. \tag{6} \]</p>

<p><strong>STEP 4</strong> Use the inverse Laplace transform to obtain \(q(t)\):</p>
<p>The inverse Laplace transform  is used to find \(q(t).\) </p>
<p> Taking the inverse Laplace transform of Equation (6) and using the linearity properties
\[ {\cal L}^{-1}\{{\cal L}\{q(t)\}\}=q_0{\cal L}^{-1}\left\{\frac{s} {(s+\alpha)^2+\omega^2}+\frac{R}{L}\ \frac{1}{(s+\alpha)^2+\omega^2}\right\}.\] Using property (P6) this can be written as \[ q(t)=q_0{\cal L}^{-1}\left\{\frac{s+\alpha}{(s+\alpha)^2+\omega^2}+\frac{-\alpha}{(s+\alpha)^2+\omega^2}+\frac{R}{L\omega}\ \frac{\omega}{(s+\alpha)^2+\omega^2}\right\}. \] Using the linearity of the Laplace
transform again \[ q(t)=q_0 {\cal L}^{-1}\left\{\frac{s+\alpha}{(s+\alpha)^2+\omega^2}\right\}+{\cal L}^{-1}\left\{\frac{-\alpha}{(s+\alpha)^2+\omega^2}\right\}+{\cal L}^{-1}\left\{\frac{R}{L\omega}\frac{\omega}{(s+\alpha)^2+\omega^2}\right\}. \tag{7}\]  </p>
<p>Using properties (P1) and (P5) \[ {\cal L}^{-1}\left\{\frac{s+\alpha}{(s+\alpha)^2+\omega^2}\right\}=\text{e}^{-\alpha t}\cos \omega t. \tag{8}\]</p>
<p> Similarly, \[ {\cal L}^{-1}\left\{\frac{-\alpha}{(s+\alpha)^2+\omega^2}\right\}=-(\dfrac{\alpha}{\omega})\{\text{e}^{-\alpha t}\sin \omega t\} \tag{9} \]</p>
<p> and \[ {\cal L}^{-1}\left\{\frac{R}{L\omega}\ \frac{\omega}{(s+a)^2+\omega^2}\right\}=(\dfrac{R}{L\omega })\text{e}^{-\alpha t}\sin\omega t. \tag{10} \]<p></p>
<p>  Substituting (8), (9) and (10) in (7) gives \[ q(t)=q_0 \text{e}^{-\alpha t}\left[\cos\omega t+\left\{-\dfrac{\alpha}{\omega}+\dfrac{R}{L\omega}\right\}\text{e}^{-\alpha t}\sin\omega t\right]. \tag{11} \]</p>


<p><strong>STEP 5</strong> Finally, show that for \(t&gt;0\) the solution is \[(t)=q_0\text{e}^{-\alpha t}[\cos \omega t+(\dfrac{\alpha}{\omega})\sin\omega t]\] where \(\alpha=\dfrac{R}{2L}\) and \(\omega^2=\dfrac{1}{LC}-\alpha^2.\)</p>
<p>Substituting \(\alpha=\dfrac{R}{2L}\) in (11) gives \[ \begin{align} q(t)&amp;=q_0 \text{e}^{-\alpha t}\left[\cos\omega t+[-\dfrac{\alpha}{\omega}+\dfrac{2\alpha}{\omega}]\text{e}^{-\alpha t}\sin \omega t\right] \\ &amp;=q_0\text{e}^{-\alpha t}[\cos\omega t+\dfrac{\alpha}{\omega}\sin\omega t]. \end{align} \]</p>]]></text>
<file name="20-4-fig-19-LRC_circuit.svg" path="/" encoding="base64">[encoded data, 13976 characters]</file>
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