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      <text>19.4.1.4 Modelling electrical circuits</text>
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      <text><![CDATA[<h5>Modelling electrical circuits<br></h5>
<p></p>
<p>Another application of first-order differential equations arises in the modelling of electrical circuits. In Section <em>Modelling with Differential Equations</em> he differential equation for the RL circuit in figure below was shown to be:</p>
<p>\(\qquad L\ \dfrac{di}{dt} + Ri = E\)</p>
<p>in which the initial condition is \(i=0\) at \(t=0\).</p>

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    <img src="@@PLUGINFILE@@/fig2.svg" alt="Figure of an RL circuit" width="300" height="122" class="img-responsive atto_image_button_text-bottom">
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<p>First we write this equation in standard form \( \{ \dfrac{dy}{dx} + P(x)y = Q(x)\}\) and obtain the integrating factor.</p>
<p>Dividing the differential equation through by \(L\) gives</p>
<p>\(\qquad \dfrac{di}{dt} + \dfrac{R}{L}\ i = \dfrac{E}{L}\)</p>
<p>which is now in standard form. The integrating factor is \( \text{e}^{\int {R\over L}dt} = \text{e}^{Rt/L}.\)</p>
<p>Multiplying the equation in standard form by the integrating factor gives</p>
<p>\(\qquad \text{e}^{Rt/L} \dfrac{di}{dt} + \text{e}^{Rt/L} \dfrac{R}{L}i = \dfrac{E}{L} \text{e}^{Rt/L}\)</p>
<p>or, rearranging,</p>
<p>\(\qquad \dfrac{d}{dt} (\text{e}^{Rt/L}\ i) = \dfrac{E}{L} \text{e}^{Rt/L}.\)</p>
<p>Now we integrate both sides and apply the initial condition to obtain the solution.</p>
<p>Integrating the differential equation gives:</p>
<p>\(\qquad \text{e}^{Rt/L}\ i = \dfrac{E}{R} \ \text{e}^{Rt/L} + C\)</p>
<p>where \(C\) is a constant so</p>
<p>\(\qquad i = \dfrac{E}{R} + C e^{-Rt/L}\)</p>
<p>Applying the initial condition \(i=0\) when \(t=0\) gives</p>
<p>\( \qquad 0 = \dfrac{E}{R} + C\)</p>
<p>so that \(C = -\dfrac{E}{R}.\)</p>
<p>Finally, \(\, i=\dfrac{E}{R}(1-\text{e}^{-Rt/L}).\)</p>
<p>Note that as \(t\to \infty, \, i\to \dfrac{E}{R}\) so as \(t\) increases the effect of the inductor diminishes to zero.</p>]]></text>
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