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      <text>19.1.4.1 RL and LCR circuit</text>
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    <li>
        <p>In this RL circuit the switch is closed at \(t = 0\) and a constant voltage \(E\) is applied.</p>
        <p></p>
        <p style="text-align: center; "><img src="@@PLUGINFILE@@/19_1_fig3_RL_circ.svg" alt="Image of an RL circuit which has a constant voltage E" width="250" height="99" class="img-responsive atto_image_button_middle"><br></p>
        <p>The voltage across the resistor is \(iR\) where \(i\) is the current owing in the circuit and \(R\) is the (constant) resistance. The voltage across the inductance is \(L \dfrac{\textrm{d}i}{\textrm{d}t}\) where \(L\) is the constant inductance.</p>
        <p> Kirchhoff's law of voltages states that the applied voltage is the sum of the other voltages in the circuit.</p>
        <p>The differential equation for the current \(i\) is \( L \dfrac{\textrm{d}i}{\textrm{d}t} + Ri = E.\)</p>
        <p>Which is the initial condition for the current at \(t=0\)?</p>
        <p>\( i = \) [[input:ans1]] [[validation:ans1]] [[feedback:prt1]]</p>
    </li>

    <li>
        <p>In the LCR circuit below the voltage across the capacitor is \(q=C\) where \(q\) is the charge on the capacitor, and \(C\) is the capacitance. Note that \( \dfrac{dq}{dt} = i.\) </p>
        <p style="text-align: center; "><img src="@@PLUGINFILE@@/19_1_fig5_RLC_circ.svg" alt="Image of an RLC circuit with constant voltage E" width="250" height="101" class="img-responsive atto_image_button_text-bottom"><br></p>
      <p>The differential equation for \(i\) is \[L \dfrac{\textrm{d}^2t}{\textrm{d}t^2} + R \dfrac{\textrm{d}^q}{\textrm{d}t} + \dfrac{q}{C} = E\]</p>
      <p>Write down the initial conditions if the initial charge is zero and the switch is closed at t = 0.</p>
      <p>\( q = \) [[input:ans2]] [[validation:ans2]] [[feedback:prt2]]</p>
      <p>\( i = \) [[input:ans3]] [[validation:ans3]] [[feedback:prt3]]</p>
    </li>
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