Skip to main content
[ "article:topic", "authorname:tatumj", "inductive reactance", "showtoc:no", "license:ccbync" ]
\(\require{cancel}\)
Physics LibreTexts

13.1: Alternating current in an inductance

  1. Last updated
  2. Save as PDF
  • Page ID
    5494

    • In the figure we see a current increasing to the right and passing through an inductor. As a consequence of the inductance, a back EMF will be induced, with the signs as indicated. I denote the back EMF by \(V = V_A - V_B\). The back EMF is given by \(V=L\dot I\). 


    FIGURE \(\text{XIII.1}\)

    Now suppose that the current is an alternating current given by

    \[\label{13.1.1}I=\hat{I}\sin \omega t.\]

    In that case \(\dot I = \hat{I}\omega \cos \omega t\), and therefore the back EMF is

    \[\label{13.1.2}V=\hat{I}L\omega \cos \omega t,\]

    which can be written

    \[\label{13.1.3}V=\hat{V}\cos \omega t,\]

    where the peak voltage is

    \[\label{13.1.4}\hat{V}=L\omega \hat{I}\]

    and, of course \(V_{\text{RMS}}=L\omega I_{\text{RMS}}\) (Section 13.11).

    The quantity \(L\omega\) is called the inductive reactance \(X_L\). It is expressed in ohms (check the dimensions), and, the higher the frequency, the greater the reactance. (The frequency \(\nu\) is \(\omega/(2\pi)\).)

    Comparison of equations \ref{13.1.1} and \ref{13.1.3} shows that the current and voltage are out of phase, and that \(V\) leads on \(I\) by 90o, as shown in figure XIII.2.

     

    FIGURE \(\text{XIII.2}\)

    Contributor