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# 13.1: Alternating current in an inductance

• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

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 $$\text{XIII.2}$$