# 13.3: Complex Numbers

I am now going to repeat the analyses of Sections 13.1 and 13.2 using the notation of complex numbers. In the context of alternating current theory, the imaginary unit is customarily given the symbol \(j\) rather than \(i\), so that the symbol \(i\) is available, if need be, for electric currents. I am making the assumption that the reader is familiar with the basics of complex numbers; without that background, the reader may have difficulty with much of this chapter.

We start with the inductance. If the current is changing, there will be a back EMF given by \(V=L\dot I\). If the current is changing as

\[I=\hat{I}e^{j\omega t},\label{13.3.1}\]

then

\[\dot I = \hat{I}j\omega e^{j\omega t}=j\omega I.\]

Therefore the voltage is given by

\[\label{13.3.2}V=jL\omega I.\]

The quantity \(jL\omega\) is called the *impedance* of the inductor, and is \(j\) times its *reactance*. Its *reactance *is \(L\omega\), and, in SI units, is expressed in ohms. Equation \ref{13.3.2} (in particular the operator \(j\) on the right hand side) tells us that \(V\) leads on \(I\) by 90^{o}.

Now suppose that an alternating voltage is applied across a capacitor. The charge on the capacitor at any time is \(Q = CV\), and the current is \(I=C\dot V\). If the voltage is changing as

\[V=\hat{V}e^{j\omega t},\label{13.3.3}\]

then

\[\dot V = \hat{V}j\omega e^{j\omega t}=j\omega V.\]

Therefore the current is given by

\[\label{13.3.4}I=jC\omega V.\]

That is to say

\[V=-\frac{j}{C\omega}I.\label{13.3.5}\]

The quantity \(-j/(C\omega )\) is called the *impedance* of the capacitor, and is \(j\) times its reactance. Its reactance is \(-1/(C\omega )\), and, in SI units, is expressed in ohms. Equation \ref{13.3.5} (in particular the operator \(-j\) on the right hand side) tells us that \(V\) lags behind \(I\) by 90^{o}.

In summary:

\[\nonumber \begin{align} &\text{Inductor : Reactance} = L\omega. \qquad &&\text{Impedance} = jL\omega. \quad &&&V \text{ leads on }I. \\ \nonumber &\text{Capacitor : Reactance} = -1/(C\omega ) &&\text{Impedance} = -j/(C\omega). \quad &&&V \text{ lags behind }I. \\ \end{align}\]

It may be that at this stage you haven't got a very clear idea of the distinction between reactance (symbol \(X\)) and impedance (symbol \(Z\)) other than that one seems to be \(j\) times the other. The next section deals with a slightly more complicated situation, namely a resistor and an inductor in series. (In practice, it may be one piece of equipment, such as a solenoid, that has both resistance and inductance.) Paradoxically, you may find it easier to understand the distinction between impedance and reactance from this more complicated situation.