Skip to main content
Physics LibreTexts

13.11: Root-mean-square values, power and impedance matching

  • Page ID
  • We have been dealing with alternating currents of the form \(I=\hat{I}e^{j\omega t}\). I have been using the notation \(\hat{I}\) to denote the “peak” (i.e. maximum) value of the current. Of course, in the notation of complex numbers, this is synonymous with the modulus of \(I\). That is to say \(\hat{I}=\text{mod }I=|I|\). I shall use one or other notation wherever it is convenient. This will often mean using ^ when describing time-varying quantities, and \(| \,\, |\) when describing constant (but perhaps frequency dependent) quantities, such as impedances.

    Suppose we have a current that varies with time as \(I=\hat{I}\sin \omega t\). During a complete period \((P=2\pi / \omega )\) the average or mean current is zero. The mean of the square of the current, however, is not zero. The mean square current, \(\overline{I^2}\), is defined such that

    \[\label{13.11.1}\overline{I^2}P=\int_0^P I^2 \,dt.\]

    With \(I=\hat{I}\sin \omega t\) this gives \(\overline{I^2}=\frac{1}{2}\hat{I}^2\). The square root of this is the root-mean-square current, or the RMS value of the current:


    When we are told that an alternating current is so many amps, or an alternating voltage is so many volts, it is usually the RMS value that is meant, though we cannot be sure of this unless the speaker or writer explicitly says so. If you wish to be understood and not misunderstood in your own writings, you will always make it explicitly clear what meaning you intend.

    If an alternating current is flowing through a resistor, at some instant when the current is \(I\), the instantaneous rate of dissipation of energy in the resistor is \(I^2R\). The mean rate of dissipation of energy during a complete cycle is \(I_{\text{RMS}}^2R\). This is one obvious reason why the concept of RMS current is important.

    Now cast your mind back to Section 4.8. There we imagined that we connected a resistance \(R\) across a battery of EMF \(E\) and internal resistance \(r\). We calculated that the power delivered to the resistance was \(P=I^2R=\frac{E^2R}{(R+r)^2}\), and that this was greatest (and equal to \(\frac{1}{4}E^2/r\)), when the external resistance was equal to the internal resistance of the battery.

    What is the corresponding situation with alternating current? Suppose we have a box (a “source”) that delivers an alternating voltage \(V\) (which is represented by a complex number \(\hat{V}e^{j\omega t}\)), and that this box has an internal impedance \(z=r+jx\). If we connect across the box a device (a “load”) that has an impedance \(Z=R+jX\), what will be the power delivered to the load, and can we match the external impedance of the load to the internal impedance of the box in such a manner that the power delivered to the load is greatest?

    The second question is quite easy to answer. Reactance can be either positive (inductive) or negative (capacitive), and so it is quite possible for the total reactance of the entire circuit to be zero. Thus for a start, we want to ensure that \(X=-x\). That is, the external reactance should be equal in magnitude but opposite in sign to the internal reactance. The circuit is then purely resistive, and the power delivered to the circuit is just what it was in the direct current case, namely \(P=I^2R=\frac{E^2R}{(R+r)^2}\), where the current and EMF in this equation are now RMS values. And, as in the direct current case, this is greatest if \(R=r\). The conclusion is that, for maximum power transfer, \(R+jX\) should equal \(r-jx\). That is, for the external and internal impedances to be matched for maximum power transfer, \(Z=z^*\). The load impedance should equal the conjugate of the source impedance.

    What is the power delivered to the load when the impedances are not matched? In other words, when \(z=r+jx\) and \(Z=R+jX\). It is \(P=\frac{1}{2}\hat{I}^2R\). The current is given by the equation \(E=I(Z+z)\). (These are all complex numbers - i.e. they are all periodic functions with different phases. \(E\) and \(I\) vary with time.) Now if \(w_1\) and \(w_2\) are two complex numbers, it is well known (from courses in complex numbers) that \(|w_1w_2|=|w_1||w_2|\). We apply this now to \(E=I(Z+z)\). [I shall use ^ for the “peak” of the time-varying quantities, and \(|\,\, |\) for the modulus of the impedances] We obtain \(\hat{E}=\hat{I}|Z+z|\).



    • Was this article helpful?