I am studying the fundamentals of electric circuits and have been stuck with the proof of the (instantaneous) electric power used in circuit analysis: $$ p \left({t}\right) = v \left({t}\right) i \left({t}\right) \tag{1}\label{p_common} $$ In the book Elements of Electromagnetics by Sadiku, they provide the following form of electric power (which I presume is valid for any material, not only conductors): $$ p \left({t}\right) = \int_V {\vec E \left({t}\right) \cdot \vec J \left({t}\right) dv} \tag{2}\label{p_vol} $$ They also say that for conductors with uniform cross section, $dv = ds \cdot dl$, and the integral can be separated as: $$ \begin{align} p \left({t}\right) &= \left( {\int_L {E\left({t}\right) dl} } \right)\left( {\int_S {J\left({t}\right) ds} } \right) \\ p \left({t}\right) &= v \left({t}\right) i \left({t}\right) \end{align} $$ The implicit conclusion is that Eq. $\ref{p_vol}$ must be used whenever the material is NOT a conductor with uniform cross section. However, Eq. \ref{p_common} is used universally in circuit analysis, even for nonlinear element like diodes and transistors. So my question is: is Eq. \ref{p_common} valid for any circuit element, regardless of its nature? if yes, how can we prove Eq. \ref{p_common} for any general material? and if not, is Eq. \ref{p_vol} the proper way to calculate the power of any general circuit element?

PD: I would be grateful if someone shared me a thorough proof of Eq. \ref{p_vol}, the one given in the book of Sadiku omits some steps so I did not understand it completely.


I'd derive eq 2 like this…

The work done per unit time on a charge q moving with velocity $\vec{v}$ in an electric field $\vec{E}$ is $q\vec{E}.\vec{v}.$

If there are $n$ of these charge carriers per unit volume, then the total work done per unit time on the charge carriers in volume dV is $$\text{Power} = q\vec{E}.\vec{v}\ n\ dV.$$ But the current density, $\vec{J}$, is a vector of magnitude equal to the charge per unit area crossing a small imaginary surface per unit time and direction that in which the charges are moving. It follows from this that$$\vec{J}=nq\vec{v}.$$ Therefore$$\text{Power} = \vec{E}.\vec{J}\ dV.$$ This is a very general formula; it will even cope with the case of $\vec{J}$ not being in the same direction as $\vec{E}$, as I imagine might arise in certain crystalline conductors. It will also work if there's more than one species of charge-carrier.

I believe that (1) applies only when $\vec{J}$ is in the same direction as $\vec{E}$, which you might expect if $\vec{E}$ gives the charge carriers a mean drift velocity, $\vec{v}$, in the same direction as itself (or the opposite direction for negative carriers), as in simple theories of conduction, for example Drude's theory. In this case we can write the last equation simply in terms of the magnitudes, E and J, of the vectors $\vec{E}$ and $\vec{J}.$ ... $$\text{Power} = EJ\ dV.$$ Now $E$ is the magnitude of the potential gradient, and $J$ is the current per unit area, so$$\text{Power} = \frac{d\Phi}{dx} \times \frac{I}{A}\ dV.$$ But $A\ dx=dV$ so $$\text{Power} =I\ d\Phi$$ in which $d\Phi$ is the pd across the length of conductor in question.

For a conductor of finite length (including a device like a diode) the current at any time will be the same all along the conducting path (except for alternating currents of frequency high enough for phase differences not to be negligible). For the same current throughout the conducting path, the last equation integrates to $$\text{Power} =I\ \Delta \Phi.$$ This equation can be derived much more simply: it follows immediately from the definitions of pd and current!

  • $\begingroup$ So, stricly speaking Eq. (1) cannot be used to calculate the power dissipated by a diode, for example? I mean, Eq. (1) is used indiscriminately for all kinds of circuit elements, it surprises me that I have not seen any comments about such approximation in circuit books $\endgroup$ – Daniel Turizo Jul 24 '18 at 21:07
  • $\begingroup$ I'm pretty sure that you can use eq. 1 for a diode, that is put a voltmeter across the diode and an ammeter in series and $VI$ will give you the power dissipated in the diode. What makes you think you can't use this equation? $\endgroup$ – Philip Wood Jul 24 '18 at 21:31
  • $\begingroup$ The fact that the electric field is probably not uniform over the cross section, and thus the integral of Eq. 2 cannot be separated. Another example would be a conductor with alternating current, in such case the electric field is not uniform over the cross section due to the skin effect, yet the power is still calculated using Eq. 1. $\endgroup$ – Daniel Turizo Jul 25 '18 at 1:16
  • $\begingroup$ Just consider the wire going into the junction and the wire coming out. If the pd between the ends, A and B, of these wires is $\Delta \Phi$, then the energy converted per unit charge going from A to B is $\Delta \Phi$, but the current, $I$ is the charge flowing per unit time between A and B, so the power (energy converted per second) between A and B is $VI.$ This applies whatever is going on inside the diode! $\endgroup$ – Philip Wood Jul 25 '18 at 7:58
  • $\begingroup$ I think you're probably right about Power = $IV$ not applying for a diode at very high frequencies ($\gtrapprox$ 1 GHz ?) because the current at any instant will not be constant through the device. $\endgroup$ – Philip Wood Jul 25 '18 at 16:52

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