The electric power produced by a current $I\in\mathbb{R}^+$ and a voltage $V\in\mathbb{R}^+$ is $$ P = IV. $$ Now the current is given as an (alternating) current density $J(\mathbf{x},t)=\Im(e^{i\omega t} j(\mathbf{x}))$, $j:\Omega\to\mathbb{C}$, across the cross-section $\Omega\subset\mathbb{R}^2$ of a conductor. What is the generalization of the power in this case?

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    $\begingroup$ You've given us a nicely generalized description of the current, which is the number of electrons (let's say) flowing past a point in a certain direction per second. Now you need to tell us how much energy each one is gaining or losing. That is, you need to tell us the analog of the voltage. And presumably that will just be the energy difference from one point in the conductor to another. So you'll basically get back to $P=IV$. $\endgroup$ – Mike Jul 7 '13 at 13:59
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    $\begingroup$ Are you a mathematician, or math student, by any chance? A physicist probably wouldn't specify that $I\in\mathbb{R}^+$. Not that there's anything wrong with that, it's just a little unusual when talking about basic E&M. $\endgroup$ – David Z Jul 7 '13 at 22:35

In sinusoidal steady-state analysis, both voltage and current are represented by complex numbers $V$ and $I$, known in the trade as phasors:

\begin{gather} v(t) = Re(Ve^{j \omega t}) \quad V=V_m e^{j \theta_v} \\ i(t) = Re(I e^{j \omega t}) \quad I= I_m e^{j \theta_i} \end{gather}

Here $V_m, I_m , \theta_v, \theta_i $ are all real, and I'm using $j$ for the imaginary unit to avoid confusion with current $i$ (again common EE-speak).

Then the complex power $P$ is defined as:

$$ P = \frac{1}{2} V I^* $$

(with * indicating the complex conjugate).

With this definition, if you go through the math of multiplying the two time-domain waveforms to get the instantaneous power $p(t)=v(t)i(t)$, you find that the average power $P_{av}$ (the power averaged over a cycle) is real and given by

$$ P_{av} = Re (P) = Re\left(\frac{1}{2} V I^*\right) $$

Updates in response to comments:

Other measures of power (the names are standard but not the symbols, sorry):

Apparent power $P_{app}$ is what you get if you measure the root-mean-square voltage and current separately and multiply them:
$$P_{app} = \frac{V_m}{\sqrt{2}}\frac{I_m}{\sqrt{2}} = \frac{1}{2} V_m I_m =\frac{1}{2} |VI|=|P|$$ Obviously, $P_{app} \ge P_{av}$.

Reactive power $P_r$ is the third leg of the triangle formed by $P$ and $P_{av}$:
$$P_r = Im(P)$$ In power transmission lines, reactive power is undesirable because the current required to support it creates resistive losses in the line without contributing to the average power delivered. (It's up to the load to minimize the reactive elements that draw reactive current.)

Finally: Reference directions must be chosen for both the voltage and current. In principle these can be chosen independently. However, it is most common to choose associated reference directions: if $v(t)$ is the voltage across a circuit branch, measured from a "+" terminal to a "-" terminal, and $i(t)$ the current through the branch, then the reference directions are associated if a positive current enters the branch at the "+" terminal and leaves at the "-" terminal. Then $v(t)i(t)$ is the power delivered to the branch at time $t$.

Note that associated reference directions mesh nicely with element constitutive relations, like Ohm's law for a resistor: $v_R=i_R R$; if non-associated directions were chosen instead, we would need branch current $i=-i_R$ (or an equivalent inversion of the voltages).

  • $\begingroup$ In principle, there's nothing that keeps $P_{\text{av}}$ from being negative here. Could you add a word about how to interpret this? $\endgroup$ – Nico Schlömer Jul 8 '13 at 8:15
  • $\begingroup$ I browsed through en.wikipedia.org/wiki/AC_power and found that there are different notions of power for AC systems. Am I right when saying that $1/2 V I^*$ is the complex power, $Re(1/2 V I^*)$ the real power, and $|1/2 V I^*|$ the apparent power? $\endgroup$ – Nico Schlömer Jul 8 '13 at 13:46
  • $\begingroup$ Yes. I've added to my answer... $\endgroup$ – Art Brown Jul 9 '13 at 0:57

I was looking for an answer to the same question in general, and what I found might be relevant here. If we have the current within the volume of a conductor as proposed in the question, then we can find the power (both real and imaginary reactive powers) in two ways; Using volume integral of terms that go as current squared, or surface integrals of the Poynting vector. The full details can be found in the EM bible for physicists (Jackson's Classical Electrodynamics, 3rd edition ch. 6.9, pages 264 --267), however the useful formulas are never stated clearly and separately as far as I can tell. The average power loss (real Ohmic losses) is given in general by: $P_{av}=\frac{1}{2}\int\limits_V \mathbf{j}^{*}\cdot \mathbf{E}\; d^3x$. Inside a nice linear conductor, we can write $\mathbf{E}=\mathbf{j}/\sigma$. Then we can simplify the Ohmic losses even further:

$$P_{av}=\frac{1}{2}\frac{1}{\sigma}\int\limits_V |\mathbf{j}|^2\; d^3x$$

The reactive power is much more subtle. If we call the reactive power $Q$ to follow the electrical engineering convention, then $Q=\frac{\omega}{2}\int\limits_V \left(\mathbf{E}\cdot\mathbf{D}^{*}-\mathbf{B}\cdot\mathbf{H}^{*} \right) \; d^3x$. Again, in a typical case of a well behaved linear conductor like Aluminum for example, the first term $\mathbf{E}\cdot\mathbf{D}^{*}$ is negligible compared to the second one. In other words, the energy stored in the magnetic field dominates the reactive power. Also we have $\mathbf{B}=\mu \mathbf{H}$, so we get

$$Q\approx\frac{-\omega}{2\mu}\int\limits_V |\mathbf{B}|^2 \; d^3x$$

Therefore, you will need to go back to Maxwell's equations to find $\mathbf{B}$ within the conductor from $\mathbf{j}$ (e.g. in low frequency situations $\mathbf{j}=\nabla\times \mathbf{H}$).

The same results can be obtained using the complex Poynting vector which is defined as:

$$ \mathbf{S}=\frac{1}{2}\left( \mathbf{E}\times \mathbf{H}^{*} \right) $$

to find the total complex power, we can use:

$$ P_{av}+i\,Q=\oint\limits_S \mathbf{S}\cdot \mathbf{n}\, da$$

where we integrate over the surface of the conductor. The tricky part again is to write the electric and magnetic fields in terms of the current density, in order to find the Poynting vector in terms of the current density. The advantage here is that since the Poynting vector is complex, the surface integral will give us the real and imaginary parts of power at the same time.


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