Joule effect - Aetherwisp

The Joule effect is the phenomenon whereby an electric current $I$ traversing a material of nonzero electrical resistance $R$ dissipates heat, the amount of which is dependent on the intensity of the current. For a material subject to an electric potential difference $\Delta V$ , the power output in heat is

P=I\Delta V=RI^{2}=\frac{\Delta V^{2}}{R}

The last two forms are derived with Ohm's law and work in the context of ohmic materials of electrical resistance $R$ . Meanwhile, the first form, $P(t)=I(t)V(t)$ , is absolutely universal and applicable in materials of all kinds, regardless of whether Ohm's law applies, if they are homogeneous, have a constant cross section or what their conduction mechanism is. Moreover, the presence of a magnetic field makes no difference since magnetic forces do no work.

If our material is a homogeneous wire of length $l$ and constant cross section $S$ , we can write

P(t)=I(t)V(t)=J_{f}(t)S\ E_{l}(t)l=J_{f}(t)E_{l}\mathcal{V}=\mathcal{V}\ \mathbf{J}_{f}(t)\cdot \mathbf{E}(t)

where $E_{l}(t)=V(t)/l$ is the projection of the total electric field over the curve $l$ and $\mathcal{V}$ is the volume of the material. If we divide by the volume we can find the heat emission per unit element of the material centered in $\mathbf{r}$ :

p(\mathbf{r},t)=\mathbf{J}_{f}(\mathbf{r},t)\cdot \mathbf{E}(\mathbf{r},t)

Derivation#

Consider any piece of a circuit subject to an electric potential difference $\Delta V$ and traversed by a current $I$ . In a time $dt$ , the piece is traversed by an electric charge $dq=Idt$ , so the work done by the electric field here is $dW=\Delta Vdq=I\Delta Vdt$ . Since power is the time derivative of work, we have

P=\frac{dW}{dt}=I\Delta V

If the piece is an ohmic resistor, we can apply Ohm's first law:

P=RI^{2}