The Electrical Properties of Solids — Question 7

Question 7

Local Joule heating can be obtained without a lumped resistor model

A homogeneous solid has conductivity $\sigma=4.00\times10^4\,\mathrm{S/m}$ in a uniform field $E=30.0\,\mathrm{V/m}$. Its active volume is $V_s=2.00\times10^{-8}\,\mathrm{m^3}$.

1. Derive the heating power per volume and calculate total power.

2. If conductivity doubles, compare power at fixed E and at fixed current density J.

Question 7 - Solution

Local Joule heating can be obtained without a lumped resistor model

(a) Method and calculation

$$p_v=\boldsymbol J\cdot\boldsymbol E=\sigma E^2=\boxed{3.60\times10^7\,\mathrm{W/m^3}}.$$$$\boxed{P=p_vV_s=.720\,\mathrm W}.$$

(b) Interpretation and further calculation

$$E\text{ fixed}:\quad p_v\propto\sigma\Rightarrow\boxed{p_v\prime=2p_v}.$$$$J\text{ fixed}:\quad p_v=J^2/\sigma\Rightarrow\boxed{p_v\prime=p_v/2}.$$

Physical check

The same material change produces opposite power trends depending on the electrical constraint. Field and current density cannot both remain fixed when conductivity changes.