Question 6 of 10: Resistance, current density and dissipation in a cylindrical resistor

Question 6: Resistance, current density and dissipation in a cylindrical resistor

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A solid cylinder of conducting material, radius 1.25 mm and length 6.5 mm, contacted over its two flat end faces so that those faces are equipotentials.

Find. The resistance between A and B, the vector current density and total current in the conductor, and the dissipated power by Joule's law.

[Figure not reproduced: Figure 4 (redrawn). Contact A covers the z = 0 face and contact B the z = L face, so the current is uniform and axial through the full cross-section. See the official exam paper.]

Approach. Because both end faces are equipotentials, the field inside is uniform and axial, so the resistor is the elementary $L/\sigma A$ case; the current density, current and power then follow from the applied potential difference, which the paper does not state and which is therefore carried as $V_{AB}$.

1. Establish the field geometry. The contacts cover the flat faces at $z = 0$ and $z = L$, and both are equipotentials, so the potential can only vary with $z$. Laplace's equation in one dimension, $d^2V/dz^2 = 0$, gives a linear potential and hence a uniform field $$\vec{E} = -\frac{dV}{dz}\hat{a}_z = \frac{V_{AB}}{L}\hat{a}_z$$ directed along the axis. This is the ordinary straight-through resistor, not a radial or azimuthal flow problem.
2. (a) Resistance between A and B. For a uniform field through a uniform cross-section, $$R = \frac{V_{AB}}{I} = \frac{L}{\sigma A} = \frac{L}{\sigma\pi a^2}$$ with $A = \pi(1.25\times10^{-3})^2 = 4.9087\times10^{-6}$ m$^2$: $$R = \frac{6.5\times10^{-3}}{(60.2)(4.9087\times10^{-6})} = \frac{6.5\times10^{-3}}{2.9551\times10^{-4}}$$ $$\boxed{R_{AB} = 22.00\ \Omega}$$ Note $\sigma = 60.2$ S/m is very low for a metal (copper is $5.8\times10^{7}$ S/m); this is a deliberately resistive material such as a carbon or ceramic composite, which is what makes a 22 Ω resistor out of a 6.5 mm slug.
3. (b) Current density and total current. Ohm's law in point form gives the current density directly from the uniform field: $$\boxed{\vec{J} = \sigma\vec{E} = \frac{\sigma V_{AB}}{L}\hat{a}_z = (9262\ V_{AB})\,\hat{a}_z\ \text{A/m}^2}$$ and because $\vec{J}$ is uniform over the whole disc the total current is simply $J$ times the area, $$\boxed{I = |\vec{J}|\,\pi a^2 = \frac{\sigma\pi a^2}{L}V_{AB} = \frac{V_{AB}}{R} = (45.46\ V_{AB})\ \text{mA}}$$ The identity $I = V_{AB}/R$ recovers part (a), which is the check that the two answers are consistent. Evaluated at an illustrative $V_{AB} = 1.00$ V: $|\vec{J}| = 9.262$ kA/m$^2$ and $I = 45.46$ mA.
4. (c) Power by Joule's law. The point form of Joule's law integrates the local dissipation density $\vec{J}\cdot\vec{E} = |\vec{J}|^2/\sigma$ over the volume: $$P = \int_v \frac{|\vec{J}|^2}{\sigma}\,dv = \frac{|\vec{J}|^2}{\sigma}(\pi a^2 L)$$ Since $\vec{J}$ is uniform this collapses to the circuit form, $$\boxed{P = I^2 R = \frac{V_{AB}^2}{R} = (45.46\ V_{AB}^2)\ \text{mW}}$$ so at $V_{AB} = 1.00$ V the resistor dissipates 45.46 mW. Substituting the volume $\pi a^2 L = 3.191\times10^{-8}$ m$^3$ and $|\vec{J}| = 9262$ A/m$^2$ into the integral gives the same 45.46 mW, confirming that the field-theory and circuit-theory routes agree.
5. Sanity-check the thermal side. 45 mW in a slug of surface area $2\pi a L + 2\pi a^2 = 6.09\times10^{-5}$ m$^2$ is about 750 W/m$^2$, which a small component can shed in still air, so 1 V across this resistor is a physically sensible operating point — useful reassurance that the assumed excitation is a reasonable one.