Hypothetical Current Distribution

There are already some good solutions posted for this, so I will attempt to prove (for a one-dimensional conductor) that a constant-current distribution minimizes the power dissipated (given voltage constraints).

Suppose we discretize the conductor into $N$ constant-current segments. $N$ can be made arbitrarily large. For each segment, the current is $I_k$ and the resistance is $\Delta R$ . We want to minimize the total dissipated power, subject to the constraint that the sum of infinitesimal voltage changes must equal the applied voltage $V$ . We can use the method of Lagrange multipliers with the following Lagrangian:

$L = \Sigma_{k=1}^{k=N} I_{k} ^2 \, \Delta R + \lambda \Big ((\Sigma_{k=1}^{k=N} I_{k} \, \Delta R) - V \Big)$

Taking partial derivatives with respect to the segment-specific currents and setting to zero (per standard practice) results in:

$\frac{\partial{L}}{\partial{I_{1}}} = 2 \, I_{1} \, \Delta R + \lambda \, \Delta R = 0 \implies I_{1} = -\frac{\lambda}{2} \\ \frac{\partial{L}}{\partial{I_{2}}} = 2 \, I_{2} \, \Delta R + \lambda \, \Delta R = 0 \implies I_{2} = -\frac{\lambda}{2} \\ etc. \\ \implies I_{1} = I_{2} = I_{k} \\$

Thus, a constant-current profile minimizes the power dissipation, given voltage constraints. It is interesting to note that nature seems to follow this minimization principle. Although I am now wondering how this works in circuits for which the electrical wavelength is small compared to the size of the circuit, resulting in spatial variation in the current.

$P_1 = \frac{V^2}{R}$

Consider the hypothetical current distribution and an elementary length of wire $dx$ . The resistance of this wire element is $dR = \frac{R}{L}dx$

The potential difference across this element according to Ohm's law is:

$dV = I_2(x)dR$

Integrating over the length of the wire leads to:

$V = \frac{aLR}{2}$

The power dissipated in the wire element is:

$dP_2 = (I_2(x))^2 dR$

Integrating this over the length of the wire gives:

$P_2 = \frac{aL^2R}{3}$

Using the relation $V = \frac{aLR}{2}$ , this simplifies to:

$P_2 = \frac{4}{3}\frac{V^2}{R} = \frac{4}{3}P_1$

This is the conclusion I arrive at. Correct me if I am wrong:

Among all current distributions, considering Ohm's law, the actual current distribution corresponds to the case where a minimum amount of power is dissipated in the resistor. However, I have not tested this with another current distribution.

This is kind-of, sort-of a case where the physical configuration corresponds to some kind of least action. Not in the strict sense, probably

Using Ohm's law, we can easily get a= $\dfrac{2V}{LR}$ . Using the definition of power, integrating over the length of the rod, we get $P_2$ = $\dfrac{4V^2}{3R}$ . Hence the ratio is 4 : 3