Irodov 3.191: Currents Split to Waste Least – Amit & Deepali Sinha Foundation

Problem Irodov 3.191

Two resistors $R_1$ and $R_2$ are connected in parallel. Show that the way a current divides between them is the division that makes the total heat generated per unit time in the pair as small as possible.

The two resistors R_1 and R_2 in parallel between two terminals. — The two resistors $R_1$ and $R_2$ in parallel between two terminals.

Given

$R_1$ , $R_2$ in parallel
the total current held fixed (assumed)

Find

the split that minimises the heat, compared with the actual split

Solution

Step 1The heat for an arbitrary split

Fix the total current $I$ that enters the pair, which we take as given, and imagine it divided in any way at all: $I_1$ through $R_1$ and the rest through $R_2$ . The current rule at the junction is kept, but nothing else is assumed.

A trial division of the current: I_1 through R_1 and I - I_1 through R_2, with I_1 free to take any value. — A trial division of the current: $I_1$ through $R_1$ and $I - I_1$ through $R_2$ , with $I_1$ free to take any value.

The heat generated per unit time in the pair is then a function of $I_1$ alone:

P(I_1) = I_1^2R_1 + (I - I_1)^2R_2 \tag{1}

Step 2The split that makes the heat smallest

The heat (1) is a parabola in $I_1$ opening upward. Setting its derivative to zero:

\frac{dP}{dI_1} = 2I_1R_1 - 2(I - I_1)R_2 = 0

and its second derivative is positive, so the stationary point is the minimum:

\frac{d^2P}{dI_1^2} = 2(R_1 + R_2) > 0

Solving the first condition for $I_1$ , and then $I - I_1$ :

I_1 = \frac{R_2}{R_1 + R_2}\,I, \qquad I - I_1 = \frac{R_1}{R_1 + R_2}\,I

(2)

Step 3Comparing with the actual split

The minimum condition, written for the two branch currents $I_1$ and $I_2 = I - I_1$ , says the two drops are equal:

I_1R_1 = I_2R_2

(3)

The least-heat split has equal drops, which is the split the parallel pair actually takes. That is exactly the condition the real circuit obeys: both resistors are joined between the same two junctions, so the voltage across them is the same, and Ohm’s law then gives (3), with the currents of (2). The division the circuit actually takes is the one of least heat.

Check

Substituting (2) into (1) gives the minimum heat:

P_{\min} = \frac{R_2^2R_1 + R_1^2R_2}{(R_1 + R_2)^2}\,I^2 = \frac{R_1R_2}{R_1 + R_2}\,I^2

which is $I^2$ times the familiar parallel resistance, as it must be. Any other split costs more: putting all the current through $R_1$ gives $I^2R_1$ , larger than $P_{\min}$ because $R_2/(R_1 + R_2) < 1$ . A limiting case: as $R_2 \to 0$ , the minimum sends all the current through $R_2$ with no heat at all, which is what a short across $R_1$ does.

Remark

This is a small case of a general theorem: in a network of resistors without sources inside it, the steady currents are the distribution, among all those that respect the current rule at every junction, that makes the total Joule heat a minimum (Thomson’s theorem). The loop rule, a zero net drop around every closed path, comes out as the condition of that minimum, just as the equal drops across $R_1$ and $R_2$ , the loop rule for the one loop they form, did here.