Irodov 3.190: Two Wires Make Three Parallel – Amit & Deepali Sinha Foundation

Problem Irodov 3.190

Three equal resistors $R$ stand in a row between two terminals, so the row has four junctions: the first at the left terminal, the second and the third between neighbouring resistors, and the fourth at the right terminal. A plain wire joins the second junction to the fourth, and another plain wire joins the first junction to the third. The terminals are connected to a dc source of internal resistance $R_0$ . For what value of $R$ is the thermal power generated in this load the greatest?

The load: three equal resistors in a row, with one plain wire from the second junction over the top to the last, and another from the first junction under the row to the third.

Given

source of internal resistance $R_0$
three equal resistors $R$ , wired as shown

Find

the

R

that makes the power in the load greatest

Solution

Step 1The two bypass wires put the resistors in parallel

Naming the junctions $a$ , $b$ , $c$ , $d$ from left to right, the lower wire makes $a$ and $c$ one node and the upper wire makes $b$ and $d$ another. Each resistor joins one node to the other: the first runs from $a$ to $b$ , the second from $b$ to $c$ , which is from $b$ to $a$ , and the third from $c$ to $d$ , which is from $a$ to $b$ . The terminals sit on $a$ and on $d$ , that is on $b$ . So all three are in parallel between the terminals.

The load redrawn on the source, emf E with its internal resistance R_0 shown as a resistor: the three resistors R in parallel. — The load redrawn on the source, emf $\mathcal{E}$ with its internal resistance $R_0$ shown as a resistor: the three resistors $R$ in parallel.

The resistance of the load is therefore

R_L = \frac{R}{3} \tag{1}

Step 2The power in the load as a function of

R_L

The question concerns the power in the load: the power of the whole circuit, $R_0$ included, only grows as $R$ falls. With the emf $\mathcal{E}$ of the source, the current is $\mathcal{E}/(R_0 + R_L)$ , and the power generated in the load is

P = \frac{\mathcal{E}^2 R_L}{(R_0 + R_L)^2}

It vanishes both for $R_L \to 0$ (no voltage across the load) and for $R_L \to \infty$ (no current), so it has a maximum in between, where the derivative is zero:

\frac{dP}{dR_L} = \mathcal{E}^2\,\frac{(R_0 + R_L)^2 - 2R_L(R_0 + R_L)}{(R_0 + R_L)^4} = \mathcal{E}^2\,\frac{R_0 - R_L}{(R_0 + R_L)^3} = 0

R_L = R_0

(2)

Step 3The resistor that matches the source

Setting (1) equal to (2):

\frac{R}{3} = R_0

R = 3R_0

(3)

Check

The sign of $dP/dR_L$ is that of $R_0 - R_L$ : positive below $R_L = R_0$ and negative above it, so the stationary point is a maximum, where $P = \mathcal{E}^2/4R_0$ . A numerical spot check with $R_0 = 1$ and $\mathcal{E} = 2$ : at $R_L = 0.5$ , $1$ and $2$ the power is $0.89$ , $1.00$ and $0.89$ , largest at $R_L = R_0$ as found.

Remark

The load’s internal wiring only matters through its total resistance; the matching rule $R_L = R_0$ does the rest. Read as a series chain, the same three resistors would give $3R$ and the wrong answer $R = R_0/3$ , a factor of nine off. At the optimum exactly half the emf appears across the load and half the source’s power is lost in $R_0$ , so matching maximises the power delivered, not the efficiency, which is only $50$ per cent there.