Irodov 3.197: Two Sources Feeding One Load – Amit & Deepali Sinha Foundation

Problem Irodov 3.197

An upper and a lower wire are joined by three parallel branches. The left branch holds a source of emf $\mathcal{E}_1$ in series with a resistance $R_1$ ; the right branch holds a source of emf $\mathcal{E}_2$ in series with a resistance $R_2$ ; the middle branch is a resistance $R$ . Both sources have their positive terminal towards the lower wire, and their internal resistances are negligible. For what value of $R$ is the heat power generated in $R$ greatest, and what is that greatest power?

The circuit: the sources E_1 and E_2, each with its long (+) plate towards R_1 or R_2 and the lower wire, and the load R between the two wires. — The circuit: the sources $\mathcal{E}_1$ and $\mathcal{E}_2$ , each with its long ( $+$ ) plate towards $R_1$ or $R_2$ and the lower wire, and the load $R$ between the two wires.

Given

$\mathcal{E}_1, \mathcal{E}_2$ (polarities as drawn)
$R_1, R_2$
internal resistances negligible

Find

the

R

that makes the power in

R

greatest · that power

P_{\mathrm{max}}

Solution

Step 1The current in each branch from the upper wire’s potential

Take the lower wire as zero of potential and call $\varphi$ the potential of the upper wire. In each branch let the current be counted downward. Going down the left branch the potential rises by $\mathcal{E}_1$ across the source (from its $-$ to its $+$ plate) and falls by $I_1R_1$ across $R_1$ , ending at zero; the right branch is the same with the index $2$ :

I_1 = \frac{\varphi + \mathcal{E}_1}{R_1}, \qquad I_2 = \frac{\varphi + \mathcal{E}_2}{R_2}, \qquad I = \frac{\varphi}{R} \tag{1}

Step 2The voltage across the load

No charge piles up on the upper wire, so the three downward currents add up to zero. With (1):

\begin{aligned} \frac{\varphi + \mathcal{E}_1}{R_1} + \frac{\varphi + \mathcal{E}_2}{R_2} + \frac{\varphi}{R} &= 0\\[2pt] \varphi\left(\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R}\right) &= -\left(\frac{\mathcal{E}_1}{R_1} + \frac{\mathcal{E}_2}{R_2}\right) \end{aligned}

Multiplying both sides by $RR_1R_2$ :

\varphi = -\frac{(\mathcal{E}_1R_2 + \mathcal{E}_2R_1)\,R}{R_1R_2 + (R_1 + R_2)R} \tag{2}

The sign says the current in $R$ flows upward: both sources push it the same way.

Step 3The power in the load

The heat power in $R$ is $\varphi^2/R$ . With (2):

P = \frac{(\mathcal{E}_1R_2 + \mathcal{E}_2R_1)^2\,R}{\left[R_1R_2 + (R_1 + R_2)R\right]^2}

(3)

Step 4The best load and its power

Write $a = R_1R_2$ and $b = R_1 + R_2$ , so that $P \propto R/(a + bR)^2$ . Setting the derivative to zero:

\begin{aligned} \frac{dP}{dR} &\propto \frac{(a + bR)^2 - 2bR(a + bR)}{(a + bR)^4} = \frac{a - bR}{(a + bR)^3}\\[2pt] a - bR &= 0 \end{aligned}

The derivative is positive below this value and negative above it, so it is the maximum:

R = \frac{R_1R_2}{R_1 + R_2}

(4)

At this load $a + bR = 2a$ . Substituting $R = a/b$ into (3):

P_{\mathrm{max}} = \frac{(\mathcal{E}_1R_2 + \mathcal{E}_2R_1)^2\,(a/b)}{4a^2} = \frac{(\mathcal{E}_1R_2 + \mathcal{E}_2R_1)^2}{4ab}

P_{\mathrm{max}} = \frac{(\mathcal{E}_1R_2 + \mathcal{E}_2R_1)^2}{4R_1R_2(R_1 + R_2)}

(5)

Check

Dimensions: $(\mathcal{E}R)^2/R^3$ is $\mathcal{E}^2/R$ , a power. Special case: remove the right branch ( $R_2 \to \infty$ ). Then (4) gives $R = R_1$ , and in $P_{\mathrm{max}}$ only the $\mathcal{E}_1R_2$ term survives:

P_{\mathrm{max}} \to \frac{\mathcal{E}_1^2R_2^2}{4R_1R_2\cdot R_2} = \frac{\mathcal{E}_1^2}{4R_1}

the familiar best power from one source of internal resistance $R_1$ .

Remark

Seen from the load, the two branches act as one source: its internal resistance is $R_1$ and $R_2$ in parallel, and its emf is the weighted mean $(\mathcal{E}_1R_2 + \mathcal{E}_2R_1)/(R_1 + R_2)$ , each emf weighted by the other branch’s resistance. The best load matches that internal resistance, and the greatest power is this emf squared over four times it. Had one source been turned round, its emf would enter with a minus sign and the two could cancel in the load altogether.