Irodov 3.180: Two Batteries Become One – Amit & Deepali Sinha Foundation

Problem Irodov 3.180

Two batteries, of emfs $\mathcal{E}_1$ and $\mathcal{E}_2$ and internal resistances $R_1$ and $R_2$ , are connected in parallel. Find the emf and the internal resistance of the single source that is equivalent to the pair.

The two batteries in parallel, each drawn as its emf in series with its internal resistance, like poles joined. A and B are the output terminals of the pair. — The two batteries in parallel, each drawn as its emf in series with its internal resistance, like poles joined. $A$ and $B$ are the output terminals of the pair.

A source is known by its terminal voltage as a function of the current it delivers. A single source of emf $\mathcal{E}$ and internal resistance $R_i$ delivering a current $I$ has

\varphi_A - \varphi_B = \mathcal{E} - IR_i \tag{1}

So we find the same relation for the pair and read off $\mathcal{E}$ and $R_i$ .

Given

$\mathcal{E}_1$ , $R_1$
$\mathcal{E}_2$ , $R_2$
like poles joined (assumed)

Find

the equivalent emf

\mathcal{E}

· the equivalent internal resistance

R_i

Solution

Step 1The pair delivering a current

Let the pair deliver a current $I$ out of terminal $A$ , through any external load, and back into $B$ ; call $U$ the terminal voltage $\varphi_A - \varphi_B$ . The battery currents $I_1$ and $I_2$ are taken upward through each battery, toward $A$ , as drawn. We take like poles joined, both positive terminals at $A$ , the usual meaning of a parallel connection.

The pair delivering I, with the battery currents I_1 and I_2. — The pair delivering $I$ , with the battery currents $I_1$ and $I_2$ .

The single source equivalent to the pair.

Walking up each branch from $B$ to $A$ , the potential falls by the $IR$ drop and rises by the emf, and at the top junction the battery currents add up to $I$ :

\begin{aligned} U &= \mathcal{E}_1 - I_1R_1\\ U &= \mathcal{E}_2 - I_2R_2\\ I &= I_1 + I_2 \end{aligned}

Step 2The terminal voltage of the pair

Solving the first two for $I_1$ and $I_2$ and putting them into the third:

I = \frac{\mathcal{E}_1 - U}{R_1} + \frac{\mathcal{E}_2 - U}{R_2}

Collecting the terms in $U$ and multiplying through by $R_1R_2$ :

U(R_1 + R_2) = \mathcal{E}_1R_2 + \mathcal{E}_2R_1 - IR_1R_2

Dividing by $R_1 + R_2$ :

U = \frac{\mathcal{E}_1R_2 + \mathcal{E}_2R_1}{R_1 + R_2} - I\,\frac{R_1R_2}{R_1 + R_2}

(2)

Step 3The equivalent source

Relation (2) has the form (1) for every current $I$ . The term free of $I$ is the emf, and the coefficient of $-I$ is the internal resistance:

\begin{gathered} \mathcal{E} = \frac{\mathcal{E}_1R_2 + \mathcal{E}_2R_1}{R_1 + R_2}\\[4pt] R_i = \frac{R_1R_2}{R_1 + R_2} \end{gathered}

(3)

Check

Two equal batteries, $\mathcal{E}_1 = \mathcal{E}_2$ : the numerator of $\mathcal{E}$ becomes $\mathcal{E}_1(R_1 + R_2)$ , so $\mathcal{E} = \mathcal{E}_1$ , and the internal resistances combine in parallel, as they should. A battery with $R_2 \to \infty$ is no battery at all: then $\mathcal{E} \to \mathcal{E}_1$ and $R_i \to R_1$ . Units: both terms of $\mathcal{E}$ are volts.

Remark

Written as $\mathcal{E} = (\mathcal{E}_1/R_1 + \mathcal{E}_2/R_2)/(1/R_1 + 1/R_2)$ , the equivalent emf is the mean of the two emfs weighted by the conductances of the batteries, so it always lies between them, nearer the stiffer battery. With no load ( $I = 0$ ) a current $(\mathcal{E}_1 - \mathcal{E}_2)/(R_1 + R_2)$ still circulates between the two, heating both: the reason unequal batteries are not put in parallel. If one battery is turned round, its emf enters with the opposite sign.