Irodov 3.198: Grouping Cells to Match the Load – Amit & Deepali Sinha Foundation

Problem Irodov 3.198

A battery is made of a large number $N = 300$ of identical cells, each of internal resistance $r = 0.3$ $\Omega$ , arranged in $n$ groups connected in parallel, each group holding the same number of cells in series. The battery feeds an external resistance $R = 10$ $\Omega$ . For what number $n$ of groups is the heat power generated in the external resistance greatest?

The grouping: n equal rows of N/n cells (each of emf E and internal resistance r) in series (rows 1, 2 and the last drawn), the rows in parallel, all + plates to the right, loaded by R. — The grouping: $n$ equal rows of $N/n$ cells (each of emf $\mathcal{E}$ and internal resistance $r$ ) in series (rows $1$ , $2$ and the last drawn), the rows in parallel, all $+$ plates to the right, loaded by $R$ .

Given

$N = 300$ identical cells
$r = 0.3$ $\Omega$ each
$R = 10$ $\Omega$

Find

the number

n

of parallel groups for the greatest power in

R

Solution

Step 1The battery as one source

Let each cell have the emf $\mathcal{E}$ . A group of $N/n$ cells in series has the emf $(N/n)\mathcal{E}$ and the internal resistance $(N/n)r$ . The $n$ equal groups in parallel keep that emf and divide the resistance by $n$ :

\mathcal{E}_{\mathrm{b}} = \frac{N\mathcal{E}}{n}, \qquad r_{\mathrm{b}} = \frac{Nr}{n^2} \tag{1}

Step 2The current through the load

The power in the fixed resistance $R$ is $I^2R$ , so it is greatest when the current $I$ is. With (1):

I = \frac{\mathcal{E}_{\mathrm{b}}}{R + r_{\mathrm{b}}} = \frac{N\mathcal{E}/n}{R + Nr/n^2}

Multiplying top and bottom by $n$ :

I = \frac{N\mathcal{E}}{nR + Nr/n}

(2)

Step 3The best number of groups

The current (2) is greatest when the denominator $f(n) = nR + Nr/n$ is least. Treating $n$ as continuous and setting the derivative to zero:

\begin{aligned} \frac{df}{dn} = R - \frac{Nr}{n^2} &= 0\\[2pt] n^2 &= \frac{Nr}{R} \end{aligned}

The second derivative $2Nr/n^3$ is positive, so this is the minimum of $f$ . Numerically:

n = \sqrt{\frac{300\cdot 0.3}{10}} = \sqrt{9}

n = \sqrt{\frac{Nr}{R}} = 3

(3)

The value is a whole number and divides $N$ : three parallel groups of $100$ cells in series.

Check

By (1) the battery’s internal resistance at $n = 3$ is $300\cdot 0.3/9 = 10$ $\Omega$ , equal to $R$ , the matching condition. Neighbours confirm it: the denominator $f(n)$ is $30 + 30 = 60$ $\Omega$ at $n = 3$ , against $20 + 45 = 65$ $\Omega$ at $n = 2$ and $40 + 22.5 = 62.5$ $\Omega$ at $n = 4$ , so both carry less current.

Remark

The grouping is the battery’s way of choosing its own internal resistance: series cells raise the emf and the resistance together, parallel rows lower the resistance at a fixed emf. The best choice makes the battery’s internal resistance equal to the load, and then exactly half of all the Joule heat is delivered to $R$ ; with a load much larger than $Nr$ all cells would go in one series row ( $n = 1$ ), with a load much smaller than $r/N$ all in parallel ( $n = N$ ).