$72 \mathrm{db} 4.5 \mathrm{~V}$-os, $2 \Omega$ belsó ellenállású zsebtelepet kapcsolunk össze. Hogyan tegyük ezt, hogy az $R=36 \Omega$-os külső ellenálláson folyó áram

a) maximális,

b) minimális legyen?

Mekkorák ezek az áramok? (Csak azokat a kapcsolásokat vegyük figyelembe, amelyek egyenlő számú telepet tartalmazó, sorosan kapcsolt csoportok párhuzamos kapcsolásából állnak.)

$72 \mathrm{db} 4.5 \mathrm{~V}$, $2 \Omega$ internal resistance pocket cells are connected together. How should we do this so that the current flowing through the $R=36 \Omega$ external resistance is

a) maximized,

b) minimized?

What are these currents? (Consider only those connections that consist of parallel-connected groups of cells connected in series, with an equal number of cells in each group.)
Let the internal resistance of the cells be $r_{b}$, their electromotive force $U_{0}$, and the number of cells connected in series be $s$, while the number of parallel groups be $p$ (see Fig. 1).

# $1987-11-428-1$. eps 

Using Kirchhoff's laws, the current flowing through the external resistance $R$ is:

$$
I_{R}=\frac{p s U_{0}}{p R+s r_{b}}
$$

Using the given data $\left(p s=72, R=36 \Omega, r_{b}=2, U_{0}=4.5 \mathrm{~V}\right)$, we get:

$$
I_{R}=\frac{4.5}{\frac{p}{2}+\frac{2}{p}}
$$

a) The current is maximized when the denominator is minimized under the given conditions. Using the inequality between the arithmetic and geometric means:

$$
\frac{1}{2}\left(\frac{p}{2}+\frac{2}{p}\right) \geq \sqrt{1}
$$

The equality condition is: $\frac{p}{2}=\frac{2}{p}=1$, i.e., $p=2$. In this case, we form 2 parallel groups from 36 cells connected in series. The current is $\left.I_{R}^{\max }=2.25 \mathrm{~A} . \mathrm{b}\right)$ The current is minimized when the denominator is maximized under the given conditions. The function $f(p)=\frac{p}{2}+\frac{2}{p}$ is strictly increasing both to the left (smaller $p$ values) and to the right (larger $p$ values) of the previously found minimum point. Therefore, it can achieve its maximum at one of the boundaries of the interval being considered.

For $p=1, \quad I_{R}=1.8 \mathrm{~A}$.

For $p=72, \quad I_{R}=0.125 \mathrm{~A}$.

Thus, the current is minimized when each element is connected in parallel $(s=1)$. The current is $I_{R}^{\min }=0.125 \mathrm{~A}$.

Remarks. 1. Several solvers considered the minimum current to be 0 by connecting the cells in reverse. We did not accept this as a valid solution. If we did not want to create any current, we could obviously achieve this in a much simpler way.

2. Finding the minimum and maximum points using different methods (differentiation, computer) does not result in physically different solutions.