[variable]
+-------[resistor]---------+
| |
| [current]
| [ meter ]
| |
+-----[ variable ]-------+
[power supply]
where:$$ \require{enclose} \def\place#1#2#3{\smash{\rlap{\hskip{#1pt}\raise{#2pt}{#3}}}} % \bbox[10pt]{\enclose{box}{\phantom{\Rule{250pt}{75pt}{0pt}}}} % \place{-275}{70}{\enclose{box}{\bbox[5pt,lightblue]{ \begin{array}{c} \text{resistor} \\ \text{set resistance:}~R \end{array} }}} % \place{-270}{0}{ \enclose{box}{\bbox[5pt,lightblue]{ \begin{array}{c} \text{power source} \\ \text{set voltage:}~V \end{array} }}} % \place{-55}{30}{ \enclose{box}{\bbox[5pt,lightblue]{ \begin{array}{c} \text{current meter} \\ \text{measured current:}~I \end{array} }}} $$
$R$ is the resistance of the variable resistor;
$V$ is the voltage difference of the variable power source;
$I$ is the current measured by the current meter.
Since we can select both $V$ and $R$ and we know Ohm's law,$$ I=\frac{V}{R}, $$we can, in principle $I=\frac{V}{R}$, we can use this circuit to divide numbers for us:
Select some division problem you want to perform, $\frac{a}{b}=?$.
Set the voltage source to $V=a~\mathrm{V}$.
Set the resistor to $R=b~\mathrm{\Omega}$.
Measure $I=?~\mathrm{A}$ to get the result!
The thing's that Ohm's law, $V=IR$, is given in terms of uses real-number values, $\left\{V,I,R\right\}{\in}\mathbb{R}$. If we believe it to be literally truethat these values actually have infinite precision, then we can select two values ofperform multiplication or division with infinite precision and multiply them together in finite time; this is a feat that a Turing machine can't perform (unless we allow for zerowith finite-time operations).