The clarified version
As far as I understand, Wigner considers a "miracle" the fact that it is even possible to find a mathematical equation that describes a natural phenomenon.
It is not exactly what I was wondering about though. Lets say such an equation has been found. What exactly does it describe?
Do we treat the phenomenon itself as just a black box that happens to "output" numbers that fit into the equation?
This idea is supported by the fact not every intermediate step in solving the system of equations has an obvious physical interpretation.The system of equations mirror the internal "structure/working" of the phenomenon?
On the other hand, this is supported by the following example. Kirchhoff's rule "the algebraic sum of currents in a network of conductors meeting at a point is zero" clearly follows from the fact no additional charges enter or leave the circuit.Is it a mix of the both options above?
Maybe throughout the history it has been discovered empirically that coming up with equations and then solving them works for physics, but no one really knows why and how it works?
An answer along these lines is perfectly fine with me too. I just have not seen the way/method math is used in physics discussed anywhere -- and so wonder if I'm missing something obvious to everyone else.
**The original question**
My question is a general one. But to explain what it is asking let's first a look at "solving" of an electrical circuit using Kirchhoff's laws as an example.
(And this example takes a considerable amount of the question's real-estate but please keep in mind it is only an example to base the actual question on).
So, the circuit below:
is solved in the following way:
According to the first law we have
$$i_{1}-i_{2}-i_{3}=0$$
The second law applied to the closed circuit s1 gives
$$-R_{2}i_{2}+{\mathcal {E}}_{1}-R_{1}i_{1}=0$$
The second law applied to the closed circuit s2 gives
$$-R_{3}i_{3}-{\mathcal {E}}_{2}-{\mathcal {E}}_{1}+R_{2}i_{2}=0$$
Thus we get a linear system of equations in
$${\begin{cases}i_{1}-i_{2}-i_{3}&=0\\-R_{2}i_{2}+{\mathcal {E}}_{1}-R_{1}i_{1}&=0\\-R_{3}i_{3}-{\mathcal {E}}_{2}-{\mathcal {E}}_{1}+R_{2}i_{2}&=0\end{cases}}$$
Which is equivalent to
$${\begin{cases}i_{1}+-i_{2}+-i_{3}&=0\\R_{1}i_{1}+R_{2}i_{2}+0i_{3}&={\mathcal {E}}_{1}\\0i_{1}+R_{2}i_{2}-R_{3}i_{3}&={\mathcal {E}}_{1}+{\mathcal {E}}_{2}\end{cases}}$$
Assuming
$$R_{1}=100,\ R_{2}=200,\ R_{3}=300{\text{ (ohms)};\ {\mathcal {E}}_{1}=3,\ {\mathcal {E}}_{2}=4{\text{ (volts)}}}$$
the solution is
$${\begin{cases}i_{1}={\frac {1}{1100}}\\[6pt]i_{2}={\frac {4}{275}}\\[6pt]i_{3}=-{\frac {3}{220}}\end{cases}}$$
$i_{3}$ has a negative sign, which means that the direction of $i_{3}$ is opposite to the assumed direction (the direction defined in the picture).
