Jtdylktuy

I know about a problem about resistance of infinite resistor ladder like this:

The solution is the equality
$$ R_l=R+R+frac{1}{frac{1}{R}+frac{1}{R_l}},$$

which is true as if we remove three leftmost resistors, we end up with exactly same ladder. We get good result because the total resistance is finite, as may be proved by representing the total resistance as a recursive sequence and proving its convergence.

I'm curious whether somebody could give me an example if a non-trivial ladder similar to this one, but such that its total resistance is infinite? I mean that if we use similar trick, we end up with something absurd.

Original question about this was asked here, but If this innapriopriate place to ask, let me know.

Well, whe we look trough it part for part:

• The first three most left resistors:
  $$text{R}_{1spacetext{in}}=text{R}+text{R}+text{R}=3cdottext{R}tag1$$


• The first six most left resistors:
  $$text{R}_{2spacetext{in}}=frac{text{R}cdotleft(text{R}+text{R}+text{R}right)}{text{R}+left(text{R}+text{R}+text{R}right)}+text{R}+text{R}=frac{11}{4}cdottext{R}tag2$$


• The first nine most left resistors:
  $$text{R}_{3spacetext{in}}=frac{text{R}cdotleft(frac{text{R}cdotleft(text{R}+text{R}+text{R}right)}{text{R}+left(text{R}+text{R}+text{R}right)}+text{R}+text{R}right)}{text{R}+left(frac{text{R}cdotleft(text{R}+text{R}+text{R}right)}{text{R}+left(text{R}+text{R}+text{R}right)}+text{R}+text{R}right)}+text{R}+text{R}=frac{41}{15}cdottext{R}tag3$$

So, we get (in general):

$$text{R}_{spacetext{n}spacetext{in}}=frac{text{R}cdottext{R}_{spaceleft(text{n}-1right)spacetext{in}}}{text{R}+text{R}_{spaceleft(text{n}-1right)spacetext{in}}}+text{R}+text{R}tag4$$

So, when $text{n}toinfty$, we get:

$$text{R}_{spaceinftyspacetext{in}}=frac{text{R}cdottext{R}_{spacespaceinftyspacetext{in}}}{text{R}+text{R}_{spaceinftyspacetext{in}}}+text{R}+text{R}spaceLongleftrightarrowspace$$
$$text{R}_{spaceinftyspacetext{in}}=left(1pmsqrt{3}right)cdottext{R}spaceimpliesspacetext{R}_{spaceinftyspacetext{in}}=left(1+sqrt{3}right)cdottext{R}tag5$$

Not exactly what you are looking for (as achille explains in the comment, that would be impossible), but along those lines. Just imagine removing the resistors along the horizontal sections in your ladder. You are left with
$$ R_l=frac{1}{1/R+1/R_l} $$
which doesn't have any real roots for $R_l$. In a way $R_l = 0$ "works", but mathematically speaking it is absurd, as you put it. Physically, what we have are just infinitely many $R$'s in parallel, so the "ladder's" resistance is more simply calculated as $R_l = R/n rightarrow 0$ as $n rightarrow infty$.

If you know how capacitors behave in DC circuits, then imagine the same example with capacitors of capacitance $C$ and the net capacitance of the ladder will be infinite (that's because capacitances, when combined, behave as would the reciprocals of resistances).