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Question:
You Can't Resist
Consider an infinite network of resistors, where each resistor has the
same resistance R, as in the following diagram. In the diagram, the
network extends infinitely off to the right.
Find, with proof, the effective resistance between nodes A and B. You
will need to know how to find the effective resistance of resistors
connected in series and in parallel. The following web link gives the
formulas:
http://www.physics.uoguelph.ca/tutorials/ohm/index.html
Note: This looks like a physics problem, but in fact, it is a math problem!
It's just a math problem in the world of physics. Other than the
information about how to find effective resistances of connected resistors,
you don't need to know anything about physics.
Answer:
There are two properties of resistors that are useful in this problem.
1. Resistors in series add linearly. i.e. R(total) = R(1)+R(2)+R(3)+...
2. Resistors in parallel add inversely. i.e. 1/R(total) = 1/R(1) + 1/R(2) + 1/R(3) +...,
which can be simplified to R(total) = (R(1)*R(2))/(R(1)+R(2)) for the case of only two resistors.
So we have this infinite series of resistors, of which I will arbitrarily
give a value of 1 to each (we can add in the 'unit' R at the end), and a
resistance between nodes A and B, assigned P(AB). If it were a finite 'row'
then we would be able to use the two properties above to help us, so let's
do that for a bit. We go out some arbitrary great length and cut the wires
to the rest of the infinite resistors. Now we have immediately three
resistors in series in parallel with one other (if you look at the figure in
the problem and consider the last box in the iteration, then the upper,
right and lower resistors are in series, which are in parallel to the
resistor on the left of the box). We can add the three in series together to
obtain a resistor of value 3 in parallel with a resistor of value 1. Adding
these:
3*1/(3+1) = 3/4
We get another box, only now the right side has value 3/4 instead of 1. We
can repeat this process numerous times, getting values like 11/15 and 41/56,
but this won't help in finding the correct answer. The fact that this is an
iterative process will.
The general equation for the next resistor value in the series is given by:
(2+x)*1/(2+x+1) = (2+x)/(3+x)
taking x to be the previous vertical resistor and the values of the rest to
be 1. Now let's use this to consider the original infinite chain of
resistors.
Another matter that needs attention is whether this infinite chain of resistors has some limit,
or if it merely oscillates between several values like a sinusoidal wave or if it blows up like a tangent.
The first thing we need to notice is that this series is monotonic down, that is that the next value is
two plus the last value divided by two plus the last value plus one: f(x+1) = (2+f(x)) / (2+f(x)+1),
thus if it has a limit, it will be lower than any of the values computed from a finite chain.
Next consider the case where we remove all of the horizontal resistors.
If we follow the same logic as before in in computing g(x+1), we get:
g(x+1) = g(x) / (g(x) + 1), which is also monotonic down for the same reasons
now let us assume that if both series start with the same f(x), i.e. f(x) = g(x) and that the first
series is larger than the second:
f(x) / (f(x) + 1) < (2 + f(x)) / (3 + f(x)), so by cross multiplying
3*f(x) + f(x)*f(x) < 2 + 3*f(x) + f(x)*f(x)
0 < 2, which is true
After one iteration, the first series will be larger than the second one. Furthermore, since the first series
now uses a larger value than the second series, so the gap between the two can only become larger.
Thus the first series is always greater than the second, so if we can find a limit for the second series,
this will be a lower bound to the first.
We are given that the one divided by the sum of any number of resistors in parallel is equal to the sum of one
divided by each of the resistors. If we consider the second series, all it is is an infinite number of identical
resistors summed in parallel. Each has a value of 1, so the sum of an infinite number of ones divided by 1
approaches infinity, as it does for any other finite value for the resistor. And if one divided by our equivalent
resistor approaches infinity, our equivalent resistor approaches zero, which is a finite limit. Thus since our
original resistor chain is monotonic down and bounded from below, it then must approach some finite value
as more of the infinite chain of resistors is considered. Now we find that value.
We have an equation to find any resistor value granted we know the previous
one. Let's use this to find the resistance between A and B, P(AB) with
respect to the resistance across the resistor immediately to the right of it,
P(A'B').
P(AB) = (2+P(A'B'))/(3+P(A'B'))
Now consider P(A'B'): There is still an infinite chain of resistors
extending from it, so, and this is key, P(AB) = P(A'B') when the entire chain is considered.
Substituting P(AB) = P(A'B') = r into the equation we get:
r = (2+r)/(3+r), thus:
3*r+r^2 = 2+r, and
r^2 + 2*r -2 = 0
Solving with the quadratic equation we get:
r = (-2 (+or-) sqrt (4 + 4*2*1))/2
r = (-2 (+or-) 2*sqrt(3))/2, and finally
r = -1 (+or-) sqrt(3)
Now since we cannot have negative resistances we can eliminate the (-)
solution, leaving
r = sqrt(3) - 1
So finally we get that
P(AB) = R*r = R(sqrt(3) - 1)
-- Stefan Dorsett
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