Monday, January 14, 2008

Resistances and Golden Ratio

We are given infinite amount of unit resistors (each resistor has a
resistance of 1 Ohm). We can connect these resistors in series or/and in
parallel resulting in a resistor network. For example, you can get a
resistance of 2/3 Ohm by connecting two 1/3 Ohm Resistor networks in series.
To get a 1/3 Ohm Resistor network, we can connect three unit resistors in parallel.
One puzzle is to realize a resistance whose value is the exact golden ratio.

The infinite ladder resistor (with end points A and B) network having a unit resistor
from point A to say Point Y and connect a unit resistor from Y to B. Now repeat the
above construction starting at Point Y. This leads to an infinite ladder network.
and the resistance between points A and B will be the golden ratio.
(as the resistance between A and B is the same as the resistance between Y and B)

To see this, consider the resistor network of unit resistor in serial connection
with a parallel resistor network of 1 Ohm and x Ohms. Let us denote the
resistance between A and B is
x = 1 + x/(1+x)

or x-1 = x/(x+1)

(The author proposed a variation of this problem to IEEE Potential in 1982.
A selected collection of problems including the variation
problem 3-17 appears in
"The Unofficial IEEE Brain-buster Gamebook : Mental Workouts for the Technically
Inclined" by D. R. Mack, Wiley-IEEE Press, 1992.)

With duality (changing resistors in series to parallel and parallel to series) and
scaling, we can realize resistances of all integer powers of the golden ratio as
well as the integer powers of reciprocal of golden ratios!

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