This is a classic problem, which you may have encountered (or seen a variation on it) if you are into these kinds of things.

What is the equivalent resistance between A and B in this infinite series of resistors? (I stole the picture from somebody else, since the problem is common enough I knew the internet would have a diagram ready for me.)

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July 20, 2009 at 11:55 pm

[…] Arcsecond Playing on the Sea-Shore, Rough Pebbles Welcome « New Problem: Infinite Resistors […]

July 21, 2009 at 12:26 am

Ok, I’ll have a bash, using the usual infinite series trick:

R = (R in parallel with R2) in series with R1

R = 1/(1/R + 1/R2) + R1

Multiply out

R = 1/(R2/R.R2 + R/R.R2) + R1

R = 1/[(R2+R)/(R.R2)] + R1

R = (R.R2)/(R2+R) + R1

R = (R.R2)/(R2+R) + [R1(R2+R)]/(R2+R)

R = [(R.R2)+R1(R2+R)]/(R2+R)

R(R2+R) = [(R.R2)+R1(R2+R)]

R^2 + R.R2 = R.R2 + R1.R2 + R1.R

R^2 – R R1 – R1.R2 = 0

Solve quadratic for R

R = [R1 +/- sqrt(R1^2 + 4.R1.R2)]/2

So if r1 is 2 and r2 is 1.5 (to pick an example entirely at random), then R is 3 ohms. 3 in parallel with 1.5 is 1, in series with 2 is 3 oms again. QED.

July 21, 2009 at 2:41 am

Thanks, Paul. I forgot the picture left and unspecified. Looks like you got the general answer.

July 21, 2009 at 2:56 pm

Paul is close, but it can be slightly reduced. Consider that the sqrt term will always be greater than R1, so subtracting it from R1 never makes any sense. Negative resistance is a no-no, as is two distinct solutions to any resistance equation in general.

R = [R1 + sqrt(R1^2 + 4R1R2)]/2