Resistor Problem
There's a question on the Google Aptitude Test about resistors:
10. On an infinite, twodimensional, rectangular lattice of 1ohm resistors, what is the resistance between two nodes that are a knight's move away?
I don't think that the answer is as complicated as what people are making it, since every node is the same value. Here's what I think the solution is:
A knight's move is a combination of 2 squares in one dimension, and 1 square in the other dimension. So, if you imagine all of these 1ohm resistors in a grid:
! ! ! ! ! ! ! ! !
+++++++++
! ! ! ! ! ! ! ! !
+++++++++
! ! ! ! ! ! ! ! !
+++++++++
! ! ! ! ! ! ! ! !
+++++++++
! ! ! ! ! ! ! ! !
then you're trying to find the total resistance from one intersection to another one that is, say, 2 down and 1 to the right (i.e., resistance from A to F):
AB
! !
CD
! !
EF
Despite the fact that this is an infinite grid, all resistors outside of this area will not have any effect on the system between the points A and F.
Breaking down the network:
ABD = 2 ohms
ACD = 2 ohms
The total resistance from AD, therefore, is 1 ohm because:
Rt = 1/(1/R1 + 1/R2) = 1/(1/2 + 1/2) = 1/1 = 1
DF = 1 ohm
So, the total resistance following path of the network ABCDF is 2 ohms.
But, there is another network in parallel: ACDEF It's value is also 2ohms.
In all, going from point A to F is like having two 2ohm resistors in parallel, which we showed above to be 1 ohm total.
I'll probably build this circuit and take physical measurements, but as I type this, I think that the answer is 1ohm.
I'd gladly admit defeat if someone who actually studied this in school has a different explanation. ;)
