My guest blogger is Levy Ulanovsky, a maven of physics puzzles. Here is one of his favorite puzzles:

There are n points in 3-dimensional space. Every point is connected to every other point by a wire of resistance R. What is the resistance between any two of these points?

Ditto Leo. I only know very basic physics, but I think this can be puzzled out from Ohm’s Law, in the form V/I = R.

Let us fix the power source and let n vary. By analogy to pressure, the voltage everywhere in the circuit is inversely proportional to the total length of wire, which is n(n-1)/2. Meanwhile, the effective current between any pair of nodes is inversely proportional to the average travel time between the two nodes. Since the current forks in n-1 equally attractive directions at each junction, this travel time is proportional to n-1.

So, between any pair of nodes, V/I is in linear proportion to n-1/[n(n-1)/2] = 2/n.
If the resistance in each wire is R, then the resistance between any pair of nodes is 2R/n (since this gives the right constant of proportionality when n=2).

I imagine a real physics maven could explain this more succinctly and persuasively!

Here is an approach I would imagine a physicist could use. Let us connect 2 points of our grid to a battery. Then all the rest of the points will have the same potential because of the symmetry, and the current in all the wires connecting them will be zero. So these wires can be removed, and all we are left with is one wire of resistanse R and n-2 wires of resistance 2R connecting the terrminals of the battery. Now use the fact that the resistance of several resistors in parallel connection is equal to the harmonic mean of the resistance of all the resistors to get the formula.

Here is another one.If you throw a ball upwards, will it take more time to reach the maximum height or to fall down from it? If we disregard the air reistance, the times will be the same. But what if we take it into account? It is an “applied project” in Stewart’s 4-th edition Calculus, pages 602-603 and includes the drag force propotional to the velocity and some dreadful calculations. Can you tell without any calculations?

## Leo:

I got 2R/n.

17 August 2009, 3:30 pm## Austin:

Ditto Leo. I only know very basic physics, but I think this can be puzzled out from Ohm’s Law, in the form V/I = R.

Let us fix the power source and let n vary. By analogy to pressure, the voltage everywhere in the circuit is inversely proportional to the total length of wire, which is n(n-1)/2. Meanwhile, the effective current between any pair of nodes is inversely proportional to the average travel time between the two nodes. Since the current forks in n-1 equally attractive directions at each junction, this travel time is proportional to n-1.

So, between any pair of nodes, V/I is in linear proportion to n-1/[n(n-1)/2] = 2/n.

If the resistance in each wire is R, then the resistance between any pair of nodes is 2R/n (since this gives the right constant of proportionality when n=2).

I imagine a real physics maven could explain this more succinctly and persuasively!

18 August 2009, 1:22 pm## misha:

Here is an approach I would imagine a physicist could use. Let us connect 2 points of our grid to a battery. Then all the rest of the points will have the same potential because of the symmetry, and the current in all the wires connecting them will be zero. So these wires can be removed, and all we are left with is one wire of resistanse R and n-2 wires of resistance 2R connecting the terrminals of the battery. Now use the fact that the resistance of several resistors in parallel connection is equal to the harmonic mean of the resistance of all the resistors to get the formula.

22 August 2009, 4:15 pm## misha:

Oops! Sorry, it’s not the harmonic mean, it’s m times smaller, where m is the number of resistors in the parallel connection.

22 August 2009, 4:22 pm## misha:

Here is another one.If you throw a ball upwards, will it take more time to reach the maximum height or to fall down from it? If we disregard the air reistance, the times will be the same. But what if we take it into account? It is an “applied project” in Stewart’s 4-th edition Calculus, pages 602-603 and includes the drag force propotional to the velocity and some dreadful calculations. Can you tell without any calculations?

13 November 2009, 10:25 pm