Solving the Physics Riddle: A Current Across a Single Resistor

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Tom McCurdy
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Today in physics lecture our physics teacher left us with a riddle.

If you had an infinite grid of resistors say with 1 ohm of resistance set up between every node of the grid and then you connected a battery that supplies a current of one amp through across a single resistor, what would be the current across that single resistor.
 
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Is this like that question game on "Whose Line Is It Anyway?"
 
Swiper no swiping!
 
by the way the answer is 2/pi i believe
 
Tom McCurdy said:
Today in physics lecture our physics teacher left us with a riddle.

If you had an infinite grid of resistors say with 1 ohm of resistance set up between every node of the grid and then you connected a battery that supplies a current of one amp through across a single resistor, what would be the current across that single resistor.
You do that by calling the effective resistance R and each resistor r.
Then (R+2r) and r in parallel equal R. (Draw a picture to see that.)
This gives 1/R=1/r+1/(R+2r), and I get R=(sqrt{3}-1) ohm.
Your phrase "a current of one amp through across a single resistor, what would be the current across that single resistor" is too confusing to determine what you mean by the current.
 
Achuz, I don't think your reasoning is correct.

Tom McCurdy said:
... a battery that supplies a current of one amp through across a single resistor, what would be the current across that single resistor[?]

Clearly, the answer to the teacher's question (as posed, a riddle) is "one amp" (notice the battery is specifically configured to supply one amp across "that" resistor -- wording implies in situ!).

A more interesting question is to interpret that the battery instead provides one volt (ie. one amp across the one ohm in isolation), that is, asking how to calculate the effective resistance of the infinite grid - in which case a factor of pi in the result wouldn't sound unreasonable.
 
cesiumfrog said:
Achuz, I don't think your reasoning is correct.



Clearly, the answer to the teacher's question (as posed, a riddle) is "one amp" (notice the battery is specifically configured to supply one amp across "that" resistor -- wording implies in situ!).

A more interesting question is to interpret that the battery instead provides one volt (ie. one amp across the one ohm in isolation), that is, asking how to calculate the effective resistance of the infinite grid - in which case a factor of pi in the result wouldn't sound unreasonable.

I thought it was one amp as well... but its not... I am going to try and find out if he meant the resistance across one set of nodes which Is 2/pi