As for the resistance problem, it appears to me that the number of parallel n-length paths increases faster than n, causing the resistance to approach 0. I need someone to verify me on that, though.

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For instance, a 20 Ohm resistor in parallel with a 30 Ohm resistor is equal to 1/(1/20 + 1/30) = 12 Ohms, far less than 20 Ohms. The current splits across *both* of them simultaneously.

If you simplify the stated problem to a rectangular grid containing 6 nodes and 7 resistors (2 across the top, 2 across the bottom, 3 vertically), then the resistance is, I’m fairly sure, 1/(1/3 + 1/3 + 1/3) = 1 Ohm, since there are three different 1 Ohm paths to choose from (and paths that form loops by touching the same node twice are invalid, so those are the only permitted paths).

However, when you make the grid big enough to allow 5 Ohm paths, you now have 1/(1/3 + 1/3 + 1/3 + 1/5 + 1/5 + …), which is obviously a bit smaller than 1 Ohm since it’s the reciprocal of a number slightly larger than 1.

So, to compute the overall answer for an infinite grid of resistors, you have to take the limit as the size of the grid approaches infinity. And there’s a combinatorial explosion in the number of possible paths, so it’s something you have to put some significant thought into.

For instance, a 20? resistor in parallel with a 30? resistor is equal to 1/(1/20? + 1/30?) = 12?, far less than 20?. The current splits across *both* of them simultaneously.

If you simplify the stated problem to a rectangular grid containing 6 nodes and 7 resistors (2 across the top, 2 across the bottom, 3 vertically), then the resistance is, I’m fairly sure, 1/(1/3? + 1/3? + 1/3?) = 1?, since there are three different 1? paths to choose from (and paths that form loops by touching the same node twice are invalid, so those are the only permitted paths).

However, when you make the grid big enough to allow 5? paths, you now have 1/(1/3? + 1/3? + 1/3? + 1/5? + 1/5? + …), which is obviously a bit smaller than 1? since it’s the reciprocal of a number slightly larger than 1.

So, to compute the overall answer for an infinite grid of resistors, you have to take the limit as the size of the grid approaches infinity. And there’s a combinatorial explosion in the number of possible paths, so it’s something you have to put some significant thought into.

While an author can’t and shouldn’t try to please everyone it is valuable information to learn that once dedicated fans no longer take interest.

Constructive negative feedback is much more useful than destructive positive feedback.

Um… I know this might not be the appropriate place to post a question about the most recent comic but Thevenin’s theorum states something like you can summarise any unknown circuit as a sum of it’s voltage sources and resistances. I have only Wikipedia to go on as I had no clue what it was before.

So the solution is 3 ohms, isn’t it? Electricity always follows the shortest path and thus wouldn’t traverse any of the superflouous infinite 1 ohm resistors other than the shortest path. (In theory)…

Am I missing something? I really don’t know much about physics or electronics yet. I’m sorry for being such a n00b.

- trying to learn

Jesus fucking christ.

XKCD is like my favorite webcomic!

If all you bitches don’t like it, then fuck off, seriously!

And hey Randall, if you would like to, i would like to have you come to my high school to have a wiked time and give a short presentation or something