Finding equivalent resistance

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Mayank Sinha

Joined Jun 12, 2013
3
how to find equivalent resistance across A & B if each wire is of resistance 'r' and it continues to infinity in all direction...?????

plzzz help me out??????????????
 

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t_n_k

Joined Mar 6, 2009
5,455
A commonly used technique in such problems is to consider the effect of the superposition of two equal magnitude currents I [usually 1Amp], one of which is injected (say) into node A and the other extracted from node B - with A and B being separated by a single resistor branch [i.e. adjacent]. Each injected (or extracted) node current is presumed to divide (or combine) symmetrically according to the number of branches at its respective node. The implicit assumption is that the injected or extracted current fans out symmetrically from the node by virtue of the network extremities (albeit at infinity) being held everywhere at a uniform potential with respect to the injection node [A] in one case or the extraction node in the other case. The superposition of the respective injected & extracted currents [in fact their summation] in the linking branch, leads one to a potential difference [VAB] across the resistor branch connecting node A & node B. Viewed as a lumped equivalent the effective resistance is RAB=VAB [Volts]/I [Amp].

The solution for the case of non-adjacent nodes is rather more complicated.
 
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Andreas

Joined Jan 26, 2009
82
Yep, so now the OP can copy the answer without learning a damn thing.

Thanks, lava_123. :(
You don't need to know anything these days so long as you know where to look for the answer. I recall this line being a signature from one of the Senior AAC members / moderators. It rings very true. That's the beauty of the internet...apparently (TIC).
 

t_n_k

Joined Mar 6, 2009
5,455
You don't need to know anything these days so long as you know where to look for the answer. I recall this line being a signature from one of the Senior AAC members / moderators. It rings very true. That's the beauty of the internet...apparently (TIC).
Isn't it great! Raising a generation who can't think for themselves.
 

WBahn

Joined Mar 31, 2012
26,398
You don't need to know anything these days so long as you know where to look for the answer. I recall this line being a signature from one of the Senior AAC members / moderators. It rings very true. That's the beauty of the internet...apparently (TIC).
Duly noting the (TIC) annotation, BTW.

I've seen people whip out their smart phones just before a lab to find out who the two famous actors where that kissed in some scene of some sitcom, but then those same people can't figure out how to find out what the red-red-orange-gold bands on a resistor mean despite the fact that it was in the book, in the lab handout, on a poster on the wall less than ten feet from them, on the course website, and on probably several million websites easily reachable from their smart phone!

"Idiocracy" is a documentary-in-waiting.
 

WBahn

Joined Mar 31, 2012
26,398
And it's becoming harder and harder for the people that want to learn to do so because the courses and labs are constantly being dumbed down so now its not a matter of pushing a bit beyond what you are formally exposed to, but needing to take it upon yourself to go WAY beyond. The further out you have to go, the fewer people there are that will have the desire, motivation, determination, and discipline to do so.

Jaime Escalante: Students will rise to the level of your expectations.
Wbahn: And usually no further.
 
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