Infinite Resistor Grid Puzzle

In this video, we show a simple technique to solve the "Equivalent Resistance of an Infinite Grid Puzzle." Please submit your suggestions for similar articles to Analog Arts (www.analogarts.com).

[@turb0flat437]

For the one ohm infinite grid, R across diagonal is 2/pi. Solution involves fairly difficult math. But it is very reminiscent of Buffon's needle, isn't it?

[@m000nd]

I created a series of videos on this topic. It starts with the basics but also covers the much harder diagonal case where the resistances is the 2/pi: kzbin.info/aero/PLoGRr8ff1uXESrWh6z0BNTYpc4Y-hlBOm

[@colinrebello7703]

Imagine there was a single current source say "i" where one end(current enters) was connected to one node say A and the other (current exits) was connected to node B which is adjacent to A. Because B is closest to A one would expect most of the current to flow direct from A to B ie the path of least resistance. So current entering Node A would divide into 4 unequal parts and NOT 4 equal parts. Similarly current exiting B is not 1/4 i though each branch but actually 4 unequal currents. The flaw is assuming the current divides equally into 4 parts. It cannot because we have a single current source connected from A to B which is equivalent to your having 2 current sources placed in exactly the same way and because A is closest B Route-wise the 4 divisions of current cannot be equal

[@academia7768]

+Colin Rebello Thank you Colin for your comment. I believe, in your comment and assumption, the key characteristic of this network is forgotten. Please recall that the network is called "Infinite Resistor Grid ." Therefore, the answer to your question, "where is that?," is that it is at the boundary of the network. Clearly, if we take away this feature of the network, we lose the symmetric behavior of the network and this explanation is longer valid. Please also be reminded that in the network, closeness is irrelevant. What matters is the impedance and if a current is divided into equal impedance paths, it would get divided equally.

[@Aadhyacedt]

Excellent Solution. Between A&C the equivalent resistance is 2/3, is this correct?

[@academia7768]

Very close Raja, excellent! The solution is somewhat involved. R-AC ends up to be 2/π . For further reading, please visit: sites.google.com/site/resistorgrid/

[@jaimelannister2925]

The 0.25A current which flows in is taken to be different than the 0.25A current flowing out. Why is that so? Why isn't the 0.25A entering taken as the 0.25 A, leaving. While one may argue on the basis of the principle of superposition, how come the net current is then not taken to be 2A, superimposing the 1A entering and the 1A leaving?

[@academia7768]

+Jaime Lannister Excellent question Jaime! Thank you. Let's review the solution again. 1. One terminal of the first current source, the one pumping current to the node, is connected to network at infinity. Therefore, since the device is symmetric, 1/4 th of the current flows through each resistor at the connected node of resistor "R". 2. The second current source is also connected at the infinity to the network. The other terminal of this current source is connected to the other side of the resistor "R". Again 1/4th of this current is drawn out from the connected node. 3. When both current sources are on, note that they are connected together at infinity, 1/2 the current must flow through the resistor. Then the rest of the current must flow through the rest of the circuit. Therefore, the equivalent resistance of the network across each resistor must be equal to R/2.

[@jaimelannister2925]

+Academia I am sorry, but your reply just recaptures what the video says. And it is still confusing.

[@jaimelannister2925]

+Academia I am sorry, but your reply just recaptures what the video says. And it is still confusing.

[@louisck7425]

+Jaime Lannister Hi Jaime. If you are still confused and need extended explanation, you may want to visit this page: www.mathpages.com/home/kmath668/kmath668.htm

[@jaimelannister2925]

+Louis CK Never mind. I understood it by thinking of current as electrons moving under force, and the battery as being the causing factor of that force. Basically, the forces get added up.

[@saicharan2159]

Please Clear me this doubt Academia...here you have told that the current through the resistor across A,B terminals is one half of the total current.....But here the total current is 2A But not 1A when both Conditions were superimposed.... hence,the current through resistance between A,B is not half of 2A(total current)....please give me reply fast...

[@academia7768]

Thank you Sai for the good question. Each current source injects 1/4 of its current through the network. The first current source (1A) pumps in 1/4A through the resistor at AB. Similarly, the second current source (1A)pumps out the same amount that 1/4A. Bear in mind, these current sources are tied at infinity. Now , if we connect a single 1A current source in parallel with R at AB, we would have 1/2A= 1/4A+1/4A through R at AB and the the remainder of the current (1/2A) flows through what is in parallel with the resistor . Therefore, the equivalent resistance must R/2.

[@saicharan2159]

Academia ...Friend...ThanK You For Giving Me Reply...connected at infinity means...they have effect on the circuit or not..? At the Same time what is the effect of new 1A current....?

[@academia7768]

"Infinity" is a concept, Sai. In this puzzle, it indicates the circuit is symmetric at each node. Otherwise, injecting current to a node or pumping out current from a node would not get equally divided through the resistors connected to that node. When we say the other terminal of the source is connected to the circuit at infinity, we simply want to maintain this property.

[@saicharan2159]

Academia Thank You Yarr...Nice Thing...Yarr..Please Send Me The Link Where You Have Find This Concept...Either Here Or Through Mail

[@saicharan2159]

Academia By The Way...You Good Name Please...

[@manchurianable]

how wud we solve if we need to take the equivalent res across A and C

[@academia7768]

The solution is a bit more complicated. There are multiple web pages covering this topic in detail, like: www.mbeckler.org/resistor_grid/ or www.mathpages.com/home/kmath668/kmath668.htm.

[@yuppiehi]

Answer would be 2/pi * R, or 0.6366 * R.

[@reetupareek531]

CAN yoy help me find eq. resistance across a side of a hexagon of an infinite mesh of hexagons

[@academia7768]

Here is a link: papuaweb.org/dlib/up/saa/saa-2004.pdf As you see it is a bit more iinvolved.

[@knvcsg1839]

But, really I didn't know about this superposition principle.

[@Heshamuddin-h1h]

2r/ 3 ans for your last question