BBC - 'Brain decline' begins at age 27

[ali777]

FWIW, follicle84 explained how I got to 25 ohms exactly the way I looked at it (i.e., 1/4 of 1/4 times 4 branches = 25 percent, or in this case 25 ohms).

Do the same operation twice.... That explanation uses the law of superposition. You have 1/4 coming in, and another 1/4 going out... There are two theoretical current sources, A and B.... One of them is positive and the other one is negative, but because they both flow in the same direction going through the resistor, we add the currents...

A. We have an infinite field. Inject 1A of current to one of the junctions, A. Because the field is infinite, we can safely assume that the the current will be equally divided in all the branches going out of the junction. Therefore the current going out on all the branches is 1/4A.

Who on earth would EVER think that was a "safe" assumption?

B. The same operation, but this time the current is coming out of the field. Let's say we draw 1A from one of the junctions. The current flowing through all the 4 branches coming into the junction is 1/4...

The same issue: who would ever assume that all four branches are passing EXACTLY the same amount of current (1/4 amp)?

Now, we superposition A and B, like in the figure below. Superposition means adding the currents from the two calculations. So, current injected into A goes from A to B and is 1/4A. Likewise, the current drawn from B, goes from A to B and is 1/4A.... They are the same values, but they are two different currents... so the total is 1/4+1/4=1/2

Huh?? If there are four different branches, each one carrying exactly the same 1/4 amp of current (which itself is impossible to believe), the total is 1/4 + 1/4 + 1/4 + 1/4 = 1.

Get it???

No I don't.

TBH, I'm still not sure if this answer is correct, or just stumbles onto the correct answer... I'm just explaining the assumed solution.....

the circuit is simplified to have two current sources (one of them is actually a sink, or negative source), one on each side of the resistor.... Now, think of them one at a time, the law of superposition...

Because it is an infinite field, current injected at one point will flow equally in all 4 directions. Then we superposition the two currents on our resistor. So, we end up having 1/4A going twice through the resistor....

 

[Bryan]

Do the same operation twice.... That explanation uses the law of superposition. You have 1/4 coming in, and another 1/4 going out... There are two theoretical current sources, A and B.... One of them is positive and the other one is negative, but because they both flow in the same direction going through the resistor, we add the currents...

Isn't that a little like the city where I live sending me my water bill at the end of the month and saying "Hey, you used 100 gallons of water this month, but you had 100 gallons coming in from the water main and another 100 gallons gallons going into the sewer, so we're gonna charge you for 200 gallons!"

TBH, I'm still not sure if this answer is correct, or just stumbles onto the correct answer... I'm just explaining the assumed solution.....

I think that guy's simplified solution is ridiculous. It makes no sense to me at all. Like you said yourself, I think somebody just played around with the numbers until he was able to think of some goofy explanation where it matched the expected answer!

the circuit is simplified to have two current sources (one of them is actually a sink, or negative source), one on each side of the resistor.... Now, think of them one at a time, the law of superposition...

Because it is an infinite field, current injected at one point will flow equally in all 4 directions. Then we superposition the two currents on our resistor. So, we end up having 1/4A going twice through the resistor....

The problem with that reasoning is that in one very crucial sense, it isn't an infinite field. The current is applied DIRECTLY ACROSS one of the resistors, so that one resistor is one of the four branches. I don't see any valid reason just to assume that the current going through that ONE MIDDLE BRANCH (one resistor, which is finite) is going to be equal to the current going through each of the other three branches (which comprise a complicated network of an infinite number of resistors).

If it actually turned out to be true that there were equal currents going through all four branches, then the final effective resistance measured across that middle resistor would be 1/4 that of the individual resistors, not 1/2. Since we know through other analysis that the correct answer is (supposedly) 1/2, that proves that the reasoning the guy used in the simplified answer is invalid.

 

[Old Baldy]

Yes Ali, it seems like you are creating more current than ever exists by doubling it in your superposition explanation. Bryan's analogy relative to doubling the actual amount of water someone uses in a measurement period explains my confusion better than I was able to explain it.

Bryan, I simply do not have an explanation on how to measue (or "think" about) the current through the finite resistor versus its other three branches.

That's where I am stumped. I had to stop there.

When you responded with "finite" in a previous post, it totally changed my thinking of this problem. And, at that point, I knew I would probably never be able to explain an answer adequately. Because I don't know the "spread" of the current, other than to assume, in my mind, that it is equal throughout all of the branches. Something that I have no way of proving in any way, shape or form.

I just don't know!!

Edit: I'm not saying 1/2 resistance is incorrect, I just can't get there. (As usual, a brain teaser beats me! )

Further edit: If the answer is 1/2, then I can't understand why it possibly wouldn't be zero. One, (i.e., in the 1/2 answer) is "doubling" the 1/4 resistance and the other (i.e., in the zero answer) is "removing" the 1/4.

Da** it, I can't get past 1/4 resistance, and it took me about two darn pages of members' posts and thinking about it to even get to that answer!!

 

[Bryan]

But OB, do you understand and accept my main point that the current can't be the same through all four branches AND have the answer be 1/2 at the same time??

If the current through all four branches is exactly the same, then the effective resistance is 1/4, not 1/2.

 

[Old Baldy]

But OB, do you understand and accept my main point that the current can't be the same through all four branches AND have the answer be 1/2 at the same time??

If the current through all four branches is exactly the same, then the effective resistance is 1/4, not 1/2.

Not only do I understand your main point, I can't even fathom the 1/2 resistance theory. I just can't get there in my mind.

I don't have any theories on how to measure that one finite resistor and its total effect on the infinite grid, unless I assume equal current flow into all four branches.

 

[Bryan]

Not only do I understand your main point, I can't even fathom the 1/2 resistance theory. I just can't get there in my mind.

I'm not even really arguing FOR or AGAINST the 1/2 theory, I'm just arguing against the childish argument presented by the guy on that Web site. If all four branch currents are the same, then the effective resistance CANNOT be 1/2. It MUST be 1/4. Since the guy himself ended with a result of 1/2 while assuming all four currents are the same, his overall reasoning is invalid.

 

[Old Baldy]

I understand Bryan, believe me I do.

How would you, or if you were challenged to, get to 1/2 resistance? Or do you have suspicions that 1/2 resistance just might be as incorrect as zero resistance?

In other words, how do you measure more or less than 1/4 resistance (i.e., in your mind)?

 

[Bryan]

Well, I don't really have any thoughts about the "1/2" answer, one way or the other. Now that we've dispensed with that silly "simplified" argument, I'll have to tackle the first one (the more complex answer). I'll get back to you on that. Watch this space!

 

[HughJass]

After reading this thread I'm convinced puzzles aren't the best way to ward of dementia