- #36
Sat-P
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Can someone just solve this question and tell me the answer?
Equivalent resistance of this 6-sided circuit
- Thread starter Sat-P
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In summary, the equivalent resistance of a 6-sided circuit can be determined by analyzing the configuration of resistors, whether they are in series or parallel. By applying the appropriate formulas for series (R_total = R1 + R2 + ...) and parallel (1/R_total = 1/R1 + 1/R2 + ...) combinations, one can simplify the circuit step by step to find the total resistance between two points. The final value depends on the specific arrangement and values of the resistors in the circuit.
- #37
SammyS
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Yes, essentially.Sat-P said:
More specifically:
By considering symmetry, we expect that current through one of the other three resistors (those being 3-0, 4-0, or 5-0) connected to node 0 will have the same current as the resistor 3-0 has, Furthermore, if the current through resistor 3-0 flows into node 0, the the current through the resistor in question, must flow out from node 0.
- #38
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Yes! Do you see how that follows from symmetry?Sat-P said:
And what about the remaining radial pair, 3-0 and 0-4?
- #39
SammyS
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Oh my!haruspex said:
I see that we've had a bit of an editing job in Post# 32 by OP.
I never saw that reference to 0-5 .
- #40
DaveE
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Perhaps you should review the Homework forum guidelines.Sat-P said:
Yes, I could solve this problem. But, what good is that for you? Would you be able to solve it next year by yourself? The only purpose of problems like this is education. They virtually never appear IRL.
- #41
BvU
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A good idea ! (the naming in #1 is rather confusing)haruspex said:
And I have this mental deformation that wants to redraw circuits with components at right angles, like in school books:
Does that help you analyze your circuit
?##\ ##
- #42
Sat-P
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Earlier, I thought that current through resistor a0 and 0b will be the same but I was wrong, when I interpreted this mathematically I got that the current through the resistor 0b will be more than that through the resistor a0, and then I thought that current through f0 and c0 might be mirror images after passing a plain in between ab and ed, but here it is; the magnitude of current passing through them was'nt same so I couldn't just take them out, there might be another reason that the current flowing through resistor 0b has contributions of current entering from the radial resistors, so I just couldn't take this one out. Now, the current passing through resistor e0 and d0 looks symmetric, but it is not, again because of the magnitude of the current passing through it. Now, I have an idea, don't know whether it is crazy or something, but I gues if I flod the grid from cf like ab coincides ed, but I now it won't work because in the symmetry the terminals must be symmetric, so any hints what I should do next
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- #43
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It must be, by symmetry.Sat-P said:
Please show the details.Sat-P said:
- #44
Sat-P
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I am sorry for the misinterpretation, but the current through resistor 0b will be more i.e 0.65 i (approx) and current through resistor a0 is 0.334 I (approx). You can see the results in the image above. I've edited the above post.
- #45
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The image is too blurry to read with any confidence, but it looks to me that you are assuming the currents along the three resistors connected to a are equal. That's not how electricity works.Sat-P said:
Say all the resistors are 1 Ohm and the voltages are +1 at a and -1 at b.
The voltage at 0 must be 0, so the current along ab is 2A and along a0 and 0b it is 1A.
- #46
Sat-P
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Can you please clarify why the current through the resistors connected to a should not be equal?haruspex said:
- #47
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See my edit to my previous post.Sat-P said:
- #48
Sat-P
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How can you say that the current through ab is 2 Ampereharuspex said:
- #49
BvU
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It's the law (Ohm's law, to be specific).Sat-P said:
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- #50
Sat-P
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Got itBvU said:It's the law (Ohm's law, to be specific).
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- #51
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So can you solve it now using the splitting up of node 0?Sat-P said:
- #52
Sat-P
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Even after I split the node 0ab , the other radial resistors are causing me trouble, if I can get a way to simplify them then it will be all good
- #53
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In post #32 you recognised (guessed?) that, by symmetry, the current along 2-0 is the same as along 0-5. That allows you to split ##0_{2345}## into two nodes, ##0_{25}, 0_{34}##.Sat-P said:
- #54
BvU
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##\ ##
- #55
Sat-P
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Sir, I've tried both the methods you gave
The answers I am getting are different, please check is there something I did wrong or misinterpreted what you said
The answers I am getting are different, please check is there something I did wrong or misinterpreted what you said
- #56
BvU
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What are we looking at ? (with a pain from craning our neck) ?
I can distinguish you replace the rightmost three with ##{2\over 3} R##, so the rightmost five give ##{8\over 3} R##. OK.
But why ##R/2## ?
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I can distinguish you replace the rightmost three with ##{2\over 3} R##, so the rightmost five give ##{8\over 3} R##. OK.
But why ##R/2## ?
##\ ##
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- #57
Sat-P
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Oh, I just saw the mistake. Let me do it again
- #58
Sat-P
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After solving this one I got 11R/20 so is it the answer, did I do it right
- #59
DaveE
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Agrees with mine, but I didn't double check.Sat-P said:
Checking your own work is a key skill in the physical sciences. Out in the real world they don't want to pay 2 people to do one job, and they don't want to pay one person to do the job incorrectly.
- #60
BvU
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The answer isn't really the important element. Whther you did it right can only be confirmed if you post your work.Sat-P said:
Free tip: work it out neatly ...