Resistance of this circuit containing a Circular section

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The discussion focuses on calculating the resistance of a circuit with a circular section and straight paths. Participants clarify that there are two parallel paths from points C to D, each with a resistance of 1/6 ohm. When these two resistors are connected in parallel, the total resistance is calculated to be 1/12 ohm. The conversation highlights the difficulty in visualizing the circuit layout and understanding the lumped model representation. Ultimately, the participants agree on the resistance values and their implications for the circuit analysis.
Aristarchus_
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Homework Statement
A piece of wire has resistance R. The wire is cut into three parts of equal length and connected
together as shown in the figure. What will be the resistance between A and B?
Relevant Equations
Solution is 1/12 + 2/3 = 3/4ohm.
1659873059820.png

I understand that the two separate parts make 2/3, but where is 1/12 ohm coming from?
 
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How did R turn into ##1\Omega##?
Call the two points where the straight sections meet the circle C, D. There are two parallel paths (electrically speaking) from C to D. What is the resistance of each?
 
haruspex said:
How did R turn into ##1\Omega##?
Call the two points where the straight sections meet the circle C, D. There are two parallel paths (electrically speaking) from C to D. What is the resistance of each?
1659873869220.png

But would we calculate then resitance in parallel(1/6 and 1/6)?
 
Yes exactly, what total resistance you get if you connect two resistors of 1/6 in parallel?
 
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Delta2 said:
Yes exactly, what total resistance you get if you connect two resistors of 1/6 in parallel?
right! Hmm... However, it is hard to picture the circuit in the way you described it. I could not come up with that sketch on my own...
 
Aristarchus_ said:
right! Hmm... However, it is hard to picture the circuit in the way you described it. I could not come up with that sketch on my own...
Yes agreed from the image of the wire with the circle in the middle your mind just doesn't think the corresponding lumped model of the two resistors of 1/3 in series with the two resistors of 1/6 which are in parallel.
 
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Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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