- #1
danielsmith123123
- 26
- 4
- Homework Statement
- Find the equivalent resistance
- Relevant Equations
- Find the equivalent resistance
Grade 11 physics and equivalent resistance
- Thread starter danielsmith123123
- Start date
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The discussion centers on a calculation error in determining equivalent resistance, where one participant consistently arrives at 45.3 ohms instead of the correct 25.6 ohms. The resolution requires sharing intermediate steps to identify the mistake. Another participant emphasizes the importance of carefully distinguishing between series and parallel components in the problem. Ultimately, the original poster acknowledges a lack of common sense in their approach. The conversation highlights the significance of methodical problem-solving in physics calculations.
- #2
Delta2
Homework Helper
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Can't tell where your mistake is unless you post your intermediate steps that lead you to a (false) answer of 45.3 ohms, but I did it my self and I found it to be 25.6 ohms indeed (25.555 to be more precise).
- #3
danielsmith123123
- 26
- 4
Yes I figured it out I wasn't using my common sense, thank youDelta2 said:
- #4
Delta2
Homework Helper
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Not exactly common sense, but it is kind of easy problem if you carefully identify the components in series and the components in parallel.danielsmith123123 said:
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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