Graphing the Power dissipated in a resistor

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The discussion focuses on graphing power dissipation in a series circuit with a constant voltage source and a variable resistor, as well as a constant current source with a variable resistor. The approach involves using the power equation to define constants for each series, allowing the relationship between resistance (R) and power (P) to be established. There is a query about the validity of this method, prompting a suggestion to verify the approach. The conversation emphasizes the importance of confirming the correctness of the method used. Overall, the participants are engaged in ensuring the accuracy of the graphing technique for power dissipation.
guyvsdcsniper
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Homework Statement
Graph the power dissipated in each resistor as a function of its resistance.
Relevant Equations
P=VI
I am solving #5 in the attached image.

So I am graphing the power dissipated by a series circuit which : (a) Contains a constant voltage source and a variable resistor, R (b) Contains a constant current source and a variable resistor.

It makes sense to me to just use the power equation that allows me to define the constant from each series and the power will be determined by what relationship R shares with P in said equation.

Just wondering if I have the right approach.

IMG_098E9BFE4AFE-1.jpeg
 
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Looks good to me. Why are you doubting your method? How could you verify yourself that you are correct?
 
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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