Percent Change Of current when the switch is closed

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The discussion centers on calculating the percent change in current when a switch is closed in a circuit with a 1.50 V battery and internal resistance. Various calculations were attempted, yielding results of -3.86%, 4.02%, and others, all deemed incorrect. The issue arises from the practical observation that bulb A dims slightly when the switch is closed, despite theoretical expectations. The internal resistance of the battery and the resistance of the bulb are critical factors in understanding this behavior. A circuit diagram is requested to clarify the problem further.
ChrisWM
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Homework Statement
By what percent does the current through A change when the switch is closed?

I found the value of the current to be .174 A when the switch is closed and value of the current when the switch is open to be .181 A.
Relevant Equations
final-initial/initial
(.174A-.181A)/.181A=-3.86% but it says it wrong, and I did (.181A-.174A)/.174A =4.02% but this was wrong too. I've tried 3.87%,3.86%,-3.87%,-3.67%,4.02%, and -4.02% but all were wrong. I'm really not sure what to do here.
 
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Is there a circuit diagram?
 
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1590556060734.png
Here you go!
 
Here is the actual problem; For an ideal battery (r = 0 Ω), closing the switch in (Figure 1)does not affect the brightness of bulb A. In practice, bulb A dims just a little when the switch closes. To see why, assume that the 1.50 V battery has an internal resistance r = 0.30 Ω and that the resistance of a glowing bulb is R = 8.00 Ω.
 

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cnh1995 said:
Is there a circuit diagram?
 

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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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