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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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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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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