Analyzing Current Flow in Parallel and Series Battery Configurations

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SUMMARY

The discussion focuses on analyzing the current flow in two identical batteries with an electromotive force (emf) of 12 V and an internal resistance of 0.200 Ω, connected to an external resistance R, either in parallel or in series. When R is set to 2.00 times the internal resistance (R = 2.00r), the current i can be determined using the equation ε = ir + (2i)R. Participants emphasize the importance of understanding the symmetry in parallel configurations and suggest drawing the circuit for clarity.

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  • Understanding of Ohm's Law
  • Knowledge of circuit configurations (series and parallel)
  • Familiarity with electromotive force (emf) and internal resistance
  • Ability to manipulate algebraic equations
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  • Calculate current flow in series and parallel battery configurations
  • Explore the effects of varying external resistance on circuit performance
  • Learn about Kirchhoff's laws for circuit analysis
  • Investigate battery capacity and discharge characteristics
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Homework Statement


Two identical batteries of emf ε = 12 V and internal resistance r = 0.200 Ω are to be connected to an external resistance R, either in parallel or in series. If R = 2.00r Ω, what is the current i in the external resistance in the (a) parallel and (b) in the series arrangements?


Homework Equations





The Attempt at a Solution



By symmetry, when the two batteries are connected in parallel the current i going
through either one is the same. So from ε = ir + (2i)R with r = 0.200 Ω and R = 2.00r, we
get ... etc.

I have no idea what the above means. What is going on?

While I appreciate the help, can you please give more than a sentence or two. I have found shortness = unhelpful response (not necessarily true, but there does seem to be a correlation).
 
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auk411 said:

Homework Statement


Two identical batteries of emf ε = 12 V and internal resistance r = 0.200 Ω are to be connected to an external resistance R, either in parallel or in series. If R = 2.00r Ω, what is the current i in the external resistance in the (a) parallel and (b) in the series arrangements?


Homework Equations





The Attempt at a Solution



By symmetry, when the two batteries are connected in parallel the current i going
through either one is the same. So from ε = ir + (2i)R with r = 0.200 Ω and R = 2.00r, we
get ... etc.

I have no idea what the above means. What is going on?

While I appreciate the help, can you please give more than a sentence or two. I have found shortness = unhelpful response (not necessarily true, but there does seem to be a correlation).
Sorry, but this will be rather short.

Where did you get the following information, if you don't understand it? (I think that's what you indicated.)
"By symmetry, when the two batteries are connected in parallel the current i going
through either one is the same. So from ε = ir + (2i)R with r = 0.200 Ω and R = 2.00r, we
get ... etc."​
It looks like you could plug-in the values for r, R, and ε, to find the current, i .
 
The information would appear to be a quote from a solution, perhaps from a textbook example problem.

It might help if auk411 were to draw the circuit in question and label the currents.
 
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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