Physics: Why do we subtract in one case and in other add?

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Discussion Overview

The discussion revolves around the reasoning behind the addition and subtraction of internal resistance in battery circuits, particularly in the context of power generation and losses. Participants explore different scenarios involving batteries with varying internal resistances and how these affect the overall power calculations.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • Some participants suggest that in the case of a battery generating power, losses due to internal resistance should be subtracted from the effective power output.
  • Others argue that in a scenario where power is stored in the battery, the applied power is partially converted to heat, indicating a need to consider both addition and subtraction of internal resistance.
  • One participant proposes that the current remains constant in the circuit, leading to potential drops across internal and external resistances, and questions whether the treatment of internal resistance differs based on the battery's configuration.
  • A later reply challenges the understanding of internal resistance in the context of an 8.0-V battery, asserting that there is indeed internal resistance present that must be accounted for.
  • Participants note that the question regarding power conversion in the 8.0-V battery differs from the previous scenario, emphasizing the need to clarify how power is absorbed by the battery system.

Areas of Agreement / Disagreement

Participants express differing views on how to handle internal resistance in battery circuits, with no consensus reached on the correct approach. Some agree on the need to subtract internal resistance in certain contexts, while others challenge this notion and highlight the complexity of the scenarios presented.

Contextual Notes

There are unresolved assumptions regarding the definitions of effective power and the specific configurations of the batteries being discussed. The discussion also highlights potential misunderstandings about the role of internal resistance in different battery setups.

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Yakin, let's try to do this in small steps. Do you understand you have different situations in each one? Can you explain what happens in each?
 
Hi yakin! :)

In (b) the battery generates power (plus), but the effective power is reduced by the losses in its internal resistance (minus).

In (c) the applied power is partially stored in the battery (plus) and part is converted to heat in the internal resistance (plus).
 
Well i know that current would be constant in the circuit. There would be potential drop across the internal and external resistances. Is it because current flows from positive to negative and in 16.0-V battery, the internal resistance(connected to +ve terminal comes in a way) dissipates some energy therefore we subtract the internal resistance from EMF, and in 8.0-V battery there is no internal resistance after negative terminal so we add the internal resistance. Am i right?
 
yakin said:
Well i know that current would be constant in the circuit. There would be potential drop across the internal and external resistances. Is it because current flows from positive to negative and in 16.0-V battery, the internal resistance(connected to +ve terminal comes in a way) dissipates some energy therefore we subtract the internal resistance from EMF,

Yes.
The effective power of the battery, is what is generated outside the battery.
That means that the power dissipated in the internal resistance must be subtracted from the power that is generated from the ideal voltage source.
and in 8.0-V battery there is no internal resistance after negative terminal so we add the internal resistance. Am i right?

That doesn't sound right. There is an internal resistance.

The question in (c) is slightly different from (b). It asks how much power is converted inside the 8 V battery.
That is, how much power is absorbed by the combination of the voltage source and the internal resistance.
 

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