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

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The discussion revolves around understanding the differences in power calculations for batteries with internal resistance. In scenario (b), the effective power generated by the battery is reduced by the energy lost due to internal resistance, necessitating a subtraction from the EMF. Conversely, in scenario (c), power is both stored and dissipated as heat within the battery, leading to a different approach in calculations. The key distinction lies in how internal resistance is treated based on the battery's configuration and the flow of current. Overall, understanding these principles is crucial for accurately analyzing battery performance in circuits.
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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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