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

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SUMMARY

The discussion clarifies the principles of power generation and dissipation in batteries, specifically addressing the differences between a 16.0-V battery and an 8.0-V battery. In the case of the 16.0-V battery, the internal resistance reduces the effective power output, necessitating subtraction from the electromotive force (EMF). Conversely, for the 8.0-V battery, the internal resistance contributes positively to the power absorbed, leading to an addition in calculations. The distinction lies in how internal resistance impacts power dynamics in different battery configurations.

PREREQUISITES
  • Understanding of electromotive force (EMF) in batteries
  • Knowledge of internal resistance in electrical circuits
  • Familiarity with power dissipation and generation concepts
  • Basic principles of circuit analysis
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  • Study the effects of internal resistance on battery performance
  • Learn about power calculations in series and parallel circuits
  • Explore the concept of energy conversion in electrical systems
  • Investigate the role of voltage sources in circuit design
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Students of physics, electrical engineers, and anyone interested in understanding battery performance and circuit analysis.

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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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