Is Work Done by a Voltage Source Positive or Negative When Current Opposes EMF?

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Discussion Overview

The discussion revolves around the nature of work done by a voltage source when the current flows in opposition to the electromotive force (EMF). Participants explore the implications of this scenario in terms of energy transfer and power calculations, touching on both theoretical and conceptual aspects.

Discussion Character

  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • Some participants assert that the work done by the voltage source is negative when the current opposes the EMF.
  • Others argue that when current enters the positive terminal of the voltage source, the source gains energy, suggesting that its power is positive and questioning whether the work done should also be positive.
  • One participant clarifies that while the voltage source gains energy, the work done by the source itself remains negative, emphasizing the distinction between work done on the source and work done by the source.

Areas of Agreement / Disagreement

Participants express differing views on whether the work done by the voltage source is positive or negative, indicating a lack of consensus on the topic.

Contextual Notes

Participants reference concepts such as energy gain, power as a derivative of energy, and the integral of power in their arguments, but the discussion does not resolve the underlying assumptions or definitions related to work and energy in this context.

hackhard
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Is the work done by a constant voltage source positive or negative when current is opposite to emf?
 
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Dale said:
Negative
when current enters positive terminal of voltage source, voltage source charges up , so it gains energy , its power(energy derivative) is positive
shoudnt work done (power integral)be positive?
 
Work done by the voltage source is negative because it receives energy.
Work done by whatever current source you have, is positive.
 
hackhard said:
when current enters positive terminal of voltage source, voltage source charges up , so it gains energy , its power(energy derivative) is positive
shoudnt work done (power integral)be positive?
Work done on the source is positive. You asked about work done by the source, which is negative.
 

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