Problem about power absorbed & dissipated by voltage sources

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Discussion Overview

The discussion revolves around a problem related to the power absorbed and dissipated by voltage sources, specifically focusing on the application of the passive sign convention and the interpretation of current direction in relation to voltage sources. Participants are examining a specific problem and its solution as presented in a referenced figure.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions why there is not an additional negative sign in computing power for voltage source B, while there is one for voltage source A, given the direction of current flow.
  • Another participant explains that if current is coming out of the positive terminal of a voltage source, the power delivered is calculated as V*I, but in this case, the current is negative, indicating that the source is absorbing power.
  • A participant emphasizes the importance of consistent nomenclature in the diagram, noting that the current direction indicated in the diagram must be adhered to when discussing power calculations.
  • It is clarified that for source B, the power output is calculated using the terminal voltage and the current, which is opposite in direction to the current denoted in the diagram.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of current direction and its impact on power calculations. There is no consensus reached regarding the application of the passive sign convention in this specific context.

Contextual Notes

Participants rely on the definitions of current direction and power calculation methods, which may not be universally agreed upon. The discussion highlights potential ambiguities in the problem setup and the diagram referenced.

s3a
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Homework Statement


Problem in Fig 2-18 and its solution:
prob11.jpg


Homework Equations


Passive Sign Convention, I think.

The Attempt at a Solution


Let's say i is counter-clockwise. Then, it's going from - to + of v_b. Why is there not an additional negative sign in computing p_b, and why is there one in computing p_a (where i being counter-clockwise means the current is negative, despite the image having the current be clockwise)? (I'm trying to follow the work as done in the pdf I linked to.)

Any input would be GREATLY appreciated!
 

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The way I remember learning it: if current is coming out of the positive terminal of a voltage source, then the voltage source is delivering V*I. But in this case I is negative, so it is actually absorbing power.
 
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Hi s3a. Instead of links to off-site storage, we prefer that essential images and other details be attached as an image to your post so that all information gets stored in one location, viz., here on the PF site. I have fixed your post.
 
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s3a said:
Let's say i is counter-clockwise. Then, it's going from - to + of v_b.
The symbol i is already in use on the diagram, so you can't decide to use it for something else, and on the diagram you can see that i is their clockwise current.

As scottdave reminds us, power delivered by a source = terminal voltage x current output
where current output is that emerging from the source's output terminal, the one here marked + .

Sticking with the nomenclature of the diagram, for source B: Pout = vb x (–i)
because that source's output current is opposite in direction to what is already denoted as i.
 
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Sorry about that, and thanks for the help! I get it now!
 

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