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Problem about power absorbed & dissipated by voltage sources

  • Thread starter s3a
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  • #1
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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Answers and Replies

  • #2
scottdave
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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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  • #3
NascentOxygen
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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.
 
  • #4
NascentOxygen
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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.
 
  • #5
s3a
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Sorry about that, and thanks for the help! I get it now!
 

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