Problem about power absorbed & dissipated by voltage sources

In summary, the conversation discusses the use of passive sign convention in solving a problem involving a voltage source and its output current. The main question is why there is not an additional negative sign in computing power for source B, while there is one for source A. The conversation concludes that this is because the output current for source B is opposite in direction to the already denoted current on the diagram.
  • #1
s3a
818
8

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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  • #2
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
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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  • #4
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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  • #5
Sorry about that, and thanks for the help! I get it now!
 

1. What is power absorbed by a voltage source?

Power absorbed by a voltage source is the amount of energy that is taken in by the source to maintain a constant voltage output. It is the rate at which the voltage source converts energy into electricity.

2. How is power absorbed by a voltage source calculated?

The power absorbed by a voltage source can be calculated by multiplying the voltage output of the source by the current flowing through it, using the equation P = VI, where P is power, V is voltage, and I is current. This calculation assumes that the voltage source is ideal and has no internal resistance.

3. What is power dissipated by a voltage source?

Power dissipated by a voltage source is the amount of energy that is lost or converted into heat within the source itself. It is the difference between the total power output of the source and the power absorbed by the source.

4. How is power dissipated by a voltage source calculated?

The power dissipated by a voltage source can be calculated by subtracting the power absorbed by the source from the total power output. This can also be calculated by multiplying the current flowing through the source by the voltage drop across the internal resistance of the source, using the equation P = I2R, where P is power, I is current, and R is the internal resistance.

5. What is the significance of understanding power absorbed and dissipated by voltage sources?

Understanding power absorbed and dissipated by voltage sources is important in the design and operation of electrical systems. It allows for efficient use of energy and helps prevent overheating and damage to equipment. It also helps identify potential issues with voltage sources and ensures their proper functioning.

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