Find the total power developed in the circuit given the readout

[PhDeezNutz]

Homework Statement

Given the reference polarities for current directions along with the current/voltage readout.....find the power developed in the circuit. (See picture below)

Relevant Equations

$p = vi$ if reference current is in direction of decreasing potential ($+$ to $-$)
$p = - vi$ if reference current is in the opposite direction of decreasing potential ($-$ to $+$)

Here is the problem with my power calculations in blue. My reasoning is that + terminal is higher potential, − terminal is lower potential and if the reference goes in the direction of decreasing potential then p=vi and if it goes against the decreasing potential then p=−vi. I've applied this reasoning and I have concluded that this circuit doesn't develop power. Which seems counterintuitive given that the problem statement is "how much power does the circuit develop?".

Again I am concluding that the power extracted is less than power delivered and therefore power is being destroyed and not developed?

Do I have serious misunderstandings of passive sign convention or is this a trick question?

Any help is appreciated in advance.

 

[berkeman]

Sorry, I'm not able to read the image well enough to help yet. Are those all resistors in the drawing? If so, where are any voltages or current inputs coming from?

Can you post a better copy of the schematic please? Thanks.

 

[PhDeezNutz]

Sorry, I'm not able to read the image well enough to help yet. Are those all resistors in the drawing? If so, where are any voltages or current inputs coming from?

Can you post a better copy of the schematic please? Thanks.

Hopefully that is easier to read

I figure for element A $p=-vi$ since reference current is going against the potential drop hence $p_a =- vi = 918W$

I figure for element B $p = vi$ since reference current is going with the potential drop hence $p_b = vi = -810W$

Following this pattern

I get

$p_a = -vi = 918W$
$p_b = vi = -810 W$
$p_c = vi = -12 W$
$p_d = -vi = 400W$
$p_e = -vi = 224W$
$p_f = vi = 1116W$

adding up all the negatives as power extracted and all the positives as power delivered, I get

$p_{extracted} = 822 W$
$p_{delivered} = 2658 W$

Less power is extracted from than delivered to so I am inclined to think power is destroyed and not developed which is counter to the problem statement. I must have some conceptual misunderstandings.

 

[berkeman]

Well if it's a physical circuit, no power is "created" or "destroyed". If it is a closed circuit, the power delivered should equal the power dissipated. I'll try to take a look...

 

[PhDeezNutz]

Well if it's a physical circuit, no power is "created" or "destroyed". If it is a closed circuit, the power delivered should equal the power dissipated. I'll try to take a look...

I think this is a question where conservation of energy is supposed to be violated and we are supposed to interpret it.

 

[berkeman]

Already on "a" I think I get a different result. If the current is flowing from - potential to +potential, that is a power source. If positive power is dissipated in a resistor for example, negative power is the power delivered to the circuit. Why is your answer for "a" positive?

 

[berkeman]

The diagram shows the directions of current and voltage for "a" implying that it is a power source, and they happen to negate *both* of those directions with the data in the table. So it is still a power source element...

 

[PhDeezNutz]

Already on "a" I think I get a different result. If the current is flowing from - potential to +potential, that is a power source. If positive power is dissipated in a resistor for example, negative power is the power delivered to the circuit. Why is your answer for "a" positive?

Sh** I made a mistake. Thanks for pointing it out. I think that one sign error (all the rest are right I think) changes the whole solution.

I then get $1740W$ are delivered and $1740W$ are extracted in total. Therefore no power is developed.

Do you agree? This seems sensible.

I can't believe i messed up on multiplication. I am embarrassed.

 

[berkeman]

Yeah, it looks like only your sign for "a" is wrong.

 

[PhDeezNutz]

I am so embarrassed haha. oh well it happens to the best of us.

 

[berkeman]

I can't believe i messed up on multiplication. I am embarrassed.

No need to be. The problem is being confusing on purpose, but it seems like you have it figured out well now.

 

[berkeman]

BTW, the power balance is always a good way to double-check your work.

 

[PhDeezNutz]

No need to be. The problem is being confusing on purpose, but it seems like you have it figured out well now.

Appreciate it man. I don't say it enough but posters like you and @hutchphd are cornerstones of this great community.

 

[BvU]

negatives as power extracted and all the positives as power delivered

It helped me a great deal to first redraw the pluses and the minuses as well as the arrows in the diagram in such a way that all values in the table are positive.

$\$