Passive Sign Convention??

Thread Starter

Sgt.Incontro

Joined Dec 5, 2012
50
Hi all,

Doing revision (Eng student), and whilst I have grasped many of the harder topics, passive sign convention has confused me a little it seems.

For example, I (most likely wrongly) disagree with the result for Example 1.3 of the link below.

My reasoning is that the current reference direction is NOT in the same direction as the voltage drop, therefore it must be P= -iv (i.e power is being extracted from the element?)

I assume that this (my reasoning) is wrong?

Can anyone explain this?

Thanks. :)

Link: http://en.wikiversity.org/wiki/Passive_sign_convention

((First post, so cut me some slack if I have done something wrong, thanks.))
 

Wendy

Joined Mar 24, 2008
23,421
Welcome to AAC!

While what you say is technically correct, in most cases when we design something we are only interested in the absolute values, the direction of current flow is basically already a known quantity. Things like resistors don't care, put too much current through them and you will have smoke, possibly a red glow.

We try to cut everyone slack here, this is a no flame zone. It is one of my several jobs to keep it that way (and ban spammers). There are no stupid questions.
 

Thread Starter

Sgt.Incontro

Joined Dec 5, 2012
50
Welcome to AAC!

While what you say is technically correct, in most cases when we design something we are only interested in the absolute values, the direction of current flow is basically already a known quantity. Things like resistors don't care, put too much current through them and you will have smoke, possibly a red glow.

We try to cut everyone slack here, this is a no flame zone. It is one of my several jobs to keep it that way (and ban spammers). There are no stupid questions.
Thank you Bill :)

So are you implying that the given source is indeed incorrect, and I am correct? (I.e the Wikiversity link)

((I know they have shown the components as resistors, but strictly speaking, they should be devices, because power can never be extracted from a resistor.))

Anyway, putting that aside, I really want to be certain that my view on this subject is correct, come exam time, so if you could clarify your view on the given example slightly more, I would really appreciate it.

Thanks once again.
 

MrChips

Joined Oct 2, 2009
30,806
There are a number of flaws on that web page and you would be well advised to ignore that concept.

Resistance in a "normal" resistor is a scalar quantity and does not have a sign.
There are some semiconductor devices, such as a tunnel diode, that do exhibit negative resistance but this is not what the web site is referring to.

Power is \(P = \frac{V^2}{R}\)

Hence power dissipated by a resistor is always positive, independent of the direction of the current.
 

ErnieM

Joined Apr 24, 2011
8,377
For example, I (most likely wrongly) disagree with the result for Example 1.3 of the link below.

My reasoning is that the current reference direction is NOT in the same direction as the voltage drop, therefore it must be P= -iv (i.e power is being extracted from the element?)

I assume that this (my reasoning) is wrong?
I see nothing wrong in your reasoning. Figure 1.3 is drawn incorrectly.

As 1.3 is drawn, your reasoning is completely correct, and that resistor is supplying power. Thus it ain't no resistor!

And welcome to the forums!
 

crutschow

Joined Mar 14, 2008
34,432
A power value is usually given without a sign. If current is flowing out the positive voltage side of a component that has a voltage difference across its terminals, then that element is supplying power (e.g. a battery). If current is flowing into the positive side of a component that has a voltage difference across its terminals, then that element is absorbing (or dissipating) power (e.g. a resistor).
 

WBahn

Joined Mar 31, 2012
30,057
Hi all,

Doing revision (Eng student), and whilst I have grasped many of the harder topics, passive sign convention has confused me a little it seems.

For example, I (most likely wrongly) disagree with the result for Example 1.3 of the link below.

My reasoning is that the current reference direction is NOT in the same direction as the voltage drop, therefore it must be P= -iv (i.e power is being extracted from the element?)

I assume that this (my reasoning) is wrong?

Can anyone explain this?

Thanks. :)

Link: http://en.wikiversity.org/wiki/Passive_sign_convention

((First post, so cut me some slack if I have done something wrong, thanks.))
I would run away from that site at full speed. There are a number of statements there that just make no sense or are flat out wrong, including Ex 1.3.

The site makes the notion of the passive sign convention much more confusing than it needs to be.
 
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