# Passive Sign Convention

1. Sep 25, 2013

### Zondrina

1. The problem statement, all variables and given/known data

This is a problem I made myself, I was curious to see if I'm understanding this convention properly.

2. Relevant equations

$P=IV$

3. The attempt at a solution

Okay, from what I understand, the power (-40W) being negative implies that the element is supplying power.

That means that the current is entering the negative end of the terminal of the element and leaving through the positive end if I'm not mistaken.

Since the current is entering from the negative terminal, this means we should use the equation:

$P = -IV$
$-40W = -(-2A)(V)$
$V = -20V$

Now if that happens to be correct ^ I'm curious about something else. Rather than draw another terrible drawing, I'll try to explain.

Lets say the power is now P = 40W and leave the current as -2A. Change the reference points around so that A is positive and B is negative.

This means that the element is absorbing energy. That means that the current is entering the positive end of the terminal of the element and leaving through the negative end.

So the equation to be used would be:

$P = IV$
$40W = (-2A)(V)$
$V = -20V$

Is this also correct?

2. Sep 26, 2013

### CWatters

I don't believe your drawing is consistent. If the power is -40W then yes current should be flowing into the -ve and out of the +ve but that's not what you show.

You have an arrow showing the current going into the -ve but it's marked -2A. If you simplify the diagram by reversing the direction of the arrow and changing the polarity to +2A you can see what you've actually shown is current coming into the +ve.

3. Sep 26, 2013

### Zondrina

Hmm after staring at the drawing in post 1 for awhile, I realize now that the entire drawing itself is inconsistent in general because of the -2A and the direction I've defined it in.

If I had made the current +2A from the start when I drew it ( without changing the direction of the arrow ), it would've made sense without changing anything else in the drawing. The +2A would travel from the negative reference point A to the positive reference B and the element is supplying power. That means that the current enters the -ve and exits the +ve.

So the equation P = -VI would be used resulting in V = 20V.

I suppose making up your own examples isn't always so easy.

Last edited: Sep 26, 2013
4. Sep 27, 2013

### CWatters

One issue is that you label the terminals +ve and -ve which is somewhat definitive . It might have been better to have put a voltage arrow on the load device labelled V. That would allow V to be positive or negative.