# Homework Help: Passive sign convention

1. Aug 28, 2014

### orangeincup

Problem statement
I'm trying to understand how passive sign convention works, and how to read the voltages/currents direction in a circuit.
For different set of voltages and currents I'm trying to calculate power. I'm using the picture below as a reference.
a) i=10A v=125V
b) i=5a v=-240V
c) i=-12A v=480V
d) -25A v=-660V

Equations
p=vi or p=-vi

Attempt at solution
I don't understand how to get the polarities.
I'm having a hard time understanding how I can use the passive sign convention. My definition says if the current is in the direction of the reference voltage drop across the terminals, p=iv.

When I look at the picture below I'm confused how I could tell if there's a voltage drop or not? Whether the current arrow is pointing left or right, wouldn't it be pointing into the positive terminal regardless?

I have the solution from the book below but I don't understand. If you're standing at A, looking at B, why is the current going into the negative terminal? Also, if the current arrow was pointing right, does that mean the current is now going to the positive terminal?

Solution from book
Assume we are standing at box A looking toward box B. Then, using the
passive sign convention p = −vi, since the current i is flowing into the −
terminal of the voltage v. Now we just substitute the values for v and i into
the equation for power. Remember that if the power is positive, B is absorbing
power, so the power must be flowing from A to B. If the power is negative, B
is generating power so the power must be flowing from B to A.

Better picture: https://www.physicsforums.com/attachment.php?attachmentid=69908&d=1400523028

2. Aug 28, 2014

### Zondrina

Consider the first case with $i = 10 A$ flowing from $B$ to $A$ and the known voltage.

The current is in the direction of the reference voltage drop as your book says. Knowing this calculate the power.

Edit: Huge typo there my bad.

Last edited: Aug 28, 2014
3. Aug 28, 2014

### orangeincup

How do I know that's the voltage drop though? My problem is in every example I find the voltage is labelled parallel to the conducting wire. In this example, the voltage polarities are labelled vertically so I have no idea how to read it.

4. Aug 28, 2014

### Zondrina

The current is going from a higher potential to a lower one. So the power must be negative.

5. Aug 28, 2014

### orangeincup

If the current was pointing right, wouldn't this also be true?

6. Aug 28, 2014

### Zondrina

Is the current in the direction of the reference voltage or not? If you are standing at $A$ looking at $B$, then the current is in the opposite direction to the reference voltage.

If the current was going from $A$ to $B$, then it would be $+iV$.

7. Aug 28, 2014

### orangeincup

What do you mean by reference voltage? It shows the voltage on the picture but I've said I don't know what it means since it's perpendicular. To me, it looks like left OR right both go from positive to negative.

8. Aug 28, 2014

### Zondrina

The solution has assumed that you are standing at box $A$ looking toward box $B$. So the current is coming from $B$ towards $A$ while the reference goes from $A$ to $B$.

9. Aug 28, 2014

### orangeincup

So, would my reference assume the voltage on either side look like this?

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10. Aug 29, 2014

### ehild

No. The horizontal lines represent wires, and there is no potential drop across an ideal wire. (The very small potential drop can be ignored.) So the upper terminals of the blocks are at the same potential and it holds to the lower terminals, too.
The blocks represent some electronic devices. For simplicity, you can take one a resistor, the other a battery. You have to decide, which is the battery and which is the resistor.
The current flows through the whole circuit. Continue it inside the blocks.
You know that the current through a resistor flows from the positive potential to the negative one. Look at the picture: the current flows from top to bottom in the left device and the potential is positive at the top, negative at the bottom. The current flows from positive to negative: That device behaves like a resistor. Is the resistor active or passive element?

The arrow of the current and the +,- signs for the voltage mean the reference, what is taken as positive. If the current is given a negative value, the arrow will change to the opposite direction. If the voltage is given a negative value, the + sign changes to - and the - sign changes to +.

Draw the currents and voltages for all cases, and find the block where the current inside flows from + to -: that is the passive element.

ehild

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11. Aug 29, 2014

### milesyoung

Just for a time, forget about the solution given and trying to figure out what the polarities are and in which direction current is flowing.

Now look at this image (ignore the big arrow with p in it):

In this image, there is a direction assigned for the current $i$ and a polarity assigned for the voltage $v$. These are called reference directions/polarities, they're very important and you shouldn't confuse them with what the actual direction of current and polarity of voltage is in the circuit. We assign these references so we know what a positive or negative value of $i$ and $v$ mean.

Now, this is the important part and the point I want to make: The passive sign convention has nothing to do with what the actual direction of current and polarity of voltage is. It's all about the reference directions/polarities, which you can assign in any way you please.

The passive sign convention simply states that if the reference direction of current $i$ is into the positive terminal of the reference polarity of voltage $v$, then the power is given by $p = v i$, and it's positive when power is flowing into the element and negative when power is flowing out from it.

If you now go back to your A and B elements, you can see that references have already been assigned. These follow the passive sign convention for the A element, so what's left to do is just to calculate $p = v i$ for the values you've been given, which will tell you if the power is flowing into the A element or out from it.

Edit:
Reading your question again, I realize that ehild's attachment was probably just what you were looking for, i.e. this:

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12. Aug 29, 2014

### orangeincup

So, if I assume A is a battery the current should normally be going from the positive to the negative if it's delivering power, and p=-iv?

As for the resistor, I thought resistors were always passive elements? What does it mean when the current is flowing into the battery the wrong way? Wouldn't it mean it's being charged?

13. Aug 29, 2014

### milesyoung

If A is supplying power then when you calculate $p = v i$ it will turn out to be negative. Stick to your reference directions/polarities and what the passive sign convention tells you.

A resistor is a passive element, which means it cannot ever supply power. If A was a resistor, $p = v i$ would never be negative.

14. Aug 29, 2014

### ehild

The current flows out from the positive terminal of the battery and flows in at the negative terminal. But inside the battery, the current flows from minus to plus.
Yes, resistors are passive elements, the current flows through them from + to -.
If the current flows into the battery on the wrong way, that is inward at the plus pole and out at the - pole, then the battery is charged, (when it is a re-chargeable one) or just dissipates power, like a passive element.

ehild