# Help with polarities

#### devoured_elysium

Joined Jun 26, 2008
13
Hello.

I'm having here some difficulties with this exercise: If I consider the flow of current as it is shown in the image attached, I will get the following equation:

3K I - 2KI + 5KI - 12 = 0

Now, If I consider the flow of current the opposite as is shown, I will get this different one:

12 + 5KI + 2KI+3KI = 0

Which one is right? And why? If I was doing my test, in the first one I would get a 2mA intensity current, while in the later one I'd get -1.2mA. I thought that switching the flow of current would only make the signal change, not the magnitude of the current! How is this?

When I'm solving exercises I use the following rule: When I am going through a loop, If I find :
a) a tension source with a + in front of me, it counts as positive, otherwise, negative.
b) an resistence, if I am going in the direction of the current, it counts as positive, otherwise, negative.

What am I doing wrong in my approach? Thanks!

#### JoeJester

Joined Apr 26, 2005
4,361
Draw the circuit with the meters installed.

You will see that Kirchoff's Voltage Law works no matter what direction you choose.

2kI will be opposite polarity of the other two IR drops.

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#### devoured_elysium

Joined Jun 26, 2008
13
Excuse me, I don't understand what you mean by meters.

#### studiot

Joined Nov 9, 2007
4,998
If I consider the flow of current...............What am I doing wrong in my approach?
First and foremost you are not considering current at all in your approach!!

Current analysis (Kirchoff or otherwise) refers to the junctions or nodes.

Voltage analysis refers to the loops.

So let us walk around the loop as you have shown, starting anywhere and returning back to the starting point.

The law you are trying to follow (Kirchoffs voltage law) states that the sum of the voltage drops around any closed loop equals zero. So here we go.

Starting at the +ve terminal of the 12 volt battery and working clockwise around the loop.

Passing through the 3k resistor in the direction of I we have +3kI.
The left hand end of the resistor is more positive than the right hand end so the voltage drops going from left to right (clockwise).
Then passing the second source we are going from negative to positive - a rise in voltage from left to right
we have minus 2kI

Then passing the 5k resistor we have + 5kI, again a clockwise drop in voltage.

There is no voltage change back to the negative terminal of the battery, but we see another voltage rise passing from the negative battery terminal to the positive. Thus we have minus12

In summary

3kI - 2kI + 5kI -12 = 0

If you proceed in the other direction (anticlockwise) around the loop you will see rises across the resistors and drops through the voltage sources so the equation will be

12 - 5kI + 2kI -3kI = 0

Which is the same equation multiplied through by -1

Hope this helps.

#### studiot

Joined Nov 9, 2007
4,998
You do not need and should not use one convention for resistors and a second for voltage sources.

I have reckoned voltage drops as positive and rises as negative but you could also choose the opposite.

#### JoeJester

Joined Apr 26, 2005
4,361
Kirchoff's Voltage Law states the sum of the voltages around the circuits is zero.

Attached is a pic going both directions illustrating the voltmeters in the circuit so you can see the IR drops.

In both cases, the 2kI IR drop (voltage) is of opposite polarity of the other two.

12V - (3kI) + (2kI) + (5kI) = 0

Going from the negative terminal of the battery to the positive terminal

+12 - (+6) +(-4) +(+10) = 0

Going from the positive terminal of the battery to the negative terminal.

-12 - (-6) + (+4) + (-10) = 0

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#### devoured_elysium

Joined Jun 26, 2008
13
You guys rock! Thanks all! Now I got it!

#### devoured_elysium

Joined Jun 26, 2008
13
Well, actually I don't. What studiot says in the first place is what I usually do. What I don't get is what is the method to know if I count resistences as positive or negative.

#### devoured_elysium

Joined Jun 26, 2008
13
My question is: If in the test I'd by some reason put the current flow in that way, how would I know how to solve the problem? Seems to me that you are answering like I did for the clockwise way, and then just saying that for the anti-clockwise one is just the opposite - ok, that is obvious. What I want to know is how to think if I by accident get to have to do this exercise this way. I'm a noob at this  edit: Is it that I can define both resistences and tension sources whichever way I want if I define that resistences have opposite signs of tension sources?

#### studiot

Joined Nov 9, 2007
4,998
You do not need and should not use one convention for resistors and a second for voltage sources
It's worth repeating.

Because you should be considering voltage rises and falls, nothing else. Especially not components.

Work your way around the loop.
As you pass each component the voltage either rises or falls (drops)
Reckon rises as positive and falls as negative
Add (subtract) them up successively until you get back to where you started.

Remember you only have one loop in this question so the current will be the same at all points. When you have several connected loops the current varies in different parts of the loop.

#### devoured_elysium

Joined Jun 26, 2008
13
I still do not know when to consider that a tension source is rising or falling! Nor a resistence!

#### studiot

Joined Nov 9, 2007
4,998
Choose any point in the circuit.

Call the voltage zero
pass through any voltage source from negative to positive, the voltage must rise by definition.
pass through any voltage source from positive to negative the voltage must fall..

Pass through a resistor and the voltage is given by IR.
In a single loop you know which way the current is flowing so you can get the IR sign correct. When the loop is part of a larger mesh of loops you don't know which way the current is flowing so you have to 'guess'. That is you assign a direction to the current. If you are wrong you will end up with a 'negative' current. You will generate a series of simultaneous equations by working your way around each loop, using the same choice of current direction each time you pass through a resistor common to both loops. there will be enough equations to solve for the voltage drops or currents, when you will find out if your direction guesses were correct or not.

#### devoured_elysium

Joined Jun 26, 2008
13
What you mean by IR? Intensity x Resistence, right? I am putting the image here so is easier for you to see it.

Now, I start where you see the I2. I go anti-clockwise. Where I2=0, V=0. As I see a voltage source going from positive to negative, I see it is falling, so now I have

Vt = -12 (as it is falling).

Now, I have a 5K ohm resistance. As I am going in the same direction as the current, by RI, I should have 5K.(I2) which is 5k.I2, positive. Now, I know this is wrong. Why?

#### studiot

Joined Nov 9, 2007
4,998

#### devoured_elysium

Joined Jun 26, 2008
13
Typo, I mean, where you find I2, V=0

#### studiot

Joined Nov 9, 2007
4,998
In your loop, look at the 3k resistor.

The left hand end is connected to the positive terminal of a battery.

Therefore the right hand end will be more negative therfore the voltage will fall going from left to right.
Therefore conventional current will flow from positive to negative - from left to right

Your examiner has made things easy by choosing the direction of I2 to agree with this.

#### studiot

Joined Nov 9, 2007
4,998
You can choose any point and call it zero to start.

I have chosen the positive terminal of a battery because I know from experience this is a good place to start.

Work around a few loops and you will see that this is so.

If you want to see the effect ask yourself what would happen if you removed the second voltage source and the 5k resistor.

Then add back the 5k resistor.

Finally add back the second voltage source.

Also ask yourself what difference would it make if you moved the 3k resistor the other side of the second voltage source?

#### devoured_elysium

Joined Jun 26, 2008
13
Hm. What if the dependent source had the polarity inverted? In the left side of the 3K ohm you would have a positive sign, meaning that at that side the resistence would be high, but in the other side you would have the same situation! How would one untie it?

#### studiot

Joined Nov 9, 2007
4,998
If you connected one battery the other way around you would not know if your current direction guess was correct until you had completed solving the single loop equation.

Of course you could tell, in this simple case, because the larger battery which would predominate, is obvious.

If you put the two batteries together as I suggested it is easy to see this.

Incidentally I think Joe slipped up in his signs. This is very easy to do if you follow component rules. That is why it is best to assign one direction and stick to it until the end, even in simple cases.

I'm off to the land of nod now but I think you willfind working through the variations I've suggested will help make things clearer.

#### studiot

Joined Nov 9, 2007
4,998
I apologise - I should have realised immediately.

Your diagram is incomplete and insoluble as it stands.

You cannot apply kirchoff's laws in the manner you are trying.

The Voltage across the dependent voltage source in the diamond is unknown and not a function of the current I1.

The voltage Va = 2000I1 refers to a voltage generated in another part of your circuit.

Please show us the full circuit and we can help properly with the analysis.

Is this part of a Spice simulation?