# Assigning Polarities and Current Directions

1. Jan 28, 2012

### JimiJams

Hey everyone I'm really struggling with assigning polarities and current directions to circuit elements to a circuit. There's an example worked in my book (in fact many) where they'll go from having a circuit and then in the next step polarities and current directions are added to each element without explanation. I wish I could show an image of it but I'll describe the best I can:

Given circuit: There's a circuit with two loops (left and right). The left loop on the furthest left has an independent voltage source of 120V with negative on the bottom and positive on top on the upper part of the left loop is 10 ohm resistor, the right branch of the left loop there's a 50 ohm resistor, and there's an current arrow in the clockwise direction
Right loop: The left branch which is the right branch of the left loop has that previously mentioned 50 ohm resistor obviously, and the right branch of the right loop has an independent current source of 6A in the counterclockwise direction.

I hope you can picture that well. Now in solving this problem the book assigns polarities and current to each element but fails to explain how you know to apply a voltage drop/rise or direction of current so:

The 10 ohm resistor has plus on the left and neg on the right, the 50 ohm resistor has an unknown current in the clockwise direction and the right circuit loop has a plus on top of the loop and negative on the bottom of the loop.

Now I know how to solve this problem. My only question is how on earth did they know what polarities and current directions to assign. If you could break this down simply and not get too conceptual with your explanation I'd greatly appreciate it. The concept I can learn only after I understand a practical, possibly step by step, method to assigning these qualities.

Thanks a ton! Andrew

2. Jan 28, 2012

### Kholdstare

There's is a general rule,
For any source (voltage or current) the (positive valued) current will flow out of the terminal marked + and flow into the terminal marked - .
For any load (resistance, capacitance, inductance and stuffs) the (positive valued) current will flow into the terminal marked + and flow out of the terminal marked - .

3. Jan 28, 2012

### vk6kro

When you are working out problems, sometimes the polarities will be obvious but sometimes you have to make a guess.

If you work it out according to your guess, and you get a negative answer, then you know that the guess was wrong and you have to reverse it.

4. Jan 28, 2012

### JimiJams

In a single loop circuit it's easy to see that current flowing out of the positive terminal of a voltage source will flow into the positive terminals of the circuit elements in the loop. However when dealing with a multiloop circuit with more than one voltage source in the circuit it doesn't work this way. This is what's confusing me. How do I know in this situation what polarities to assign to each element (forget about current and its direction for the moment, just think KVL)?

5. Jan 28, 2012

### vk6kro

It is just the same.

If you assume a current direction, then the polarity you mark on a resistor is positive where the assumed current enters and negative where it leaves the resistor.

If the current turns out to be negative when you solve the simultaneous equations, then the direction you guessed was wrong.. No big deal, you just reverse it.

6. Jan 28, 2012

### JimiJams

So what if a resistor for instance is part of a two loop circuit with voltage sources on each loop. If you follow one loop clockwise it indicates positive to negative for the resistor but if you go clockwise in the other loop the resistor should indicate opposite polarity signs. How do you know which polarity configuration for the resistor is correct?

I mean the book we use, Electric Circuits by Nilsson (proving to be a fantastically expensive paperweight so far), just jumps from a plain circuit to a redrawn version with applied polarities with no explanation whatsoever about how to go about assigning them. And there's some complex circuits in the examples already in the second chapter. Are we just supposed to randomly assign polarities because that doesn't seem correct to me. There must be some method. And the teacher is of no help either. This is my planned major and I'm getting really frustrated.

Like suppose I do randomly apply polarities and there's 10 resistors involved in one loop and adding the voltages is not giving me a sum of zero. It seems rather insane to have to go through and flip each polarity and recalculate one at a time until you get a sum of zero. Is that the only way???

Thank you for your patience to0 by the way, it's my first electrics course so it's very new to me.

7. Jan 28, 2012

### vk6kro

No, you only assume the polarity across a resistor is one way or the other.

After that, that voltage either adds to the other voltages around a loop or it subtracts from them.

If the same resistor is in two loops, it may add to one loop and subtract in the other loop.

A loop does not imply that the current around the loop is constant. It is more related to the voltages around the loop which must sum to zero.

However you never apply random guesses to the whole circuit. It is usually obvious which way some of the currents will flow and there may be one or two currents that could go either way.

8. Jan 29, 2012

### jim hardy

"""If you follow one loop clockwise it indicates positive to negative for the resistor but if you go clockwise in the other loop the resistor should indicate opposite polarity signs. How do you know which polarity configuration for the resistor is correct?""

that gets us all in first EE course.

as Vk6 says - you do your best to assign them , with awareness you could miss a couple of them.
As he also said - the ALGEBRA will take care of it.
Be rigorous when you write your KVL equations:
I was taught 'Imagine yourself IN THE CIRCUIT, walking around it. ALWAYS take the first sign you encounter at a circuit element. "

when you grind through the algebra you will get a negative number for the ones you assigned wrong.
No sweat. You have got the magnitude and polarity. That's victory
If your algebra is kinda uncertain, focus on being very neat and orderly.
this is where neatness pays off.
It's no different than the simultaneous equations you solved in high school algebra.

9. Jan 31, 2012

### Staff: Mentor

Hi. You are misinterpreting the meaning of vk6's use of the word "wrong". It does not carry negative connotations.

You can arbitrarily assign direction of loop currents. There is nothing wrong with doing this; we do it all the time. And if it turns out that the current has a negative sign, then again there is nothing wrong with this. It just means that if you intend connecting an ammeter (of a type for which polarity must be observed) into the circuit, you must take care to connect it with the leads appropriate for that direction of flow.

You don't reverse your arrow on the circuit and then redo all the calculations. One set of calculations is all that's needed. If the question was along the lines of "determine the current flowing from resistor R1 into resistor R2" and you determine it to be negative, then you can state in words your answer as a current of certain magnitude is flowing from R2 into R1. Or you can leave it in the form "current flowing from R1 into R2 is -2.5A." Either is correct.

Before you start thinking circuit analysis sounds chaotic, let me emphasise that you must still adhere to Ohm's Law. For example, once you assign a direction to a loop current, then you no longer have discretion in assigning polarity across elements in that loop. For a resistor, v(t) = R ยท i(t) so the direction of assumed current into a resistor determines one end as being positive (where your current enters) and the other as negative (where your current exits that resistor). You must get that the right way, otherwise you are introducing a negative sign into Ohm's Law.

Where 2 or more loop currents are in opposing directions in an element common to a number of loops, then the same thing applies. Determine the nett current in the direction of the loop current for that loop (it might be I1-I2), and the direction of the loop current has predetermined which end of the resistor must be marked with positive polarity--the end that the loop current enters. Otherwise, your v-i equation fails Ohm's Law.

Often you won't know the real direction of current in every element before you start the analysis, and it doesn't matter, As the saying goes, it all comes out in the wash. You just have to be scrupulously careful in the maths all the way through.