# Homework Help: Confusion in using KVL (assigning polarities) in a problem

1. Apr 16, 2012

### ViolentCorpse

This is actually a solved example of Thevenin's principle in my textbook and I understand all of it except the part that I'm attaching to this post below; the part where KVL (Kirchoff's Voltage Law) is applied.

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

Find Eth in the circuit shown in the attached figure.

(This is not the actual question, it's just the part that I'm having problems with)

3. The attempt at a solution

The way I've learned it, conventional current flows from + to - in resistors. But in the selected loop where Kirchoff's Voltage Law is applied, current is flowing in the 60Ω resistor from - to +. It's the polarities chosen for the resistors that is confusing me.
The way I was personally doing the problem was like this:
I started from the positive end of Eth and moved counter-clockwise in the loop, assigning the resistors their polarities according to the direction of the current that I was assuming (passing from + to - in both the resistors, by this way). By this procedure, I get:

12V + 6V = Eth
Eth = 18V

But the correct answer and the equation formed in the book is:

6V + Eth = 12V
Eth = 6V

which makes sense to me, but only after seeing the polarities assigned to the resistors. My procedure yields the same result, but only if I keep the polarities as they are shown in the figure intact. My question is, why are the polarities in the example assigned the way they are and what is wrong with the way I am doing it?

Thanks a ton guys!

P.S: If it's any help, the voltages for the two resistors were found using voltage-divider rule.

#### Attached Files:

• ###### figure.JPG
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2. Apr 16, 2012

### RoshanBBQ

When someone says the voltage is positive, they mean you take it at face value, and that face value CAN be negative. Let me try to explain with an analogy. If we are doing a work calculation, we can call positive work the work that is done by the person pushing the box. So let's say I know he is exerting some force f and the distance traveled is d. I can call the "positive work" f*d (In this e.g., the movement is in the same direction of the box). However, if I then say f is negative while d is positive, this "positive work" is actually negative, indicating he is doing negative work, thereby absorbing work. The same is true if d is negative while f is positive. If they are both positive or negative, the "positive work" will be positive.

It's sort of the same thing here. When the current approaches the + sign, we call it "positive voltage" and write v = ir. But depending on the sign of i, v can be negative, simply indicating the +/- sign needs to be switched around to make it positive (if you want it positive). "Negative voltage" would be written as v = -ir with the negative sign just reversing the sign of the current since it's going "the wrong way."

In your e.g., your current is actually approaching the negative sign of the +/- 12 V across the 60 ohm resistor. Simply convert the +/- 12 V into -/+ -12V to conform it to your notion of positive voltage. Then your loop becomes -12V + E + 6V = 0. Note, the +/- 6V already has the current approaching its + sign, so it is already in the form of "positive voltage".

Thing about it this way: If the voltage drop WERE positive, that would contradict your intuition of the circuit. The 18V applied across that chunk of resistors means if we drew I1 and I2 going down the 2 branches, both would be positive values. That is, the voltage source is sort of pumping current down those branches. When writing your loop equation, your current for the 60 ohm resistor is GOING UP the branch, which is in the opposite direction of the direction we know the current is actually going. So of course the voltage drop across that resistor in negative! The current is negative! v = ir., i negative => v negative. Back to the analogy, it's sort of like saying if f = negative and d = positive, of course the work does is negative. The guy was pushing against the motion, thereby removing energy from the box.

Last edited: Apr 16, 2012
3. Apr 17, 2012

### ViolentCorpse

Hey thanks a lot! I think I understand now. It's basically that when we know for certain what is actually happening in the circuit, we can't just change that and make our own assumptions, right? It's just that in mesh analysis problems, I always assign the resistors in different loops their polarities without regard to what is happening in other loops and it always works (I choose the same direction of current for every loop, though). In this problem, I was thinking in terms of how Eth within the selected loop would "see" the configuration of the resistors and so, they looked like positive drops to me WITHIN that loop. But I sort of get it now.

Thanks a lot RoshanBBQ! Really appreciate your help! :)