KVL Chronic Polarity Problem

KVL is just not working out for me. Conceptually it makes total sense, but there's some key detail I'm missing because algebraically I'm finding that "arbitrarily" choosing current direction dramatically changes the answer - not just a negative / positive reversal.

Here's a simple example circuit diagram I found where KVL does not appear to work when I execute it by plugging in the voltages (using Ohm's law with the 2A answer they got and the given resistances) despite that very document showing how it works no matter which arbitrary direction you chose for current.

http://web.engr.oregonstate.edu/~traylor/ece112/lectures/kvl_analysis.pdf

Using that circuit diagram, if I choose current to go clockwise, I get this:

R1 = 30 Ohms X 2 A = 60V
R2 = 15 Ohms x 2 A = 30V

So,

-120V + 60V + 30V + 30V = 0

No problem, that looks great. So what if I chose current to go counter-clockwise instead? I get this:

-120V - 60V + 30V - 30V = 0

That's not right. That's -180, not 0. What in the world am I doing wrong here? I stare and stare at this, and I'm just not seeing it. That equation comes straight out of their own example, merely plugging in the actual voltage value, E, in place of the IR term.

Actually, I can simplify this whole question to a single voltage source and a resistor.

In that kind of example it would seem impossible for someone to choose the wrong current direction, but according to KVL I should be able to choose the wrong direction, regardless...

So, a 9V source and a 10kΩ resistor would produce 0 if I choose the correct current direction, and 18 if I do not.

The issue seems simple. Voltage sources don't change the math when you choose the wrong current direction, but polarity of loads *do*. And that changes how you add and subtract integers, such that I cannot get zero when I choose the wrong current direction.

What obvious point am I missing here?

You have to be very careful with the +/- signs in the problem. The voltage drop across a resistor is always creates a positive voltage drop for which ever way you choose the currents. So, in your 9V and 10kOhm example, the equations go as follows:

For a clockwise current rotation-

Current from - to + in voltage source give +9V.
Current I through a 10kOhm resistor gives V across resistor of I*10k.

So, +9V (source) - I*10k(resistor) = 0
-I*10k = -9V
I = 9V/10k

For counterclockwise current rotation-

Current from + to - in voltage source give -9V.
Current I through a 10kOhm resistor gives V across resistor of I*10k.

So, -9V (source) - I*10k(resistor) = 0
-I*10k = 9V
I = -9V/10k
minus sign shows current actually in the opposite direction which is what you see in the clockwise rotation portion.

Basically, let the math do the work for you. The direction of the current that you choose also defines the voltage drops across the resistive components. If the direction that you have chosen is incorrect, the currents and voltages become opposite of the way they were define by the initial selection of current rotation.

So, even though you sometimes "know" which way the voltages are going to end up, you can't add that information to the equations. Once you select a current direction you have to stick with it for every component.

StayatHomeElectronics said:

You have to be very careful with the +/- signs in the problem. The voltage drop across a resistor is always creates a positive voltage drop for which ever way you choose the currents. So, in your 9V and 10kOhm example, the equations go as follows:

For a clockwise current rotation-

Current from - to + in voltage source give +9V.
Current I through a 10kOhm resistor gives V across resistor of I*10k.

So, +9V (source) - I*10k(resistor) = 0
-I*10k = -9V
I = 9V/10k

For counterclockwise current rotation-

Current from + to - in voltage source give -9V.
Current I through a 10kOhm resistor gives V across resistor of I*10k.

So, -9V (source) - I*10k(resistor) = 0
-I*10k = 9V
I = -9V/10k
minus sign shows current actually in the opposite direction which is what you see in the clockwise rotation portion.

Click to expand...

Hey, thanks much for the reply. I get hung up on details like this and find it difficult to move forward without a resolution. I appreciate the help.

The part I bolded and highlighted is the point of contention, I think. According to what I've read, current direction does not change how you interpet the voltage source as a -9V drop or +9V rise.

I know it's a pain, but this link I pasted is actually a very simple example of this. They demonstrate KVL, both for a clockwise and counter-clockwise current direction *but* maintain going in a clockwise direction to add up the drops and rises after assigning the polarities, in both cases.

That causes the voltage sources to remain the same in either case, while the resistive components change polarities - and that changes the math.

If I assume a clockwise current *and* I'm traversing the loop in a clockwise direction, then I will see a voltage rise of +9 V and then a drop of -9V, so 9 - 9 = 0. Yay, KVL checks out:



But, if I assume a counter-clockwise current *and* I'm still traversing the loop in a clockwise direction, then I get this:



Now, that's a +9V rise, plus another +9V rise (going from - to +, just like the voltage source). +9 + 9 ≠ 0.

That's also the case if I traverse the circuit the opposite direction, I'll get a -9V drop (+ to -) on the resistor and then another -9V drop on the voltage source. -9 - 9 ≠ 0.

I'm not seeing how KVL checks out, using voltage values and an incorrect current assumption.

The problem is actually simple...if you assume the wrong current direction, then the resistive component's polarity is identical to the voltage source polarity - they are both dropping or rising in voltage, never opposite from each other in order to generate a difference of 0.

Write your equations in terms of assumed current directions. Solve them: don't you find that the answer turns out to be negative in any case where your initial assumption was wrong?

How likely do you think it is that old Gustav will be proved wrong after all these years?

Adjuster said:

Write your equations in terms of assumed current directions. Solve them: don't you find that the answer turns out to be negative in any case where your initial assumption was wrong?

Click to expand...

Of course not, otherwise I wouldn't be posting a question on the matter - ha!

How likely do you think it is that old Gustav will be proved wrong after all these years?

Click to expand...

Not even possible. Conservation of energy is not up for debate. That's why I keep phrasing this asking where *I'm* going wrong.

There's a very simple step, some fundamental thing I'm doing wrong here, but I can't pinpoint it. It can't be this difficult. I know the concept is simple, but when I go through the *steps* to assign polarites and assume current directions, it doesn't work out.

That's why I retreated to a basic example of a voltage source and a resistor. And I drew them as I interpret the change in polarity on the resistor above, while the cell's polarity remains the same. So...now, why isn't KVL adding up to zero in this simple example using the same steps as a complicated example?

zippoinc said:

But, if I assume a counter-clockwise current *and* I'm still traversing the loop in a clockwise direction, then I get this:

Now, that's a +9V rise, plus another +9V rise (going from - to +, just like the voltage source). +9 + 9 ≠ 0.

Click to expand...

When you go around the loop in this manner you get +9 + I*R = 0, not +9 + 9 = 0.

Assuming the I*R is +9V is not correct. You have to calculate the current, which will be negative since it actually flows in the opposite direction of the assumed current, and then multiply it by the resistance.

That will yield +9V + (-i)*R = +9V + -9V = 0.

Ah, I think I see the light. I think where I'm missing the point is that once the actual voltages have been determined, you can't just defy reality and figure them backward and expect KVL to work. KVL works as an equation with an incorrect current direction because you're setting that IR term on the other side of the equal sign, causing the polarity to be worked out algebraically.

I don't think I'm saying that quite right. But it makes more sense now.

Thanks for your help.

Your welcome.

That is the key, let the math do the work for you. Once you start the math you gave to stick to it.