Mesh Current Method, Sign Convention - Page 1
Forum section: electronics / beginners

author: rstofer
author_rank: Super Contributor
timestamp: December 10, 2019, 11:27:01 pm
content: Back in the early '70s, I was taught that: When I went through a battery from - to + that was a gain and was considered positive. If I went through a resistor I assumed I was going from a higher voltage to a lower voltage and that was considered negative, a loss in voltage.So, for a simple battery in parallel with a resistor I would have +Vbat - I1*R1 = 0 or Vbat = I1 * R1 which seems pretty good, voltage isn't piling up anywhere.Now I see many tutorials where the sign convention is reversed. Going up through the battery is considered negative and going down through the resistor is considered positive. I realize it doesn't matter because it will all work out in the end. If the assumed current direction is wrong, it will have a minus sign. No big deal!The problem is, I didn't get the memo! Is everybody on board with the newer sign convention? Is that really the way it is being taught today? What is the justification for considering the battery voltage as negative when clearly there was a gain?

author: sibeen
author_rank: Frequent Contributor
timestamp: December 11, 2019, 12:01:22 am
content: I had to think about this and grinned to myself when reading this as I suspect that I change which way I define the direction depending upon which day of the week it is. This may be a result of my electrical education. I started out with a four year apprenticeship as an electronics tech in the late 70s. The interesting tidbit is that military techs, at least in the English speaking portion of the world, get taught electron flow. I've never been able to track down the why of that and it's always interested me as to why that occurred. A few years later I decided to go down the university route and was taught thecorrectconventional current flow.Still to this day when I'm looking at a circuit I'll revert to electron flow and this occasionally causes consternation with people I'm showing the workings to as "I'm doing it wrong" only to be a tad bemused when my answer turns out to be the correct one.As to the OP, I didn't realise that any change had been made, but then I haven't looked.

author: RoGeorge
author_rank: Super Contributor
timestamp: December 11, 2019, 12:11:27 am
content: No idea what exactly you describe. Do you have a link?There is an international convention that current, I (A), flows from + to -. In electric and electronic this is the only convention in use.Sometimes, in physics there is a distinction about "electrons current", but that's something else than what we usually call "current" (or more pedantic "conventional current", I).

author: rstofer
author_rank: Super Contributor
timestamp: December 11, 2019, 12:44:21 am
content: I don't have a workable link. My question is really "How would you write the loop equations?" for a battery in parallel with a resistor.I would start in the lower left corner and draw the loop in a clockwise rotation. I would go up through the battery and consider that a gain of Vbat. I would continue over to the resistor and see the voltage at the top of the resistor as higher than at the bottom, justifying my clockwise rotation. I would compute the voltage drop as the current I times the resistor R and assign it a negative sign because I am entering at the + terminal.When I was done I would have +Vbat - (I * R) = 0.Apparently, the new math is the opposite: Consider the battery as negative and, since the current is entering the resistor at the + terminal, consider that voltage as positive. The equation would be -Vbat + (I * R) = 0Either way, the equation simplifies to Vbat = I*R -> the voltage in is equal to the current flowing through the resistor times the value of the resistor. Solving for I = Vbat / R.Like I said, the same answer but the convention seems backwards from what I was taught.The only reason I bring it up, knowing that the proper application of either approach will yield the same answer, is that my grandson is taking a circuits course next semester and I need to get it right,https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/circuit-elements/a/ee-sign-convention?modal=1

author: RoGeorge
author_rank: Super Contributor
timestamp: December 11, 2019, 01:24:19 am
content: Now I see what you mean, you were talking about the voltage arrow, right?I didn't know there are two conventions, American and Europeanhttps://tex.stackexchange.com/questions/37182/voltage-sign-convention-european-vs-american-in-circuitikzhttps://tex.stackexchange.com/questions/298538/european-convention-of-voltage-in-circuitikzAbout the signs in a loop, it shouldn't matter. We can pick arbitrary directions for any arrows in the circuit. After solving the circuit and finding the numerical values, wherever is a -x, it means we picked the wrong direction for our arrow, wherever is a +x, it means we were lucky and picked the correct direction for the arrow. This is how I remember it.Like the electron current vs. conventional current wouldn't have been already enough confusion, the Khan Academy link you posted have the voltage arrow reversed from how I would draw it.

author: rstofer
author_rank: Super Contributor
timestamp: December 11, 2019, 01:26:04 am
content: Quote from: sibeen on December 11, 2019, 12:01:22 amI think the military is still thinking about vacuum tubes and perhaps electron flow is easier when dealing with tubes.Once institutionalized, it will never change.I like current flow because I can follow the arrows on transistors and diodes. Somehow, that is comforting.

author: rstofer
author_rank: Super Contributor
timestamp: December 11, 2019, 01:32:20 am
content: Quote from: RoGeorge on December 11, 2019, 01:24:19 amNow you see why I'm confused. The RIGHT answer will be whatever the Prof says it is. I just need to be more flexible in my thinking.I believe we had an example of the American vs European approach in one of Simon's homework problems a long time back (2 years?). As an aside, I don't recall ever drawing a voltage arrow when doing mesh equations. I just labeled the assumed + and - ends of components and wrote the loop equation based on my assumed circuit path.I'm going to drag out my textbook and make sure I'm not mis-remembering but it sure seems right that a battery is a voltage gain and a resistor is a voltage loss. But I graduated in '73 so what do I know?

author: RoGeorge
author_rank: Super Contributor
timestamp: December 11, 2019, 01:47:31 am
content: Indeed, whatever the teacher say.Just to be clear, I know it like in the attached fig. 17.1 (a), (b) or (c), AKA CEI/IEC 60375 standard. Khan Academy link was using the other convention, like in fig 17.1 (d).

author: rstofer
author_rank: Super Contributor
timestamp: December 11, 2019, 02:59:59 am
content: 17.1(b) works for me!I went back to my text and they take a shortcut with the earliest problems: They set the battery as positive on the left hand side of the '=' and then sum all the drops through resistors as positive on the right hand side. That's fine, because if the drops were pulled to the left hand side they would all be negative and that's what I recall.In the end, it doesn't matter and the Prof is always right so I'll just have to wait and see what they teach.

author: gcewing
author_rank: Regular Contributor
timestamp: December 11, 2019, 03:26:54 am
content: Most circuits will contain more resistors than batteries, so some people may find it easier to think in terms of voltage drops rather than voltage gains. That way most of the voltages will be positive rather than negative.