If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Labeling voltages

Labeling voltages on a schematic is not a matter of "right" and "wrong". It simply establishes how the voltage appears in the analysis equations. Created by Willy McAllister.

Want to join the conversation?

  • duskpin ultimate style avatar for user Kate Ripley
    Why was the current labeled as flowing out of the positive side of the battery? Doesn't positive current always flow into the positive side?
    (8 votes)
    Default Khan Academy avatar avatar for user
    • leaf blue style avatar for user wgrace
      Just think like charges repel. Positive charge is always going to flow away from the positive terminal and into the negative one. Remember the conventional way of labeling current is the path a positive charge would take. Hope that's helpful.
      (9 votes)
  • piceratops seedling style avatar for user anilsonmonteiro
    At , If I is going out of the positive terminal why the result is not negative?
    (5 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user cbj.groenewegen
    @ or so you could have been more clear and specific that the plus side of the V1 is still considered to be 1V less then the minus side of V1. Although this is for many counter intuitive it is still correct yes. I needed to replay this part 3 times to exactly listen what you said and where you were pointing too. Just wanted to add for more clarity.
    (4 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user _phidot_
      Welcome to the world of many-teachers vs many-style-of-explanation thing. It is not easy sharing a topic to a 30 people class, let alone to a 8 billion khan class. huhu..

      Anyway, I kind of like your question/post. It was challenging to be a good teacher, the kind that all-people-can-understand. I wish we can all be one of them. /(^_^)
      It was also very challenging to be a good student, the kind that are so fast to reach the gist(point) of a lecture/class. I am a very slow learner, and not as hardworking, let alone doing revision. I envy them very-very much.

      Nevertheless, let's learn something (if not everything) with all our hearts. And may we become wiser, and our kids (and other people kids) become smarter.
      (8 votes)
  • male robot hal style avatar for user Jordan1408
    Where does the V1 and V2 come in?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user _phidot_
      From nowhere.

      For this video/section, the purpose of the video is to share how to label our required variable/parameters when it was not given/labeled in the problem/real_circuit. It is to share that we CAN work out any circuit and get every parameters value ( V, I, R, power, equipment_rating) but we need to be systematic in approach and consistent in naming/labeling parameters. This is VERY useful when you are troubleshooting an electronics/power system that you don't have much info to start with. :)
      (5 votes)
  • duskpin seedling style avatar for user Voncarlo Giovanni D. Pasco
    Generally, what is the most used assumption in circuits: the electron flow [which is the current comes from the negative terminal] or the conventional flow [in which the current comes from the positive terminal same as the voltage (for simplicity)]?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • leaf red style avatar for user emarantia ngomane
    lets say i had "assumed" a starting direction for the voltage and when calculating the current, get the current to be negative, does this mean that the direction is opposite to my originally assumed direction?
    also when i substitute my current solution to get V1 or V2, do i just use V=IR with no signs or do i follow the signs i had indicated previously, like if my assumed current flow at the begining got a negative sign for a certain voltage, will i have V=-IR when proving or just V=IR?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • leaf green style avatar for user _phidot_
      IMHO :
      does this mean that the direction is opposite to my originally assumed direction? > Yup.
      will i have V=-IR when proving or just V=IR? > Just V = IR.

      Since this is a question of current direction, then we technically discussing a vector, in which, I don't have a clear way of typing it. :p I would rather draw. :| Sorry.

      So, I'll use an analogy. let say, we are an alien.. seeing the blue dot (earth, as human call it). we (alien) wanted to study, how fast does this dot move around the star (human call it the sun).
      The answer have two part, one scalar, one vector. One is the amount of the blue dot displacement change, and one is the direction of the rotation. We need both.

      The displacement change amount can be calculated, independent of where we are flying from. But for the rotation around the star, we may get 'clockwise' or 'anticlockwise'. The direction calculated depends on which way are we viewing this dot and star. (human : it's anticlockwise if we view from the earths' north and clockwise if we view from the earths' south)

      Both are true answer(for the alien), but we(alien) can only calculate/observe one at any one point of reference.

      Similarly, (back to human) putting V1 and V2 means we assume the direction of the current flow. If we calculated a positive value, means it flow in our assumed direction. if we get a negative value, means it flow in the opposite of our assumed direction.

      hope that helps. :)
      (3 votes)
  • aqualine ultimate style avatar for user Mateo Ceballos Querol
    why is the second diagram negative one volt? resistors don't have polarity
    (2 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Greg Janusz
      For future readers with the same question: I believe it is clearer if you imagine someone else handed him the schematic with the voltages already marked—which label the voltages entering and leaving the resistor, not polarities of the the resistor itself. (Nothing changes the fact that the resistor is causing a voltage drop.) And now he is using the KVL to analyze a loop.

      The signs around the resistor do not alter the effect of the resistor (which is revealed by how many ohms it is). The signs simply affect whether the voltage is a rise or drop in calculations, but note that a negative number for a rise is the same as a drop, and a negative number for a drop is the same as a rise. So when you do the calculation, if you see a situation where you want to reverse the signs around the resistor, you can instead just flip the sign of the value of number of ohms. It doesn't change the role of the resistor in the calculation, it just affects the math we do.
      (2 votes)
  • old spice man green style avatar for user Gunther Schmit
    At you use Kirkoff Law, meaning the sum of voltages=0. Is this a proprety we can apply in every and any circuit? or is there some kind of requirement?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Tony Tacos
      I'm looking at this serie on circuit analysis to remember what I learnt before, but you got to wonder, since now all I saw was some really basic notations, and all of a sudden he uses K Law without even an introduction ? Imo spend less than 10 videos about notations and at least one to present this important law ...
      (2 votes)
  • old spice man blue style avatar for user Galba
    That it works out mathematically makes sense since the arithmetic and algebra are very simple but WHY would you ever depict current flowing into a negative terminal and thereby have to reverse the sign of the current ? It seems like this is just complicating things unnecessarily by deliberately introducing something that is wrong and deciding to mathematically compensate for it.
    (1 vote)
    Default Khan Academy avatar avatar for user
    • purple pi purple style avatar for user APDahlen
      Hello Galba,

      In the near future you will encounter systems where the direction of current flow cannot be determined by inspection. To solve using "nodal" and "mesh" analysis you will need to guess. Sometimes we get it correct, sometimes not. It is only after the simultaneous equation are solved that the direction of current is known.

      Know that current can enter a voltage source from the positive terminal. Think of this as charging a battery.

      Please leave a comment below if you have any questions.


      (3 votes)
  • male robot donald style avatar for user Matteo Morena
    I still struggle to understand why the sign convention puts differences of potential in the opposite direction of the current for passive components. In my head I think of the current i as a kind of vector, so V would have to be a vector too. Because V=iR, and R>0, V would have to point into the same direction. As an additional problem, this way of thinking still works with Kirchhoff's voltage law (ΣV=0). So why would V have to point into the opposite direction?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • spunky sam orange style avatar for user Willy McAllister
      I feel your pain. This is one of those humps you have to get over at the beginning. Fear not, it is a small hump. Your reasoning about vector current and voltage is valid, but it happens to not be the one we use. Instead, be sure you've seen this video on the sign convention: https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/modal/v/ee-passive-sign-convention.

      With circuits we don't bother with a vector notation for current, because there are only two possible directions, this way and that way. This distinction can be taken care of with just the sign of the current, so we don't need vectors.

      The definition of current we use is "positive current is the direction positive charge moves (or would move if it was present)." That's the first decision. We use this even though we know negative electrons are moving in wires.

      Next, we want Ohm's Law to give the right answer for the voltage polarity when there is a current flowing in a resistor. Suppose you have a battery connected to a resistor. We define current to flow OUT of the positive battery terminal on its way to the resistor. If you measure the voltage with a voltmeter, the more positive voltage is on the end of the resistor closest to the + battery terminal. So we label the voltage that way, with positive voltage sign next to where the current is coming INTO the resistor.

      This is the sign convention we use at KA and in almost every EE text I've ever seen. It is totally arbitrary that we do this, but we are super consistent about it.

      It is possible to use the opposite convention, which means we define current to flow in the direction electrons move. That's what you've described in your question. This is also valid, but it is not commonly used. The only example I've ever seen is in the training material used by the US Navy and other branches of the US military. (search for "NEETS").
      (3 votes)

Video transcript

- [Voiceover] In this video, I want to do a demonstration of the process of labeling voltages on a circuit that we're about to analyze. This is something that sometimes causes stress, or confusion, and I want to just basically try to get out of that stressful situation, so the first thing I want to do, is remind ourselves of the convention, the sign convention for passive components, so we said, If we have a resistor that we draw this way, and we label the voltage plus or minus V on it, then when we label the current arrow, we want to label the current arrow so it goes into the positive terminal of the component, in this case a resistor, so just another quick example, if I draw the resistor sideways like this, if I label the voltages, the minus one on this side, and the plus one on this side, then when I go to apply the current arrow, I put the current arrow into the positive voltage sign, so this here is the sign convention for an individual component. How do we label the voltage, and the current together to be consistent? Now we're going to to over and analyze a circuit. I've drawn a circuit here. It's two identical circuits, and we're going to do it two different ways, with two different voltage labels, so the first thing we do, of course, when we analyze a circuit, is we set up the variables that we want to talk about. We'll do this side first. I'll label this as i, and that's a choice I can make, and then I'm going to label the voltages too, and I'm going to choose to label the voltages like this, plus and minus V, and we'll call this V1, and this will be plus or minus V2, so first, let's carry through and do the analysis of this circuit, and I'll give some component values to this. We'll call this 10 ohms, and this one's 20 ohms. Now I'm going to do an application of Kirchhoff's voltage law, around this loop, and we'll see how it turns out. All right, so Kirchhoff's voltage law will start at this node here, and will go around the circuit this way, and we'll do some equations. We'll say, plus three volts, when we go through this device, we get a voltage rise of three, then we get a voltage drop, because we go from plus to minus, we get a voltage drop of minus V1, and then we get another drop of V2, minus V2, equals zero. That's our KVL equation for this circuit over here, so let's keep going with this analysis. Three volts, minus V1 is i times R1, i times 10, and V2 is i times 20 ohms, and that equals zero, so let's keep going. Three minus i times 10, plus 20, equals zero, and that means that i equals minus three over, minus three goes to this side, 10 plus 20 is 30, and the minus sign goes with the 30, minus 30, so i equals plus amp, so we solve for i, and let's just pick out V1. Let's solve for V1, and we said earlier that V1 was i times R1, so V1 equals i, which is .1 amps, times 10 ohms, equals one volt. If we do it for V2, that equals .1 amps, times 20 ohms, and that equals two volts, and I can do one last little check, I can go back, and I can see KVL, I could do a check to see if this equation came out right, so three volts, minus V1 is one volt, and V2 is two volts, and that equals zero, and I get to put a check mark here, because yes, it's equal to zero. That was a real quick analysis of a simple two resistor circuit, and we got all the voltages and currents. Now I'm going to go do the same thing again, but this time I'm going to do the voltage labels a little bit odd. What I'm going to do here, is I'm going to say two plus. I'm going to define V1 to be in that direction, and we'll keep V2 like it was before, and now I need a current variable, and I'm going to call my current variable i here, just like we did before. Now at this point, you might say, Willie, you did it wrong. You did it wrong. That's not the right voltage label, but I want to show you that I'm going to get the same answer, even though I did it this unusual way, so let's do the same KVL analysis on this circuit, and what I want to show you, is that the arithmetic that we're about to do takes care of the science just fine. All right, so KVL on this circuit says that we'll start at the same place, and go around the same direction, so this says they're three volts, a voltage rise of three volts, we go in the minus sign, and out the positive sign, so that's a voltage rise. Now we get over to R1, and we go in the minus sign of R1, and out the plus sign, so that's plus V1, that's different than we had last time, right? Last time we had minus V1 here, see, and now we have plus V1. This is going to work out okay though. Now we go in the plus sign of V2, and we come out the minus sign, so that's a voltage drop, so we do a subtraction, and that equals zero. All right, we've got different equations, but we've got different definitions of V1, so now I want to write these V terms in terms of the resistance value, and the current value, and this is where we use the sign convention carefully, so now we need to include a term to represent R1, to use Ohm's law here. Now we have to be careful, this is one point we have to be careful, the current is going in the negative terminal of R1, so we're going to say VR1 equals negative i, times R1. Does that make sense? If we define our current variable to be going in the negative sign, then Ohm's law picks up this negative sign, to make it come out right, so down here, we plug in minus iR1, which is minus i times 10, that's a difference, and then we go through V2, and V2 is the same as it was in the other equation, minus V2 is i times R2, which is 20, and that all equals zero. Even right now, if you look at this equation, you see this minus sign, that snuck in here, because of our good use of the sign convention for passive componets. That makes this equation look just like this one here, so let's continue with the analysis. Just need a little bit more room, three volts minus i times 10 plus 20, equals zero. Now we have the same equation as before, so we're going to get the same i, i equals, three goes to the other side, and becomes minus three, 10 plus 20 is 30, with the minus sign. It goes over, same as before, so those came out the same, so now let's go check the voltages, see if we can compute the same voltages, and what he have to notice here, is our reference direction, the original reference direction we have for V generated a minus sign when we used Ohm's law, so we keep doing that, it's okay, so V1, equals minus iR1, equals minus .1 amps times 10, and that equals minus one volt, and V2 equals i times R2, which is equal to .1 times 20 ohms, and that equals plus two volts, and now the difference, we see the difference here. Here is the difference that showed up. V1 in this circuit has the positive voltage at the top, it came out with a value of plus one volt, and when we flipped over V1, negative one volt here, means that this terminal of the resistor, is one volt below this terminal of the resistor, and that means exactly the same thing in this case, as it does in this case, so these two things mean the same thing, and of course, the voltage on number two came out the same, so the purpose of the demonstration was just to show you, that no matter which way you name the voltages, somewhere in here, like right here, Kirchhoff's voltage law will take care of keeping the sign right, and you end up with the same answer at the end, so when you're faced with the problem of labeling a circuit like this, don't stress out about trying to guess ahead, what the sign of the voltage is going to turn out. You just need to pick an orientation, and go with it, and the arithmetic will take care of the positive signs, and the negative signs.