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

Application of the fundamental laws (setup)

We can solve circuits by the direct application of the fundamental laws: Ohm's Law and Kirchhoff's Laws. (Part 1 of 2). Created by Willy McAllister.

Want to join the conversation?

  • mr pink red style avatar for user dindibo4
    I don't understand how is it possible that there are two power supplies that provides current in two different directions.
    The voltage source is providing current that goes clockwise and when it reaches R2, it goes downwards, but the current source is providing current that goes clockwise as well, and when it reaches R2 it goes upwards.
    How is it possible that there are two electric currents that are moving through one wire at two opposite directions?
    (11 votes)
    Default Khan Academy avatar avatar for user
    • spunky sam orange style avatar for user Willy McAllister
      Ah! You might be trying to intuitively solve the circuit before the detailed analysis. If your intuition gets you in a bind, as it has here, its important to do two things.

      1. Continue on through the full formal analysis to find the currents and voltages in every element. The math will reveal the truth about where those two currents flow.

      2. After you are done, take a moment to compare the analytical result with your original thoughts that led to perplexity. Where does the analysis say the currents actually flow? In particular, follow the current coming out of each source and trace how it finds its way back to the other side of the source.

      As you gain experience with circuit analysis, your initial intuitions will become more and more refined. But that path to a well-informed intuition is not easy. You have to pass through spots like this where things seem to be nonsense. The math will reveal the truth, and if you take the time to reflect, it will also inform your senses for the next time.
      (17 votes)
  • mr pink green style avatar for user Masrur Hossain Rakin
    At he said that now we have four equations. My question is how can i understand that which law has to use on which node ?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Robert Opalko
    At , "there is a current source" I_s of having 3mA. But later in the video at the KCL applied for node B and I_s is shown coming out of node B. If the source of the current is after node B, how is there the same current/value I_s coming out of node B? Wouldn't the current I_s not be 3mA regardless of a "current source"? A "current source" sounds like it is something generating current independent of what is coming out of node B....?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • spunky sam orange style avatar for user Willy McAllister
      Don't think of currents as being "before" or "after" a node. Rather, currents flow "into" or "out of" nodes. Node b is the junction between three circuit elements, R1, R2 and I_s. I gave variable names to two of those currents, i1 flowing "into" node b, i1 flowing "out of" node b. The third current (flowing out of node b) is easy to identify, since this wire is connected directly to a constant current source. I_s, the current in this wire has to be I_s = 3mA.

      One form of KCL tells us: Sum(currents in) = Sum(currents out) or i1 = i2 + I_s.
      Current source I_s guarantees a constant 3mA current in its wire. That means the other two currents, i1 and i2 will somehow become values to make the KCL equation come true.

      When you write out a KCL equation there isn't a cause-and-effect happening. The left side of the equation is not forcing the right side to do something. Every term in the equation can wiggle around to make it come true. In this case, there is no wiggle in the value of constant current I_s, but the other two currents are able to adjust.
      (2 votes)
  • blobby green style avatar for user Sofia Kelsey Pinzon
    Could you simplify the circuit by combining the resistors?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • leafers seed style avatar for user Mochamad Alghifari
    1. in real worlds, where does the Is (current source) come from? is it doesn't have a resistance?

    2. KCL is talking about the "eternity of current/electrons" / "conservation of current/electrons". We know that, the Is (current source) doesn't come from I1. So, why does the KCL is can be applied?

    sorry for my bad grammar
    (1 vote)
    Default Khan Academy avatar avatar for user
  • leafers seed style avatar for user mike miky
    Hi,
    As i understand, at node B, the current I1 "splits" into I2 and Is, right?
    But then, we have a current source that is supposed to generate some more current. So, my question is- after node B, the Is shouldn't be a reminiscence of I1? and then that current source isn't "adding" more current to the circuit?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • spunky sam orange style avatar for user Willy McAllister
      As you do this circuit analysis it is tempting to think ahead and predict how the currents are flowing. Try not to think ahead too far. Just focus on the small details you know from KVL or KCL. If you do that the overall answer will emerge at the end.

      You are correct that i1 splits up into i2 and Is. In fact you can specify this exactly with a KCL equation: i1 = i2 + Is.

      The idea of something happening "after" node B is not a clear idea. Nodes don't have a "before" and "after". What you are sure is true is the KCL equation.

      The current source determines the current in itself and in the wires connected to it (one of which is connected to B). That information is perfectly captured in the KCL equation.

      Don't try to squeeze more knowledge out of the KCL equation. To fully solve the circuit you have to come up with another independent equation dealing with the left side of the circuit. Then, together, these equations provide enough constraints to determine everything (all the v's and i's).
      (2 votes)
  • blobby green style avatar for user Kyle Delaney
    So the components in the circuit are the branches, right? I remember you explaining that a node includes all the wires between the components and is not just the junction where those wires meet. But if the point where the wires are joined is where current splits off into multiple currents, doesn't that mean one node will have multiple different current values? Why isn't a distinction made between the junction point and the multiple wires attached to it? Why are they all treated as one thing?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • old spice man green style avatar for user Willy McAllister
      A node can be simple, like a junction between two resistors, or, it can look like an octopus with lots of paths leading elsewhere.

      A node always has a single voltage. (We are assuming the wires are Ideal, they have zero resistance).

      A simple node will have a single current. All the current flowing in has to flow out. See Kirchhoff's Current Law.

      A complicated node with many paths may have many different currents in each section. The one thing we know for sure is that all the current flowing into the node adds up to all the current flowing out of the node. (Again, this is Kirchhoff's Current Law.)
      (2 votes)
  • leaf green style avatar for user SJNewton123
    I understand most of the concepts that go into formulating the equations, except for a few things. First of all, how can i sub 1 and i sub 2 exist, if currents always enter through positive terminals, because they are shown on the circuit as moving away from the voltage source's positive terminal. Also, the last equation is a little confusing to me: i sub 1 = i sub 2 + I sub s. Could somebody explain this equations a little more please? Thanks.
    (1 vote)
    Default Khan Academy avatar avatar for user
  • leafers ultimate style avatar for user shreeyash
    what does a current source mean
    (1 vote)
    Default Khan Academy avatar avatar for user
    • spunky sam orange style avatar for user Willy McAllister
      A constant current source produces voltage and current. It is designed to keep its current constant.

      You are more familiar with a constant voltage source, which produces voltage and current, but it keeps its voltage constant. The most familiar constant voltage source is a battery. A battery uses the properties of chemical reactions to generate a constant voltage.

      There isn't a chemical version of a constant current source, but it is possible to build a circuit that has this property.
      (2 votes)
  • blobby green style avatar for user pbarton131
    So for V1 and V2, we are measuring the potential difference across the resistors? I feel like the concept of voltage is tricky; my physics book emphasized that we are always measuring change in voltage (electric potential) between two points (using a triangle, 'delta V' symbol), and that the voltage at a particular point makes no sense. So we are in effect measuring the voltage 'drop' at R1 and R2, yes? (on a side note, I find it odd that voltage seems to be the only case where the SI unit shares the symbol of the quantity its measuring:'Amperage' or current is I, resistance is R, but the units are A, and big Omega respectively).
    (1 vote)
    Default Khan Academy avatar avatar for user
    • old spice man green style avatar for user Willy McAllister
      The term 'voltage' is tricky. Voltage is the honorary name of 'electric potential difference'. Notice that 'difference' is part of the definition of voltage.

      It helps me understand voltage to think about it like height or elevation. Both of these spatial terms are based on differences (differential measurements). There's always two points involved. Your height is the change of elevation between your feet and your head. The elevation of a mountain is the number of meters it takes to go from the reference elevation (sea level) to the top of the mountain.

      When we measure voltage there are always two points involved (it is always a difference measurement). Sometimes the two points are clearly stated, like "the voltage across resistor R1 is 5v. Your voltmeter probes touch the ends of the resistor R1. This is referred to as an element voltage because it is measured across a circuit element.

      Notice I used the word "across" for element voltages. I didn't say "The voltage at R1." That "at" doesn't quite make sense. You probably mean "The voltage across R1" because it clearly indicates where to touch your voltage probes.

      Other times you might identify a circuit location while the other point is implied or understood. "The voltage at Node B is 10v." So you know Node B is one of the points. What is the other point? Where does the second voltmeter probe touch? The second point is the "reference node", also called the "ground node". When voltage is described this way it is called a node voltage. Your voltmeter probes touch Node B and the reference node.

      With node voltages you properly use the word "at". "The voltage at Node B." tells you exactly where to touch your first voltage probe.

      Your observation about the mixed up electric unit names is correct. These names are historical and involve a lot of honorary names and even terms borrowed from French. It all depends on who discovered something first. My advice: get used to it.
      (1 vote)

Video transcript

- [Voiceover] Alright, now we're ready to learn how to do circuit analysis. This is what we've been shooting for as we've learned our Fundamental Laws. And the fundamental laws are, Ohm's law, and Kirchhoff's laws, which we learned was Kirchhoff's Current Law and Kirchhoff's Voltage Law. This says that the sum of all the currents going into a node adds up to zero. And the Voltage Law says the sum of the voltages going around a loop adds up to zero. And the other thing we learned was the sign convention. So, we had a passive sign convention, which was, if I label the voltage like that, then I label the current flowing into the positive end, the positive end and the current go together. So, these are the tools that we have for analyzing a circuit and I've drawn a circuit here. So, what we see is, we have a voltage source, we'll call that VS, and we'll give it a value of 15 volts. We have a resistor, R1, that has a value, we'll give that a value of four K ohms. And another resistor, R2 is sitting right here, and we'll give it a value of 2 K ohms, 2,000 ohms. And finally, there's a current source sitting over here. And this has a value of three milliamps flowing downwards. So, the point of circuit analysis means what we wanna do is, we wanna find all the currents and all the voltages in this circuit using our Fundamental Laws. So first, let's identify the nodes in the circuit. We'll give them letter names. We'll call this node A, and that's the junction between R1 and Vs. We'll give this one a name, this is a node that's distributed, all three of these components joined right about here, we'll call that node B. And that's one of the nodes. And down at the bottom here, there's the third node, which is this distributed node here. It connects the voltage source, the resistor, and the current source, and this is all, we'll call that node C. Now, we could also label some voltages on here. We'll label the voltages, we'll do that in orange. We'll call this V1 and we'll give it a plus sign here. And the minus sign here, that's the voltage across R1. We'll give R2 a name, we'll get the voltage here, we'll call this plus, minus V2. That's the voltage across R2, it's also the voltage across the current source. B, it's the voltage between node B and node C. And now let's label the currents. Let's label the currents in this circuit. We have a current coming through R1, we'll call that i1. There's another current going through R2 and we'll call that i2. There's a current going here through Is and we actually already know that, we'll just call that Is. And that's all the currents, that's all the unknown currents in our circuit. If we stare at this, we're not gonna be able to tell, on just inspecting this, what's going on, we have to come up with some solution technique. And we're going to solve this by, what's called, Application of the Fundamental Laws. And here's our laws over here. And the whole technique is based on solving simultaneous equations. That's the skill of circuit analysis, circuit analysis means setting up and solving simultaneous equations. So, let's write down some things we know about the circuit. And first thing we're gonna apply is, we'll apply Ohm's law to the two resistors. So, we can say V1 equals i1 times R1. And we can say that V2 equals i2 times R2. Alright, that's not a bad start, we have two equations, how many unknowns do we have? I2 and V2 are unknown. I1 and R1 are unknown. So, we have four unknowns and we have two equations, so, we're halfway there. And now, we need to come up with some more equations. And I'm just gonna stare at the circuit and what I'm gonna do is, I'm gonna apply KVL around this loop here, that's gonna be my KVL loop. And after we get that equation, what we're gonna do is, we're gonna write KCL, we're gonna write the Current Law at node B. That's the two equations we're gonna develop here. So, let's do KVL. Let's start at this node, at this corner of the circuit and go around clockwise around. And what we'll see first is a voltage rise, plus Vs. Next, I see a voltage drop, so it's minus V1, going around this way, minus V1. I'm at this point, now I'm gonna go through R2 and I see a voltage drop of V2, minus V2 equals zero. That's Kirchhoff's Voltage Law around this loop. One rise, two drops, one rise, two drops. Now, we're gonna move up here, and do Kirchhoff's Current Law at this node right here. So let me draw a few more arrows. This is i1, it comes right through that resistor. And this is Is. So Kirchhoff's Current Law, let's count the currents going into node B, current going into node B is i1. Currents go out, we'll set that equal to the sum of the currents going out of node B. So, that's equal to i2 plus Is. So look, now I have my four equations, here's my four simultaneous equations. And this is what we have to solve. So let me move the screen up and we'll start doing that. Let's do a double check, we have four unknowns, V1, i1, V2, i2. Here's V1 and V2, and here's i1 and i2. And I have four equations, so we oughta be able to do this.