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

Resistors in parallel

Resistors that run alongside each other. Created by Sal Khan.

Want to join the conversation?

  • blobby green style avatar for user Sanjula Mansingh
    I have a very basic doubt.. its said that the current flowin through 2 resistances in series is same.. but shouldnt the initial current decrease after passing through an R, n decrease furthr on passing through the 2nd one???
    (59 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user ajmayberry
      Its helpful to think of an analogy to visualize it better. Lets say you have a water pump (BATTERY) and it has the potential (lets call this the VOLTAGE) to pump out 100 gallons/sec through a uniform pipe with a 2.256 ft diameter. This diameter happens to give a Cross-Sectional Area of 4 pi ft^2. We connect this water pump to a uniform pipe, this time with a CS-Area of 2 pi ft^2, and it goes around in a complete loop, ending at the back of the water pump (thus completing a CIRCUIT). Assume there is no air in the pipes and the pipes are completely full of water. The rate at which the water goes around is called the CURRENT. If you were to examine different sections of the pipe you would find the avg current to be a constant 50 gallons/sec; notice the current decreased by half as CS-Area decreased by half (Also true for ELECTRICAL CURRENT). Now lets say we add a SERIES of restrictions, one after another. The first restriction is 1 pi ft^2 and a second is 1/2 pi ft^2. What happened to the flow of water? Well, the 1st restriction slows the current to 25 ft/s and the 2nd one slows it to 12.5 ft/s. What is the overall rate of the entire CURRENT? Believe it or not, its 12.5 ft/s no matter where you measure it. Even though it can go through the 1st at 25ft/s the water has to wait for the the 12.5 ft/s water to go through the 2nd resistive pipe. This is how my physics textbook explains Direct Current. The relationship of cross-sectional area also applies to many other aspects in engineering and science. Try to apply this analogy to parallel circuits and things will start to click.
      (16 votes)
  • blobby green style avatar for user rdbrid
    Why is it that the total resistance of the circuit is less than each constituent resistor in parallel? Wouldn't it seem that it would be at least as much as the lesser resistor, but probably more?
    (25 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Kevin BizNasty Bezanson
      so essentially the circuit has more options for the electrons to flow, its like a traffic jam. say its rush hour and there is only on lane open heading into Chicago or NYC. at 7 am they open a second lane for cars heading in. when there is an alternate path through this pile up of cars (or in this case a couple of resistors) the cars can go fast. hope that helps
      someone!
      (46 votes)
  • leaf green style avatar for user Matt
    I'm confused at .. i1 = i1 + i2.. does he mean i1 = i2 + i3? and then he says i1 + i2 = i1 on the other side.. (the parallel currents are labelled i2 and i3 right? which do add up to i1 right) thanks in advance
    (13 votes)
    Default Khan Academy avatar avatar for user
  • stelly orange style avatar for user Krishna Phalgun
    what is the difference between Voltage and Voltage Drop ? // thanks in advance
    (5 votes)
    Default Khan Academy avatar avatar for user
    • duskpin ultimate style avatar for user Alex.Piotrowski
      To put simply Voltage is the amount of energy each coulomb (6.24*10^18 electrons) has and thus voltage equals to joules per coulomb. So whenever we talk about voltage, it is with respect to the circuit. The source of voltage comes from the battery

      Voltage Drop is how much energy is lost when current goes through a component. it could be a resistor or the wire itself. The flow of electrons collides with the lattice of the material and some of its kinetic energy is converted to heat. The electron has lost energy and we call that the voltage drop.

      There is a pretty cool law called kirchhoff law, where at the end of the circuit, the voltage must be 0 (i.e. voltage Drop = voltage) . This simply has to do with the principle of the conservation of energy. If you are interested just google kirchoff's law., electronics is very interesting.

      I hope this helps, best of luck
      (16 votes)
  • leaf grey style avatar for user charles
    At in the video the resistance comes out to be 4 ohms. While the mathematics behind this makes perfect sense, the physics doesn't. The smallest resistor on the entire circuit is 5 ohms. I understand that most of the electrons will want to pass through this, but wouldn't that mean that the total resistance should never come out to be less than the smallest resistor?
    (9 votes)
    Default Khan Academy avatar avatar for user
  • orange juice squid orange style avatar for user Christina Zhang
    At , can someone explain why there is less resistance when there are more resistors without using math? Thank you!
    (8 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user vaibnakgupta100
    I cant understand why does the potential difference across resistors arranged in parallel are same?
    and what exactly doe this potential drop signifies
    (8 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Ali
      Another way to look at it is by tracing your fingers along the path from the battery to your load like a resistor or lamp. Put your index fingers on the positive and negative terminals on the diagram. Imagine your fingers are connected to a voltmeter which measures the voltage across the battery terminals. Now trace your fingers along the path WITHOUT lifting them from the page until you get to your first resistor or lamp. All along the path, the voltage drop is the same, so you would expect to get the same voltage across the resistor or lamp. Now again move your fingers to the second resistor or lamp without lifting them off the page. Again the voltage across the path will not change and when you reach your resistor or lamp, the voltage will be the same. This is a simple way to find out if your resistors are wired in parallel. If you can trace the path without lifting your fingers off the page, they are in parallel. If you have to lift your finger off the page to reach another resistor, they are not in parallel.
      (0 votes)
  • blobby green style avatar for user dineshnainwal140773
    i have a doubt that why it is said that the current flows from positive terminal to negative terminal but the electrons flows from negative terminal to positive terminal.
    (3 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Andrew M
      To avoid dealing with negative signs all the time, we define something called conventional current as a postive current and we say that it flows from positive to negative. Just pretend that there are positive charges that move in the opposite directions of the electrons. It makes no difference. Electrical engineers never worry about which way the electrons are going. They just deal with conventional current.
      (7 votes)
  • male robot johnny style avatar for user Mr. Chai
    I've always wondered this; with the equation 1/Rtotal = 1/R1 + 1/R2 + ... why can't you take the inverse of both sides and just get Rtotal = R1 + R2... like in a series circuit? Why doesn't that work out mathematically?
    (1 vote)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Rebecca A.
      It has to do with common denominators in fractions. If you have like in the example 1/20 + 1/5 = 1/Rtotal, 1/20 +1/5 is 5/20 reduced to 1/4. Then you can take the reciprocal of 1/Rtotal = 1/4 giving R total =4.

      If you were to take the original formula and just invert it (like you are suggesting) to R total = R1 + R2 + R3 you would have R total= 20 +5 =25. Which is obviously wrong. Because R1 and R2 are different values, when adding them up as fractions of 1/R you have to find a common denominator.

      Hope this helps!
      (7 votes)
  • purple pi purple style avatar for user Abhinav Java
    if three bulbs are connected in series and the switch is closed the bulb just after positive terminal glows first but Sal said it is just a convention, the truth is that electrons move from negative to positive terminal, so shouldn't the bulbs start glowing starting from negative terminal?
    (4 votes)
    Default Khan Academy avatar avatar for user

Video transcript

Last video, we saw what happens when we have resistors in series. Now let's see what happens when we have resistors in parallel. All right, let me pick a new color. New color will be magenta. There's my battery: positive, negative. There's my ideal conducting wire. Here's my ideal conducting wire, but then-- and this is new-- it branches off, and I have two resistances. I have one here. And that's another resistance. And let's say that this has a resistance R1, this has a resistance R2. And, of course, the non-intuitive convention is that the current flows from the positive to the negative terminal, but we know that the electrons are actually flowing in the other direction. And I want to keep saying that, because I think it's so important to understand what's actually happening as opposed to the convention. Well, anyway, in the previous video, we said, well, when we have devices or components in series, that the current through the entire circuit is constant. But let's think about what happens here. So we have these electrons-- actually, let's think about it from the electron flow. The electrons are flowing, flowing, flowing at a given rate, and then here they have a choice. Some of them can take this top path, some of them can take this bottom path. So if you think about it, the flow of electrons in this branch plus the flow of electrons in that branch have to add up to the flow of the electrons in this branch, right? And then they're going to meet back up, and then the flow of electrons here-- so if we think of it this way, and now I'm going to go back to the convention, this is I1. So you have these electrons flowing at a given rate. This is the current right here. They're going to branch off, and maybe half of them go-- we'll see if the resistances are equal, if both of these branches have an equal amount of capacity in terms of how fast the electrons can flow through. If they're equal, or since we're going to current in this direction, let's talk about positrons or positive charges. If the positive charges-- although I just want to keep saying that it is not the positive charges that are moving. It's the electrons. But if you say that the lack of electrons can flow equally easily between both paths, that's if the resistances were the same, we could imagine that the current, the flow, would split itself up, and then over here would meet back up. And then we would say that the current here would also be I1. But let's figure out where the current's going. Let's call this current I2 and let's call this current I3. So I think it is reasonable, and you can imagine with water pipes or anything, that the current going into the branch is equal to the current exiting the branch. Or you could even think of it that the current entering-- when the currents I2 and I1 merge, that they combine and they become Current 1, right? I mean, think about it. In a given second, if this is 5 coulombs per second-- I'm just making up numbers-- and this is 6 coulombs per second, in a given second right here, you're going to have 5 coulombs coming from this branch and 6 coulombs coming from this branch, so you're going to have 11 coulombs per second coming out once they've merged, so this would be 11 coulombs per second. So I think hopefully that makes sense to you that this current is equal to the combination of this current and that current. Now, what do we also know? We also know the voltage along this entire ideal wire is constant, so then voltage-- let me draw that in another color, in blue. So, for example, the voltage anywhere along this blue that I'm filling in is going to be the same, because this wire is an ideal conductor, and you can almost view this blue part as an extension of the positive terminal of the battery. And very similarly-- I'll do it in yellow-- we could draw this wire as an extension of the negative terminal of the battery. This is an extension of the negative terminal of the battery. So the voltage difference between here and here-- so let's call that the total voltage, or let's just call it the voltage, right? The voltage difference between that point and that point is the exact same thing as the voltage difference between this point and this point, which is the exact same thing as the voltage difference between this point and this point. So what can we say? What is the total current in the system? If we just viewed this as a black box, that this is some type of total resistance, well, the total current in the system would be the total voltage, the voltage divided by-- let's call this our total resistance, right? Let's say we couldn't see this and we just said, oh, that's just some total resistance, and that is equal to the current going through R1. This is I1. This is a 1 right here. This is current I1. What's current I1? Well, it's going to be the voltage across this resistor divided by the resistance, right? That's what Ohm's law tells us: V is equal to IR, or another way we could say it is V over R is equal to I, right? So I1 is equal to the voltage across this resistor, but we just said that voltage is the same thing as this voltage, right? The voltage here is the same thing as the voltage here. The voltage here is the same thing as the voltage here. So the voltage across that resistor is still V, and so the current flowing across that resistor is V over R1. And the same logic: what is I2? I2 is this current. What is the voltage across this device? Well, that's just V again, right? It's the same thing as the voltage across this device, so it's V over R2 by Ohm's law. Well, all these V's are the same, so we can divide both sides of that equation by V, and we get 1 over the total resistance is equal to 1 over R1 plus 1 over R2. And you could make that argument if we had an R3 here. Let's say that we had another device, and that is R3. You could use the exact same argument, and you would have a plus 1 over R3. And if you had Rn or 10 of them, you'd just keep that 1 over R4, R5, et cetera. So let's see if we can use this information we have learned to actually solve a problem, and I actually find that it's always easier to solve a problem than to explain the theory behind a problem. You'll see that with most of these circuit problems, it's actually very basic mathematics. So let's say I have a 16 volt battery plus, minus, it's 16 volts. And just to hit the point home that you always don't have to draw circuits the same, although it is nice if you're actually drawing complicated circuits, I could draw it like this. I could draw the circuit like this, and let's say that there's a resistor here. And then let's say there's a wire and then there's another resistor here, and that this decides to do some random loopy thing here and that they connect here, and that they come back here. This strange thing that I have drawn, which you will never see in any textbook, because most people are more reasonable than me, is the exact same-- you can almost view it topologically as the exact same circuit as what I drew in the previous diagram, although now I will assign numbers to it. Let's say that this resistance is 20 ohms and let's say that this resistance is 5 ohms. What I want to know is, what is the current through the system? First, we'll have to figure out what the equivalent resistance is, and then we could just use Ohm's law to figure out the current in the system. So we want to know what the current is, and we know that the convention is that current flows from the positive terminal to the negative terminal. So how do we figure out the equivalent resistance? Well, we know that we just hopefully proved to you that the total resistance is equal to 1 over this resistor plus 1 over this resistor. So 1 over-- I won't keep writing it. What's 1 over 20? Well, actually, let's just make it a fraction. So it's 1/20 plus-- 1/5 is what over 20? That's 4/20, right? So 1 over our total resistance is equal to 5/20, which is equal to what? 1/4. So if 1/R is equal to 1/4, R must be equal to 4. R is equal to 4 ohms. So we could redraw this crazy circuit as this. I'll try to draw it small down here. We could redraw this where this resistance is 4 ohms and this is 16 volts. We could say that this whole thing combined is really just a resistor that is 4 ohms. Well, if we have a 16-volt potential difference, current is flowing that way, even though that's not what the electrons are doing. And that's what our resistance is, 4 ohms. What is the current? V equals IR, Ohms law. The voltage is 16 volts. It equals the current times 4 ohms. So current is equal to 16 divided by 4, is equal to 4 amps. So let's do something interesting. Let's figure out what the current is flowing through. What's this? What's the current I1 and what's this current I2? Well, we know that the potential difference from here to here is also 16 volts, right? Because this whole thing is essentially at the same potential and this whole thing is essentially at the same potential, so you have 16 volts across there. 16 volts divided by 20 ohms, so let's call this I1. So I1 is equal to 16 volts divided by 20 ohms, which is equal to what? 4/5. So it equals 4/5 of an ampere, or 0.8 amperes. And similarly, what is the amount of current flowing through here? I2? I'm going to do this in a different color. It's getting confusing. I'll do it in the vibrant yellow. So the current flowing through here, once again, the potential difference from here-- that's not different enough-- the potential difference from here to here is also 16 volts, right? So the current is going to be I2, is going to be equal to 16/5, which is equal to 3 1/5 amps. So most of the current is actually flowing through this, and that makes sense because the resistance is less, right? So that should hopefully give you a little bit of intuition of what's going on. And less current is flowing through here, so I2 through the 20-ohm resistor is 0.8 amps is I1, and I2 through the 5-ohm resistor is equal to 3.2 amps. And it makes sense that when you add these two currents together, the 3.2 amperes flowing through here and the 0.8 amperes flowing through here, that when they merge, they merge and then you have 4 amperes flowing through there. Anyway, hopefully, I have given you some intuition on what happens when we put resistors in parallel. I will see you in the next video.