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### Course: MCAT > Unit 9

Lesson 11: Current and resistance# Resistors in parallel

Explore the intriguing world of parallel resistors in circuits. Learn how current splits and combines in different branches, and how voltage remains constant across resistors. Uncover the formula for total resistance and apply it to solve real-world problems. Created by Sal Khan.

## Want to join the conversation?

- 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???(60 votes)
- 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.(19 votes)

- I'm confused at3:08.. 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(15 votes)
- its just a mistake. Basically what he is trying to say is that the sum of the current in those 2 pathways is equivalent to the current of the wire, i1(8 votes)

- what is the difference between
**Voltage**and**Voltage Drop**?*// thanks in advance*(7 votes)- 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(20 votes)

- At9:25in 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)
- i think you meant to say...."Take the bigger resistor out. Now the resistance is just 5 ohms"(1 vote)

- At0:45, can someone explain why there is less resistance when there are more resistors without using math? Thank you!(8 votes)
- In simple terms, when resistors are in parallel, the total resistance decreases because there are more ways for the electric current to travel compared to just one resistor alone. It's like having more lanes on a road, making the journey easier for the flow of traffic (electricity), thus, less overall resistance(1 vote)

- 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)- 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)

- 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)
- 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)

- 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)
- 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)

- 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)
- How do the electrons choose what path to take? Do they just take the nearest path, or do they decide(3 votes)
- You'll have to imagine the electrons in the circuit as flowing water, instead of single, individual electrons running around the circuit freely...(2 votes)

## 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.