Mesh current method (step 4 solve) Khan Academy English (auto-generated) English

Mesh current method (step 4 solve)YouTube↗ IA↗(long shot)
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we're working on the mesh current method

- [Voiceover] We're working on the mesh current method,

of a method of analyzing circuits and in

a method of analyzing circuits.

the previous video we set up our circuit

And in the previous video we set up our circuit,

we set up our mesh currents flowing around these loops within the circuit

we set up our mesh currents flowing around these loops

and we solve for the ez currents that

within the circuit, and we solved for the easy currents.

was the the current on this side this

That was the current on this side,

current source constrains I three to be

this current source constrains i3 to be a fixed value,

a fixed value so we already know I three

so we already know i3.

and then we created two equations from

And then we created two equations

the two independent variables the two currents here to describe how the

from the two independent variables, the two currents here,

circuit works and in this video we're

to describe how the circuit works.

going to solve these two equations one

And in this video we're going to solve these two equations.

thing to remind ourselves when we define

One thing to remind ourselves,

these mesh currents all of them went in

when we defined these mesh currents

the same direction

all of them went in the same direction.

it helps to make them all go in the same

It helps to make them all go in the same direction,

direction but all clockwise in this case

but all clockwise in this case.

and that's so that when you have this

And that's so that when you have

little pattern down here of i1 minus i2

this little pattern down here

or i2 minus i3 then it becomes a

of i1 minus i2, or i2 minus i3,

familiar pattern and it helps to get the

that it becomes a familiar pattern,

signs right the other thing to point out

and it helps to get the signs right.

is the reason we're using meshes and not

The other thing to point out is

any other loop in the circuit not

the reason we're using meshes and not any other loop

overlapping loops like we could have

in this circuit, not overlapping loops

gone all the way around the outside in a

like we could have gone all the way

loop the reason that meshes in

around the outside in a loop. The reason that meshes,

particular the open areas of the circuit

in particular the open areas of the circuit,

are a good way to do this method is

are a good way to do this method

because it produces exactly the right

is because it produces exactly

number of equations to solve the circuit

the right number of equations to solve the circuit.

you don't have too few equations and you

You don't have too few equations,

don't have too many equations so let me

and you don't have too many equations.

move the screen up and we'll go to work

So let me move the screen up

on these equations so now we begin step

and we'll go to work on these equations.

4 which is to solve the equations and

So now we begin step four, which is to solve the equations.

what I want to do first is just to tidy

And what I want to do first is,

up a little bit I'm going to take the

just to tidy up a little bit,

constants and I'm going to put them on

I'm gonna take the constants

the right-hand side and I'm going to

and I'm gonna put them on the right hand side,

gather together the terms for I 1 and

and I'm gonna gather together the terms for i1

the terms for I 2 in separate places so

and the terms for i2 in separate places.

if we go at the first equation and we

So if we go at the first equation and we look for i1,

look for i1 it's mentioned here twice so

it's mentioned here twice,

we get - and it's a combination of R 1 and plus R 2 times i1 and for the i2 term let's see if I two shows up I too just shows up

so we get minus and it's a combination of r1 and plus r2 times i1. And for the i2 term, let's see if i2 shows up,

once here so it's going to be plus r2 i2 and that's going to be equal to let's

i2 just shows up once here. So it's gonna be plus r2 i2. And that's gonna be equal to,

move V over to the other side so V

let's move V over to the other side,

becomes minus V over here so that's our

so V becomes minus V over here.

So that's our first equation.

so now let's tidy up the second equation

So now let's tidy up the second equation.

I 1 terms this is the only i1 term here

The i1 terms, this is the only i1 term here.

so it's plus r2 i2 oops

So it's plus r2 i2, oops,

plus R 2 i1 that's the only i1 term now

plus r2 i1, that's the only i1 term.

if we do i 2 terms we're going to have

Now if we do i2 terms,

minus r2 i2 minus r3 I 2 and minus R for I 2 so let's do that

we're gonna have minus r2 i2, minus r3 i2, and minus r4, i2.

this way r2 plus r3 plus r4 x i2

So let's do that this way, r2 plus r3 plus r4 times i2.

now we have an i3 term in here but we

Now we have an i3 term in here,

know what I 3 is i3 we know from this so

but we know what i3 is, i3 we know from this.

I'm just I'm going to write that down right here that equals minus I there's lots of minus signs here so let's see if

So I'm gonna write that down right here, that equals minus I, there's lots of minus signs here,

we can get this right - are four times

so let's see if we can get this right.

minus I 3 is positive r4 i3 or its negative r4 i and when we

Minus r4 times minus i3 is positive r4 i3, or it's negative r4 I,

take it to the other side it'll be plus r4 I let me check that again - are four

and when we take it to the other side it'll be plus r4 I. Let me check that again.

times I minus I 3 is a positive sign I

Minus r4 times minus i3 is a positive sign,

gives it a negative sign and when we

I gives it a negative sign, and when we take it

take it to the other side it gets a positive sign good alright so this is what our equations look like now let's

to the other side it gets a positive sign. Good. Alright? So this is what our equations look like now.

Let's keep going.

I can continue using symbols for the

I can continue using symbols for the resistor values,

resistor values but it's going to get kind of complicated and it doesn't help

but it's gonna get kind of complicated

with our understanding so I'm going to

and it doesn't help with our understanding,

switch over to a specific circuit so

so I'm gonna switch over to a specific circuit.

what I did was I filled in some resistor

So what I did was I filled in some resistor values here

values here and I put five volts on the

and I put five volts on the voltage source,

voltage source 1k ohm is 4 R 1 2 K ohms

one k ohms for r1, 2k ohms is r2,

as R 2 3 K ohms is R 3 3 km 0 and

3k ohms is r3, 3k ohms is r4,

there's 2 milliamps over here so I'm

and there's two milliamps over here.

just going to write these values under

So I'm just gonna write these values

the equation here so I don't mess up 1 K

onto the equation here so I don't mess up.

2 K 2 K that's 5 volts that's 2k ohms 2k ohms 3 K ohms and our four was also 3 is 3 and I is 2 milliamps

One k, two k, two k, that's five volts. That's two k ohms, two k ohms, three k ohms, and r4 was also three, is three, and I is two milliamps.

I'm going to move the circuit down out of the way a little bit now we have room

I'm gonna move the circuit down out of the way a little bit.

for our equation let's let's fill it out

Now we have room for our equation.

again with the real values so it's going

Let's fill it out again with the real values.

to be minus 1 plus 2 is 3 I 1 plus 2 I 2

So it's gonna be minus one plus two is three i1, plus two i2

equals minus 5 we know all the resistors

equals minus five.

are K ohms and so we can actually scale

We know all the resistors are k ohms

our equations down by a factor of a

and so we can actually scale our equations down

thousand and you just use the simple

by a factor of 1,000 and just use

numbers without the case so our second

the simple numbers without the ks.

equation is 2 I 1 minus 2 plus 3 plus 3 is 8 I 2 and that equals 3 K ohms times 2 milliamps is plus 6 so this is getting

So our second equation is two i1 minus two plus three plus three is eight i2, and that equals three k ohms times two milliamps, is plus six.

close to something we can solve so I

So this is getting close to something we can solve.

think the way I want to solve this is

So I think the way I want to solve this

I'm going to take this equation here and

is I'm going to take this equation here

multiply it by 4 and then add it to this equation here and let's see what we get

and multiply it by four, and then add it to this equation here.

so 4 times minus 3 is minus 12 and we

And let's see what we get. So four times minus three is minus 12,

add it and we're going to get

and we add it and we're gonna get.

four times two is eight minus eight and I 2 drops out that was the purpose of

Four times two is eight, minus eight, and i2 drops out.

the four and that equals 4 times minus 5 is minus 20 plus 6 is minus 14

That was the purpose of the four. And that equals four times minus five is minus 20, plus six is minus 14.

so I 1 equals minus 14 over 10 or I 1 equals one point 4 milliamps and that's

So i1 equals minus 14 over 10, or i1 equals 1.4 milliamps,

cool we get to put a box around that so

and that's cool. We get to put a box around that.

as a reminder we defined I 1 to be the

So as a reminder, we defined i1 to be the mesh current

mesh current flowing around like this and I won one point 4 milliamps that's

flowing around like this in i1. 1.4 milliamps.

one of our independent variables

That's one of our independent variables.

now that we know I won let's just choose

Now that we know i1, let's just choose one

one of our original equations here and figure out what I 2 is we'll plug in 2

of our original equations here and figure out what i2 is.

times I 1 which is 1 point 4 milliamps

We'll plug in two times i1, which is 1.4 milliamps,

minus 8 I 2 equals 6 and now we have to

minus eight i2 equals six.

figure out I 2 so let's solve this minus

And now we have to figure out i2. So let's solve this.

8 I 2 equals 6 minus 2 times 1 point 4 is 2 point 8 I 2 equals 3 point 2 divided by minus 8 or I 2 equals minus 0 point 4 milliamps and we could draw a

Minus eight i2 equals six minus two times 1.4 is 2.8. i2 equals 3.2 divided by minus eight, or i2 equals minus 0.4 milliamps.

box around that too so let's put that

And we can draw a box around that too.

mesh current in and our mesh current was

So let's put that mesh current in. And our mesh current was.

that was I - and that equals minus 0.4 milliamps and what that means is that

That was i2, and that equals minus 0.4 milliamps.

the current is not as flowing in the

And what that means is that the current is flowing

negative direction so it's actually

in the negative direction,

flowing this way through these resistors

so it's actually flowing this way through these resistors.

it's actually flowing in the counterclockwise direction is the

It's actually flowing in the counterclockwise direction

positive current and from before we

is the positive current.

already knew I 3 I 3 equals -2 milliamps so let's do one last thing here let's

And from before we already knew i3. i3 equals minus two milliamps.

figure out the the currents in the

So let's do one last thing here, let's figure out the currents in the elements.

elements let's let's just figure out a

Let's just figure out a couple of them,

couple of them that one and that one

that one and that one.

so that is I r2 and this one is I are four I are two equals I to - oops I 1

So that is ir2 and this one is ir4. ir2 equals i2 minus, oops,

minus I 2 equals I 1 is 1 point 4 minus negative point 4 and so that current equals 1 point 8 milli amps and we'll do

i1 minus i2 equals, i1 is 1.4 minus negative .4, and so that current equals 1.8 milliamps.

one last one will do i r4 that's this art this is our 4 here and I are for is i2 minus i3 so that

And we'll do one last one, we'll do ir4, this is r4 here. And ir4 is i2 minus i3.

equals i2 is minus 0.4 minus i3 i3 is

So that equals, i2 is minus .4, minus i3.

minus 2 milliamps so I R 4 equals 2

i3 is minus two milliamps. So ir4 equals

minus 0.4 or 1 point 6 milliamps and

two minus .4, or 1.6 milliamps.

there you have it we used the mesh

And there you have it,

current method to solve our circuit

we used the mesh current method to solve our circuit.

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