L3 2 5Super Mesh Example with numbers

Example Mesh Current analysis with Super Mesh

In this video, we're gonna
demonstrate the super mesh technique.

As you look at this,

you'll notice that there are three meshes,
this one here to the left.

And we'll label the current
in that mesh current I1.

We have a mesh up here.

And we'll call this mesh current I2, and

we have a mesh here with a mesh
current that we'll label I3.

And further before we even get started
we'll notice that separating mesh one from

mesh three is a branch that
contains a current source.

And as we've mentioned in the past,
we don't have a.

A numerical or mathematical
relationship between the voltage across

a current source and the current itself.

That will then require us to
do a super mesh which we will

Be writing around there.

Let's get started first of all
though by writing the mesh

current equation around
this mesh number one.

Starting down here in the lower left hand
corner, we have going minus to plus.

It will be a negative 12 volts.

And then across this resistor of a voltage
drop of five resistance five times current

rate through that which is I1-12

plus the volts struck across this
film resistor which is going to be 3

times I 1- I3 and

the sum of those voltage
drop must equal 0.

Now, the Super Mesh starting here at this
point going up and around the 15 ohm, and

when we're across the 15 ohm resistor now
the voltage drop there being 15 times I2.

Now continuing on down here,

not going around this branch here because
that's where the current source is.

Continuing on down here across the 10
ohm resistor we have a voltage drop

of +10 x (I3) Plus,

now continuing around here,
across the 3 ohm resistor, +3 times.

Now the current going in
this direction is I3-I1.

I3-I1+ the voltage

drop across that 5 ohm
resistor Is five times.

And again,
because we're going from right to left,

the current going in that
direction will be I2 minus I1.

I2 minus I1,
the sum of those terms must equal 0.

And finally the third equation we need for

these three unknowns comes from
the relationship between I2, I3.

And this current source.

In this case, I2 is going in
the direction of the current source, so

we're going to have I2 minus I3,

which is the current going through that
branch in terms of the mesh currents.

It's going to equal 2 Amps.

So we have our three equations and
three unknowns.

Let's go ahead and combine terms.

Taking the first equation
we have I1 times,

we got 5 there plus 3 is 8 plus

I2 times We have a negative 5

Plus I3, we have a negative 3.

And then we have this negative 12 volts on
the left-hand side we bring to the other

side as a positive 12 volts.

And there is our first equation.

Second equation combining like terms,
we have I1.

And in this we have -3 and a -5.

That makes -8 times I1 + I2 times,

we got 15

plus five is 20 times i2 plus i3, we've

got 10i3 plus 3 is 13i3 equals, and

there are no constants there So

the sum of those terms equals 0.

Finally, our third equation is I2,

Times 1 + I3(-1) = 2.

And we have our three equations and
three unknowns so

get ready with your calculator or whatever
you choose to solve through your systems

through your equations and three unknowns.

And when you do we get the following.

I1=2.03 amps.

I two is equal to one point two eight amps

and I three is equal to negative

point seven two amps.

LEts do a couple of consistency checks
here just to make sure we didn't make any

mistakes not to prove that our work was
correct but at least to look at a couple

of calculations that would be consistent
with what we know to be true.

For example we know that i two
minus i three equals two amps.

So I2, that is 1.28

minus I3, well I3 is a -0.72 amps.

The sum of those,
they're supposed to equal 2 amps,

and sure enough that works there.

And of course we could plug those values
into any one of those equations And

proves that our solutions are correct.

But now that we know what I1,
I2, and I3 are.

Let's use them to say for

example now determine what the voltages
drop is across this current source.

We said that we couldn't numerically
calculate it based upon the current going

through it, but now that we know
the currents around it, we can write a KVL

once again around that loop, and
solve for the voltage across the current.

Starting here and going clockwise,
across that 3 ohm resistor,

we're going to have Three
times I three minus I one.

Now, going across this current source with
the voltage as referenced plus to minus.

That represents a voltage drop,
so it would be plus v.

Now, coming down across here,
that would be plus 10.

Oops, plus 10 times I3

equals that's where we started from, 0.

Now I plug in our values for I1, I2,

and I3 and we have 3 times I [SOUND] So

3 times I3, well, I3 is a -0.72 amps,

-0.72,- I1, which is

2.03 amps, + V + 10 times I3,

which is a negative 0.72 amps.

The sum of those things has to equal 0.

Solving this for V gives us

V = 15.45 volts.

All righty, now looking at this source,
we see that the current is reference into

the negative terminal of our voltage and
leaving the positive terminal.

So it's acting as a source.

And we can then say that the power being
generated by that source is equal to

negaitive i times v Is equal to negative i

which is two times v which is 15.45 and
the power

then would be equal to
an negative what is that

30.9 watts the fact that

it's negative means that it's energy being
put into the circuit from the source.

So once again, once we know all
three of the mass/f currents,

we can determine any voltage or
current or for that matter power or

any other quantity that we might
want to know within this circuit.