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.