Welcome back.
This is a lesson on Systematic Solution
Methods.
I've actually broken this lesson into two
parts, so they'll be on two different
videos.
Looking at this module, we've covered a
lot of
background material so far, and in these
two lessons.
We will be doing the Systematic Solution
Methods.
From the previous lesson, we demonstrated
the superposition principle,
for circuits with multiple sources.
Now, the lesson that we're covering today,
we're going to be introducing several
different methods.
And it doesn't matter if we've got one
source, or many sources,
there are systematic ways of solving the
problems of finding voltages and currents.
They are all based on a physical behavior
and the physical laws that we've already
covered.
Ohm's Law, the Kirchhoff's Voltage Law, or
KVL,
the Kirchhoff's Current Law or KCL.
Let's give you an overview of these
different methods.
There's some Mesh analysis and Node
analysis.
Now, these two are both algebraic in
nature.
In one case, the Mesh analysis, we're
going to be looking at Systematic
KVLs, systematic applications of the
Kirchhoff's
Volt, Voltage Law, to obtain simultaneous
equations
for currents.
And in the Node analysis, we'll be
applying the KCL.
Systematically to obtain simultaneous
equations for voltages.
So in both cases, they're algebraic.
The Thevenin and Norton equivalent
circuits are more graphical in nature.
We look at a circuit, and we look at the
schematic of that
circuit, and we try to reduce down that
schematic to a simpler form.
So it's a little bit more graphical in
nature.
So in this part of the
lesson, part one, we'll be doing the mode,
the node and Mesh
Analysis and then, the Thevenin and Norton
equivalent circuits in the second part.
Let's take a look at Mesh Analysis.
There are several steps to do, first is
to define mesh currents, one for each
noninclusive loop.
By non-inclusive, I mean a loop that
doesn't contain other loops.
For example, here we have three
non-inclusive loops.
And then we're going to do a KVL around
each loop.
And these, I've defined these currents as
what I'm going to call mesh currents.
Now these are the unknown parameters here.
I've introduced some of this intermediate
variables that I need to solve for.
So I do a KVL around each loop and then I
solve for those quantities.
Now let me go back and remind you of a
couple of things because
this is something that some students have
trouble with and that's plarity.
Just going back to Ohm's law.
If I have i current going into a resister,
the corresponding voltage is here.
V is equal to iR.
In other words, the plus side of the
voltage
is the same direction as the current
enters that element.
Now, what happens
when I've got.
Two components to my current, an i sub one
and i sub
two and they're going in different
directions.
Well, I can equate that to something with
like the equivalent
Current which is the difference.
So this is i sub 1 minus i sub 2 in
this direction and then the corresponding
voltage.
Well, that V.
Is equal to R times i1
minus i2.
Now going back to this problem,
I want to do a KVL around each loop just
keeping this, these concepts in mind.
So as I do a KVL.
I look at the loop and they and I'm doing
it in the direction of the
current, and I determine if I use a plus
or a minus based
on what the sign is of the voltage as I
approach that element.
So, as I approach this from this
direction, I see a minus first.
So I'm going to put a minus.
So looking at I1, net nash I put a minus
V1 first, plus coming around I want, well,
the currents
going into this resistor.
So that's a plus
voltage.
R1, I1 plus R2, in this case,
we've got I1 going in this direction and
then I3 going in this direction.
So the
net going this direction is I1-I3.
And that has to sum to zero.
Now, I2 is a pretty easy mesh.
This current I2 has to equal to source
current.
So now we go to I3.
Okay.
So, if I start right here, I hit that V2
first, so the plus sign, so it's going to
be a plus V2, plus, and then the voltage
plus R2, will then, I'm going in this
direction.
The current, total current will be I3,
minus I1.
Okay.
Coming around this way, I get this
voltage, which is R3 times that current.
Again, if I'm going in this direction,
it's I3 minus I2.
[BLANK_AUDIO].
But I already know what I2 is, I've solved
for that and I'll make that substitution I
sub s.
And then I've got one more element and
that's
going around this way, so that's R0 times
I3.
[BLANK_AUDIO].
And that has to equal 0.
So, I really only have two unknowns now,
I1 and I3 and
I want to write all of my equations of
terms of I1 and I3.
So, let me take this, this one right up
here.
And I'm going to rewrite it.
And I'm going to pull out these unknowns.
So it's I1 times R1, plus
R2, plus I3 times a minus
R2, has to equal V sub 1.
So I put all my known quantities on right
and all
my unknown quantities on the left
multiplied by whatever their coefficients
are.
In this case, we are assuming if you
actually had
to solve for this you would know what
these resistors are.
And you would know what this voltage
source is.
So this becomes one of the equations you
want to solve.
Now let's do the same thing with this
other equation.
I'm going to pull out my I1 and I3.
Or.
Other I1 here.
The only term involving I1 is the R2 plus
I3.
And I get an R2 plus
R3.
Plus R0.
And I put all of my known quantities on
the
right, which means that minus V2 plus R3 i
s.
Now this is my second equation.
Again, assuming that you knew all these
resistors and you
knew these sources, the only unknowns are
I1 and I3.
So you want to solve those using
a simultaneous equation solver, or by back
substituting.
For example, solving for I sub 3 in this
equation, in terms of.
All the rest of the quantities.
And then plugging that back in to I3 right
here.
There's a, a link to an online equation
solver in
the resources part of this course on the
course site.
So once I solve for I1 and I3, I can go
back and solve for V0.
V0 was the ultimate thing I might have
been interested
in here, but that's the last thing I solve
for.
I have to write all my equations in terms
of mesh currents, and
that at the very end I can go back and
solve for V0.
V0 is equal to I3 times R0.
So to summarize the mesh method.
I define the mesh currents.
One for each non-inclusive loop.
I do a KVL around each loop.
Always trying to solve for the only
unknowns
that I need which are they mesh currents.
So I have to write everything in terms of
mesh currents as my only unknowns.
And then I reduce
this down to simultaneous equations with
these unknowns.
I solve for them using some sort of
linear equation solver or back
substituting and then
at the very end I can go back and solve
for any other quantities I need.
So that's Mesh Analysis.
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Now let's look at node analysis.
In this particular thing the, the steps
are first defined at ground node.
Well, a ground node here, I usually pick a
ground node to
be one that I know that can, is connected
to several elements like
this one is connected to several elements
and the other thing that
makes my life a little bit easier is if I
pick a node.
That is next to a voltage source.
So I'm going to pick this node.
It is actually
next to two voltage sources.
And I want to pick that node as my ground.
And then I defined node voltages.
So I look at my nodes and say, what are
the voltages?
With respect to that ground.
So I'm going to call this V sub A.
That's one node V sub B is another node V
sub C is another node and I'll
define this as V sub D as another node.
And I want to solve for these
node voltages.
Now in this particular one, this one is
easy
to solve for because the only thing
between that node.
And the ground is this source.
So that's V sub 1.
And similarly, V sub d is equal to V sub
2.
So that means the only other unknowns I
have are V sub c and V sub b.
To solve for
those, I have to do a KCL at those two
nodes.
Now, I'm going to give you a little
background here to figure
out how to find the currents leaving those
or, or entering those nodes.
So for example, suppose I've got a
resistor between two node
voltages, V sub a and V sub b If I wanted
to find
the current in this direction, well that
current
i is equal to V sub a minus V sub b, all
over R.
So this current leaving V sub a is equal
to V sub a minus V sub b over R.
And the second fact that I'm going to be
using, in solving this, is that, I'm going
to be,
looking at the currents, leaving the node.
So I'm going to be summing up
all the currents leaving the node, and
that has to equal 0.
So the two unknowns I'm solving for V sub
b and V sub c.
Let's do a KCL at those two nodes.
At V sub b, at node b.
let's look at the
current leaving the node.
So, the current in this
direction is a simple one that's i sub s
and then the current leaving
in this direction is going to be V sub b
minus V sub c over R3.
And then the current going in this
direction
was V sub b, well and then this is ground
here
so this voltage is just V sub b over R,
and
then the current is over R2 plus V sub b.
Over R2.
And then the last one is the current in
this direction.
That's V sub b minus V sub 1, because
we've already solved for that node, all
over R1, has got to equal 0.
And I'm trying to solve for Vb and Vc.
So I'm going to write a, an equation that
pulls out those variables.
So V sub b, so I'm just rewriting this, V
sub b is 1 over R3 plus 1over R2 plus 1
over
R1, plus V sub c minus 1 over R3 and then
everything else is known quantities.
I sub S, we are assuming we know that and
we are assuming we know V sub 1 So
I've got a minus i sub
s plus V1 over R1.
And this becomes one of the equations I
need to solve using, for example,
linear equation solver.
Now let's look at V sub c.
And I wanted some of the currents leaving
that node.
So this, starting in this direction, this
one's easy.
That's, i of s comes in.
That means what's leaving is negative i
sub s.
Okay.
And then the current going down and
through this link through this branch is.
V sub c minus V sub d which I've already
found
to be V2 over R0 plus the current going in
this
direction is V sub c minus V sub b over
R3.
All of that has to equal 0.
Well, I want to rewrite this equation,
pulling out my unknown node voltages.
So V sub b times,
let's see, I gotta minus 1 over
R3, plus V sub c.
1 over
R0 plus 1 over R3 and everything
else in this equation is a known
quantity.
I'm going to put that on
the right, and that's equal to i sub s
plus.
V two over R0, that's
my second equation.
So now, I have
two equations into a node and I can solve
for example for V
sub c in terms of Vb and substitute that
in for Vc here.
Or I can use, once I have numbers in
here, actual values, I can use an equation
solver.
Once I solve for Vb and Vc, I can solve
for V0 as the difference between them.
V0 is equal to Vc minus V2.
So to summarize.
In node analysis, I first select the
ground node.
And again I want to find something that
connects a lot of elements, plus
is right next to voltage source if I can
do that makes it simpler.
And then I define my node voltages at
every
node and I solve for those that are
unknown.
By going back and do a KCL at every one of
those nodes.
And then I solve for those node voltages,
in this case Vb and Vc,
and at the end I go back and solve for
whatever quantity I was interested.
So the first thing I have to solve for, is
my node voltages, and then
once I do that, I solve for any other
parameters
or values in the in the system that I'm
interested in.
The next lesson will cover the details of
Thevenin equivalent circuits and Norton
equivalent circuit methods.
And gives some examples.
I do want to mention that there's extra
work to problems available for these two
lessons.
And I do also want to mention.
That we're introducing these different
technologies,
these different methodologies, for you.
And I don't care, in working the
solutions, which ones of these that you
use.
So work some of these problems, feel
comfortable
with one of them, and then use that.
And as long as you're able to solve the
problem, how you do it is fine.