Good morning, OK.
Let's get started. We have one handout today.
That's your lecture notes. There's some copies still
outside for those who haven't picked one up.
In general, what I do is, in the lecture notes,
I leave out large amounts of material.
So, this will enable you to keep your hands busy while I'm
lecturing and take down some notes and so on.
So, don't assume that everything that I talk about is
on here. Please follow along.
OK, so as is my usual practice, let me start with a quick
review of what we covered so far.
So what we did primarily was looked at this discipline that
we call the lump matter discipline, which was very
similar, very reminiscent of the point mass simplification in
physics. And this discipline,
this set of constraints we imposed on ourselves,
allowed us to move from Maxwell's equations to a very,
very simple form of algebraic equations.
And specifically, the discipline took two forms.
One is, we said that we will deal with elements for whom the
rate of change of magnetic flux is zero outside of the elements,
and for whom the rate of change of charge I want to charge
inside the element was zero. So, if I took any element,
any element that I called a lump circuit element,
like a resistor or a voltage source, and I put a black box
around it, then what I'm saying is that the net charge inside
that is going to be zero. And this is not true in
general. We will see examples where,
if you choose some piece of an element for example,
there might be charge buildup, but net inside the,
if I put a box around the entire element,
I am going to assume that the rate of change of charge is
going to be zero. So, what this did was it
enabled us to create the lump circuit abstraction,
where I could take elements, some element of the sort,
this could be a resistor, a voltage source,
or whatever, and I could now ascribe a
voltage, some voltage across an element, and also some current,
"i," that was going into the element.
And as I go forward, when I label the voltages and
currents across and through elements, I'm going to be
following a convention. OK, the convention is that I'm
going to label, if I label V in the following
manner, then I'm going to label "i" for that element as a
current flowing into the positive terminal.
It's just a convention. By doing this,
it turns out that the power consumed by the element is "vi"
is positive. OK, so by choosing I going in
this way into the positive terminal, the power consumed by
the element is going to be positive.
OK, so in general of even simply following this
convention, when I label voltages and currents,
I'll be labeling the current into an element entering in
through the plus terminal. Remember, of course,
if the current is going this way, let's have one amp of
current flowing this way, then when I compute the
current, "i" will come out to be negative.
OK, so by making these assumptions, the assumptions of
the lumped matter discipline, I said I was able to simplify
my life tremendously. And, in particular what it did
was it allowed me to take Maxwell's equations,
OK, and simplify them into a very simple algebraic form,
which has both a voltage law and a current law that I call
Kirchhoff's voltage law, and Kirchhoff's current law.
KVL simply states that if I have some circuit,
and if I measured the voltages in any loop in the circuit,
so if I look at the voltages in any loop, then the voltages in
the loop would sum to zero. OK, so I measure voltages in
the loop, and they will sum to zero.
Similarly, for the current, if I take a node of a circuit,
if I build the circuit, a node is a point in the
circuit where multiple edges connect.
If I take a node, then the current coming into
that node, the net current coming into a node is going to
be zero. OK, so if I take any node of
the circuit and sum up all the currents going into that node,
they will all net sum to zero. So, notice what I've done is by
this discipline, by this constraint I imposed on
myself, I was able to make this incredible leap from Maxwell's
equations to these really, really simple algebraic
equations, KVL and KCL. And I promise you,
going forward to the rest of 6.002, if this is all you know,
you can pretty much solve any circuit using these two very
simple relations. It's actually really,
really simple. It's all very simple algebra,
OK? So, just to show you an
example, let me do a little demonstration.
Let me build let me build a small circuit and measure some
voltages for you, and show you that the voltages,
indeed, add up to zero. So, here's my little circuit.
So, I'm going to show you a simple circuit that looks like
this, and let's go ahead and measure some voltages and
currents. In terms of terminology to
remember, this is called a loop. So if I start from the point C
and I travel through the voltage source, come to the node A down
through R1 and all the way down through R2 back to C,
that's a loop. Similarly, this point A is a
node where resistor R1 the voltage source V0,
and R4 are connected. OK, just make sure your
terminology is correct. So, what I'll do is I'll make
some quick measurements for you, and show you that these KVL and
KCL are indeed true. So, the circuits up there,
could I have a volunteer? Any volunteer?
All you have to do is write things on the board.
Come on over. OK, so let me take some
measurements, and why don't you write down
what I measure on the board? What I'll do is,
let me borrow another piece of chalk here.
What I'll do is focus on this loop here, and focus on this
node and make some measurements.
All right, so you see the circuit up there.
OK, so I get 3 volts for the voltage from C to A.
so why don't you write down 3 volts?
OK, so the next one is -1.6. And so that will be,
I'm doing AB, V_AB.
OK, and then let me do the last one.
It is -1.37. The measurements,
I guess, have been this way. So, what's written is V_AC.
But it's OK for now. Don't worry about it.
So, well, thank you. I appreciate your help here.
OK, so within the bonds of experimental error,
noticed that if I add up these three voltages,
they nicely sum up to zero. OK, next let me focus on this
node here. And at this node,
let me go ahead and measure some currents.
What I'll do now is change to an AC voltage so that I can go
ahead and measure the current without breaking my circuit.
OK, this time around, you'll get to see the
measurements that I'm taking as well.
So, what I have here, I guess you can see it this
way. What I have here is three wires
that I have pulled out from D. And this is the node D,
OK? So, I have three wires coming
into the node D just to make it a little bit easier for me to
measure stuff. OK, so everybody keep your
fingers crossed so I don't look like a fool here.
I hope this works out. So, you roughly get,
what's that, 10 mV.
OK, so it's about 10 mV peak to peak out there,
and let's say that if the waveform raises on the left-hand
side, it's positive. So, it's positive 10 mV.
And another positive 10 mV, so that's 20 mV.
And this time, it's a negative,
roughly 20, I guess, -20.
So, I'm getting, in terms of currents,
I have a -10, -10, I'm sorry,
positive 10, positive 10,
and a -20 that adds up to zero. But more interestingly,
I can show you the same thing by holding this current
measuring probe directly across the node.
And, notice that the net current that is entering into
this node here is zero. OK, so that should just show
you that KCL does indeed hold in practice, and it is not just a
figment of our imaginations. So, before I go on,
I wanted to point one other thing out.
Notice that I've written down two assumptions of the lumped
matter discipline, OK?
There is a total assumption of the lump matter discipline,
and that assumption is, in spirit, at least,
shared by the point mass simplification in physics as
well. Can someone tell me what that
assumption is? A total assumption,
which I did not mention, which you can read in your
notes in section 8.2 in the appendix, what's a total
assumption that is shared in spirit with the point mass
simplification? Anybody?
A total assumption to be made here is that in all the signals
that we will study in this course, we've made the
assumption that the signal speeds of interest,
transition speeds, and so on, are much slower than
the speed of light. OK, that my signal transition
speeds of interest are much slower than the speed of light.
Remember, the laws of motion, the discrete laws of motion
break down if your objects begin moving at the speed of light.
OK, the same token here, our lump circuit abstraction
breaks down if we approach the speed of light.
And there are follow on courses that talk about waveguides and
other distributed analysis techniques that deal with
signals that travel close to speeds of light.
OK, so with that, let me go on to talking about
method one of circuit analysis. This is called the basic KVL
KCL method. So just based on those two
simple algebraic relations, I can analyze very interesting
and complicated circuits. The method goes as follows.
So, let's say our goal is, given a circuit like this,
our goal is to solve it. OK, in this course,
we will do two kinds of things: analysis and synthesis.
Analysis says, given a circuit,
OK, what can you tell me about the circuit?
OK, so we'll solve existing circuits for all the voltages
and currents, voltages across elements,
and currents through those elements.
Synthesis says, given a function,
I may ask you to go and build circuits.
OK, so for analysis here, we can apply this method that I
want to show you. And the idea here is that,
given a circuit like this, let us figure out all the
voltages and currents that are a function of the way these
elements are connected. So, the basic KVL and KCL
method has the following steps. The first step is to write down
the element VI relationships. OK, right down the element VI
relationships for all the elements.
The second step is write KCL for all the nodes,
and the third step is to write KVL for all the loops in the
circuit. That's it.
Just go ahead and write down element rules,
KVL, and KCL, and then go ahead and solve the
circuit. So, what we'll do,
we'll do an example, of course.
But, just as a refresher, we've looked at a bunch of
elements so far, and for the resistor,
the element relation says that V is pi R, where R is the
resistance of the element here. For a voltage source,
V is equal to V nought. That's the element
relationship. And for a current source,
the element is the relation is, "i" is simply the current
flowing through the element. OK, so these are some of the
simple element rules for the devices that the current source,
voltage source, and the resistor.
So let's go ahead and solve this simple circuit.
And what I'll do is go ahead and solve the circuit for you.
OK, if you turn to page five of your notes, I'm going to go
ahead and edit the circuit here. You can scribble the values on
your notes on page five. OK, so as a first step of my
KVL KCL method, I need to write down all my
element VI relationships. So, before I do that,
let me go ahead and label all the voltages and currents that
are unknowns in the circuit. So, let me label the voltages
and currents associated with the voltage source as here.
Notice, I continue to follow this convention where whenever I
label voltages and currents for an element, I will show the
current going into the positive terminal of the element
variable, OK, after element variable voltage.
So here, I have V nought and I nought.
Let me pause here for five seconds and show you a point of
confusion that happens sometimes.
Often times, people confuse between what is
called the variable that is associated with the element
versus the element value. OK, notice that here,
capital V nought is the voltage that this voltage source
provides, while this name here, v nought, is simply a variable
that we've used to label the voltage across that element.
So, similarly, I can label v1 as the voltage
across the resistor, and i1 is the current flowing
through the resistor. So this method of labeling,
where I follow the convention, that the current flows into the
positive terminal is called the associated variables discipline.
I was trying to use the word discipline in situations where
you have a choice, OK, and of a variety of
possible choices, you pick one as the convention.
OK, so here, as a convention,
we use the associated variables discipline, and use that method
to consistently label the unknown voltages and currents in
our circuits. OK, so let me continue the
labeling here, v4, i4, i3, v3 here,
and v2 and i2, v5, and i5.
I think that's it. So, I've gone ahead and labeled
all my unknowns. So each of these voltages and
currents are the voltages and currents associated with each of
the elements. And my goal is to solve for
these. OK, so in terms of our solution
here, let's follow the method that I outlined for you.
So, as the first step I am simply going to go ahead and
write down all the element VI relationships.
OK, so as a first step, I'm going to go ahead and write
down all the VI relationships. So, can someone yell out for me
the VI relationship for the voltage source?
OK, good. So, v0 is capital V nought,
that is that the variable V nought is simply equal to the
voltage, v0. Similarly, I can write the
others. v1 is i1, R1.
v2 is i2, R2, and so on.
OK, and I have one, two, three, four,
five, six elements. So, I will get six such
equations. Step two, I'm going to go ahead
and write KCL for the nodes in my system.
So, let me start with node A. So, for node A,
let me take as positive the currents going out of the node.
So, I get i nought flowing out, plus i1 flowing out,
plus i4 flowing out, and they must sum to zero for
node A. Then, I can go ahead and do the
other nodes, let's say, for example,
I do node B. For node B, I have i2 going
out. That's positive,
i3, and i1 is coming in, so I get -i1 equals zero.
OK, so I have one, two, three, four,
I have four nodes. OK, so I would get four
equations. It turns out that the fourth
equation is not independent. You can derive it from the
others. So, I get three independent
equations out of this. I can then write KVL.
And for KVL, I just go down my loops here.
And let me go through this first loop here in this manner.
OK, and a simple trick that I use, you have to be incredibly
careful when you go through this in keeping your minuses and
pluses correct. Otherwise you can get
hopelessly muddled. Once you label it,
you need to be sure that you get all your minuses and pluses
correct. So, for KVL,
what I'd like to do is, let's say I start at C,
and from C I'm going to go to A.
For A I go to B, and from B I'm going to come
back to C. OK, that's how I traverse my
loop. And, the trick that I'm going
to follow is, as my finger walks through that
loop, I'm going to label the voltage as the first sign that I
see for that voltage. OK, so I'm going to start with
C, and I go up. I start by punching into the
voltage source element, and then punch into it,
I hit the minus sign for the V nought.
OK, so I'm just going to write down minus V nought,
plus then I go through and as I come up to A and go down to B,
I punch to the plus sign of the V1.
So, that's plus V1. And then I punch into the plus
sign of the V2, and so I get plus V2,
and that is zero. OK, good.
So, that matches what you have in your notes as well.
So, this is the first equation. Similarly, I can go through my
other loops and write down equations for each of the loops.
OK, and the convention that I like to follow is as I go
through the loop, I write down as a sign for the
voltage the first sign that I counter for that element.
OK, you can do the exact opposite, if you want,
just to be different. But, as long as you stay
consistent, you'll be OK. All right, so in the same
manner here, there are four loops that I can have,
so four equations. Again, one of them turns out to
be dependent on the others. So I end up getting three
independent equations. So, I get a total of 12
equations. I get 12 equations.
There are six elements, OK, voltage source,
and five resistors. So, there are six unknown
voltages, and six unknown currents.
So, I have 12 equations, and 12 unknowns.
OK, I can take all of the equations and put them through a
big crank, and sit there and grind.
And if I was really cruel, I'd give this as a homework
problem, and have you grind, and grind, and grind until you
get your six voltages and six currents.
OK, it works. OK, so you get 12 equations,
and this method just works. However, notice that this is
quite a grubby method. It's quite grungy.
I get 12 equations, and it's quite a pain even for
a simple circuit like this. However, suffice it to say that
this fundamental method is one step away from Maxwell's
equations, simply works. OK?
So what you'll do is the rest of this lecture,
I'll introduce you to a couple more methods.
One is an intuitive method, and another one called the node
method is a little bit more formal, but is much more,
I guess, terse Than the KVL KCL method.
Method 2. So the relevant section to read
in the course notes is section
One of the things that I will be stressing this semester is
intuition. What you'll find is that as you
become EECS majors, and so on, and go on,
or if you talk to your TAs or your professors and so on,
you will find that very rarely do they actually go ahead and
apply the formal methods of analysis.
OK, by and large, engineers are able to look at a
circuit and simply by observation write down an
answer. And usually in the past,
what we have tried to do is kind of ignore that process and
told our students, look, we teach you all the
formal methods, and you will develop your own
intuition and be able to do it. What we'll try to do this term
is try to stress the intuitive methods, and try to show you how
the intuitive process goes, so you can very quickly solve
many of these circuits simply by inspection.
OK, so this method that I'm going to show you here is one
such an intuitive method. And I'll call it element
combination tools. OK, for many simple circuits,
you can solve them very quickly by applying this method.
The components of this method are these.
I learned about how to compose a bunch of elements.
So, let's say, for example,
I have a set of resistors, R1 through RN,
in series. OK, you can use KVL and KCL to
show that this is equivalent to a single resistor whose value is
given by the sum of the resistances.
OK, so if I have resistors in series, then effectively it's
the same as if there was a single resistor whose value is
the sum of all the resistances. OK, you can look at the course
notes for a proof for derivation of this fact.
Similarly, if I have resistances in parallel,
so let me call them conductances.
A conductance is the reciprocal of a resistance.
If resistance is measured in ohms, conductance is measured in
mhos, M-H-O-S. OK, so that's the conductance
is G1, G2, and G3. And effectively,
this is the same as having a single conductance whose
effective value is given by the sum of the conductances.
OK, the conductances in parallel add,
and resistances in series add. Similarly, for voltage sources,
if I have voltage sources in series, then they are tantamount
to the sum of the voltages. And similarly,
for currents, if I have currents in parallel,
then they can be viewed as a single current source,
whose currents are the sum of the individual parallel
currents. So, let's do a quick example.
So let's do this example. So, let's say I have a circuit
that looks like this, and three resistances.
And let's say all I care about is the current,
I, that flows through this wire.
All I care about is that current.
Of course, you can go ahead and write KVL and KCL.
You will get four equations, and there are four unknowns.
And you can solve it. But, I can apply my element
combination rules, and very quickly figure out
what the current I is, using the following technique.
So, what I can do is, I can, first of all,
take this circuit. And, I can compose these two
resistances and show that the circuit is equivalent as far as
this current, I, is concerned to the
following circuit, R1.
And I take the sum of the two conductances,
OK, and that comes out to be R1, R2, R3, R2 plus R3.
And then, I can further simplify it, and I get a single
resistance, whose value is given by R1 plus R2,
R3, R3. OK, I'm just simplifying the
circuit. Now, from this circuit,
I can get the answer that I need.
I is simply the voltage, V, divided by R1 plus.
OK, so in situations like this where I'm looking for a single
current, I can very quickly get to the answer by applying some
of these element combination rules.
And, I can get rid of having to go through formal steps.
So, in general, whenever you encounter a
circuit, by and large attempt to use intuitive methods to solve
it. And go to a formal method only
if some intuitive method fails. Even in your homework,
by and large, the homeworks are not meant to
be grungy. OK, if you find a lot of grunge
in your homework, just remember you're probably
not using some intuitive method. OK, so just be cautious.
All right, so let me go on to the third method of circuit
analysis, and the third method is called the node method.
So, the node method is simply a specific application of the KVL
KCL method and results in a much, much more compact form of
the final equations. If there's one method that you
have to remember for life, then I would say just remember
this method. OK, the node method is a
workhorse of the easiest industry.
OK, if there's one method that you want to consistently apply,
then this is the one to remember.
So, let me quickly outline for you to method,
and then work out an example for you.
The first step of the node method will be to select a
reference or a ground node. This is the symbol for a ground
node. The ground node simply says
that I'm going to denote voltages at that point to be
zero, and measure all my other voltages with reference to that
point. So, I'm going to select a
ground node in my circuit. Second, I want to label the
remaining voltages with respect to the ground node.
So, label voltages for all the other nodes with respect to the
ground node. Next, write KCL for each of the
nodes write KCL. OK, but don't write KCL for the
ground node. Remember, if you have N nodes,
the node equations will give you N-1 independent equations.
So, write KCL for the nodes, but don't do so for the ground
node. Then, solve for the node
voltages. So, let's say when we label
voltages. I want to be labeling them as E
something or the other. So, solve for the unknown node
voltages. And then, once I know all the
voltages associated with the nodes, I can then back solve for
all the branch voltages and currents.
OK, once I know all the node voltages, I can then go ahead
and figure out all the branch voltages and the branch
currents. So, let's go ahead and apply
this method, and work out an example.
Again, remember, if there's one method that you
should remember, it's the node method.
OK, and when in doubt, consistently apply the node
method and it will work whether your circuit is linear or
nonlinear, if the resistors are built in the US or the USSR it
doesn't matter. OK, the node method will simply
work, linear or nonlinear, OK?
So, what I'm going to do is I'm going to build a circuit that's
my old faithful. It's our old faithful,
plus I'll make it a little bit more complicated by adding in
the current source. So, let's go have some fun.
Let's do this. So here's my voltage source,
as before. OK, what I'll do is for fun,
add a current source out there. And, you can convince
yourselves that if you go ahead and apply the KVL KCL method,
it'll really be a mess of equations.
OK, so R1, R3, R4, R2, R5.
OK, so let's follow our method and just plug and chug here.
So let's apply the first step. I select a ground node.
It's a reference node from which I'll measure all my other
voltages. OK, now without knowing
anything about the node method, try to use intuition as to
which node you should choose as a ground node.
Remember, you want to label the ground node with the voltage
zero, and measure all the other voltages with respect to that
node. OK, a usual trick is to pick a
node which has the largest number of elements connected to
it as the ground node. OK, and in particular,
you will find out later it's useful to pick a node in which
all your voltage sources, the maximum number of your
voltage sources are also connected.
OK, so in this instance, I'm going to choose this as my
ground node. OK, that's my first step.
I chose that as my ground node. And I'm going to label that as
having a voltage zero. Second step,
I'll label voltages of the other branches with respect to
the ground node. OK, so what I'll do is add this
node here. So I'm going to label that
voltage E1. These are my unknowns.
Remember, node method, because my node voltages are my
unknowns. So, I label this as E1.
I label this one as my unknown voltage, E2.
What about this one here? Is that voltage unknown?
No. I know what the voltage is
because I know that this node is at a voltage,
V0, higher than the ground node.
OK, notice that to go from here to here, I directly go through a
voltage source. And so, this node has voltage
V0. And I'll simply write down V0.
OK, try to simplify the number of steps that you have to go
through, so directly go ahead and write down the voltage,
V0, for that node. What I will also do,
is for convenience, I'm going to write down,
I'm going to use conductances. So I'm going to use GI in the
place of one by RI, and write down a bunch of node
equations. OK, so step one,
I've chosen my ground node. Step two, I've labeled my node
voltages, E, OK? I've done that with two of my
steps. Now, let me go ahead and --
OK, so let me go ahead and apply step three.
And, step three says go ahead and apply KCL for each of the
nodes at which you have an unknown node voltage.
And then that will give you your equations.
So let me start by applying KCL at E1.
So, let me write KCL at E1. I do one more thing.
Notice, I don't have any currents there.
OK, so how do I write KCL? KCL simply says the sum of
currents into a node is zero again, remember,
by the lump matter discipline. So, if I don't have currents in
there, so the trick that I adopt is that to write KCL,
I use the node voltages, and implicitly substitute for
the node voltages, divide by the elemental the
resistance, for instance, so I take the node voltages,
and divide by the resistance, get the current.
OK, so I implicitly apply element relationships to get the
node currents. So, the example that make it
clear, so I take node E1 and, again, for currents going out
I'm going to assume to have, to be positive.
So, the current going up is E1 minus V nought,
divide by R1, so I multiplied by the G1.
That's the current going up. Plus, the current going down is
E1 minus zero where the ground node potential is zero,
G2, OK, plus the current that is going to resistor R3,
which is simply E1 minus E2, divide by R3.
So, E1 minus E2, divide by R3,
or multiplied by G3 is equal to zero.
OK, see how I got this? This is simply KCL,
but to get my currents, I simply take the differences
of voltages across elements, and divide by the element of
resistance, and I get the currents.
OK, so I can similarly write KCL at E2.
So, at KCL at E2, again, let me go outwards.
So, the current going up is E2 minus V nought multiplied by G4.
The current going left is E2 minus E1 divided by R3 or
multiplied by G3. The current going down is E2
minus zero multiplied by G5. And, the current going down is
-I1. OK, you've got to be careful
with your polarities here. So all the currents going out
sum to zero. And here are the currents that
are going out at this point. So what I do next is I can move
the constant terms to the left-hand side and collect my
unknowns. So, let me write them out here.
So, let's say I get E1 here, OK, and from this equation,
I have a V nought, G1, which comes out here.
So, minus V nought G1 comes over to the other side.
And, let me collect all the values that multiply E1.
So I get, G1 is one example. I have G2, and I have G3.
And then, for E2, I have minus G3.
OK, so I'll simply express this as the element voltages
multiplied by some terms in parentheses, and I put my
external sources on the right hand side.
Similarly, I go ahead and do the same thing here.
In this instance, let me move my sources to the
right. So, I get I1 coming out there,
and I get V nought G4 coming out there.
By the way, I just want to mention to you that if you're
looking to fall asleep, this is a good time to do so
because as soon as I write down these two equations,
OK, from now on it's nap time. There's nothing new that you're
going to learn from here on. It's just Anant Agarwal having
fun at the blackboard, pushing symbols around.
So, once you write down these two node equations,
the rest of it is just grubby math.
So, let me just have some fun. So let me just go ahead and do
that. So, I moved my voltages and
currents to the other side. And let me collect all the
coefficients for E1 here. So, E1 minus G3,
and that's it, I guess.
OK, and then I'll do the same for E2.
So, I get G4, and I get G3,
and I get G5. OK, so notice here that I have
two equations, and two unknowns.
OK, the two equations are on the right hand side,
I have some voltages and currents which are my dry
voltages and dry currents. OK, so actually this is getting
quite boring. I'm going to pause here,
and talk about something else. So, you can take this and you
can put it in matrix form, so I've done that for you on
page ten. It's all matrix form.
Yeah, I know that. You can use any technique to
solve it, use algebraic techniques, use linear algebraic
methods to solve it, use a computer,
whatever you want. And, computers,
when computers analyze circuits, they write down these
equations, and deal with solving matrices.
So, when you take the linear algebra across,
how many people here have taken a linear algebra class?
How many people here have heard of Gaussian elimination?
How can more people have heard of Gaussian elimination than
took a linear algebra class? Well anyway,
so now you know why you took those linear algebra classes.
And so, if I just collected these into matrix form --
OK, so I just simply expressed those two equations in linear
algebraic form, and here's my column vector of
unknowns, and you can apply any of the techniques you've learned
in linear algebra to solve for this.
Gaussian elimination works. OK, and in computer,
people doing research in computer techniques,
or solving such equations simply deals with huge equations
like this, building computer programs that,
given equations like this, can go ahead and solve them.
OK, so let me stop here and reemphasize that what you've
done is made a huge leap from Maxwell's equations to using the
lump matter discipline to KVL and KCL, which ended up giving a
simple algebraic equation to solve, and not having to worry
about partial differential equations that were the form of
Maxwell's equations.