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For the last 2, 3 classes, we have been learning
how to obtain the first order
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differential equation. So, before going to
the electrical circuit specifically, let me
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give
some problems that you solve before coming
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to the next week’s classes. Firstly, all
the
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problem that you are done using the Lagrangian
method, where you have obtain the
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second order equations, you obtain the first
order equation for them. So, that is the think
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to be done.
But, in that assignment do not do the whole
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problem all over again, because you have
already drawn the physical system diagram
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obtain the kinetic energy, the potential
energy, the Lagrangian all that remains. So,
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in this assignment, what will do for those
problems, you just start from there, and then
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do the rest of the steps to obtain the first
order equations. In addition to that I will
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give some problems will you solve apiary from
first principles to obtain the first order
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equations.
.
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The first problem will consist of a mass,
which will be standing on a spring damper
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support on the ground and if we apply a force
that will move up and down depending on
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.the force. But, that is not is the complete
story, then is a
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lever and at the other end of it,
there is a mass hanging by a spring. So, there
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are two masses m 1 and m 2, there is the
spring constant k 1 and there is another k
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2, there is 1 r and there I have would be
also
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specify the ratio a and b.
For the sake of simplicity, assume that this
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rod will remain close to horizontal position.
So, do not really need to consider this angle
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that approximation you assume and go
ahead. So, in this case there are two masses
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and therefore, there will be two momentum
coordinates and two position coordinates.
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Four dimensional system really and then the
rest will be simply, is it simple, you do
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it, fine, so that is one.
.
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Two, I will give one electrical circuit to
be done using this method. There is a voltage
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source connected with a resistance and a capacitance
and then an inductor. Then, again
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an inductor and then again a capacitance,
I would like to add something more, fine.
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So, L
1, L 2, C 1, C 2, R 1, R 2 and E, for this
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obtain first order differential equations,
note it
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down.
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..
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Suppose, there is a cot like think, which
is able to move on the ground and this fellow
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has some mass say M 1 and it is being pulled
by a force F. And on this surface, there is
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a
friction, that is say R 1 is consolation,
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but here is a mass, that is connected by means
of a
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spring, so that is another mass M 2. And here
is the K spring constant, I suppose that
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complete the story, no, there will be some
kind of a friction.
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So, how many degrees or freedom does it have,
two position coordinates and two
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momentum coordinates. Because, now we are
talking in terms of the first order equation,
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when we do, so we will have to stock in terms
of both position as well as the momentum
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coordinates. We have already done the problem
in the class for the inverted pendulum;
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obtain the first order equation for that.
So, I am not again writing it, we are already
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done the problem, but we did it following
the Lagrangian method, so obtain the first
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order equation for that, so that is problem
number 4. So, I will write problem 5 here,
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this problem will concern the way, the loud
speakers work. So, how does the loud speakers
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work, it is a electro mechanical system,
loud speaker is an electro mechanical system,
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there is a magnet, which is energized and
that pulls a diaphragm which oscillates.
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The diaphragm pulling another stuff that is
the mechanical system, the other thing is
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electrical system and in between there is
a coupling two magnetism, which is similar
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to
the decimation problem that we did, so remember
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that. So, you have some kind of a
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.voltage source, so that is E t, the signal
which you want to be reproduced in sound.
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A
resistance and inductance will be there, because
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there is a winding, winding will always
have these things.
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And then you imagine that you apply this to
these two terminals and here there is some
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kind of a transducer that applies, that pulls
it with the force F, here is an iron piece,
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I am
just symmetrical drawing, it should not like
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that, but nevertheless. You understand that
here is something that converts the current
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signal into a force signal, to be applied
on the
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mechanical side.
What is there in the mechanical side? There
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must be some kind of a mass, which is
essentially representing the mechanical arrangement
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of the loud speaker. So, get the
sound here, fine, but not only that this fellow
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should not be hanging in the air that has
to
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be attach to some kind of a ground by means
of a spring damper arrangement. So, this is
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the mechanical ground and this arrangement
is connected to the mechanical ground by
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means of the spring and damper arrangement.
And then as this force is applied, it will
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oscillate, it will vibrate and that is what
is
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communicated into there. So, there will be
a mass here M, there is a spring constant
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K,
there is a resistance of the device RD may
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be and here is the R electrical and L electrical.
And here is the transducer is there anything
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I am missing, no and here is the current that
is I flowing, fine.
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So, how you will do it, as I have already
shown in the decimation problem, this can
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do in
two possible ways. In first, you take the
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electrical sub system and the mechanical sub
system separately, write down the separate
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t, v Hamiltonian and all that and then obtain
the differential equation separately. And
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then since you know how much is the force
depending on the current here and this also
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applies some kind of a back e m f, say e
always thus so… Whenever, there is an electrical
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to mechanical coupling, it will always
happen.
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So, it also applies a back e m f, because
here is a here is something magnetic going
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on,
there is an interaction between these two,
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that will apply a voltage here, which will
seen
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as a back e m f. So, here is a voltage that
is seen by the electrical subsystem. So, the
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total
amount of voltage applied on the electrical
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sub system is E minus small e and that one
the mechanical system is F.
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.If you want to do it as a single electro
mechanical system, then also you can do that,
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but
remember in that case the F is proportional
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to I and therefore, it it will result in a
velocity
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dependant potential. So, you cannot take the
original Lagrangian equation that we
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derived, we have to take the enough for the
first principles, we have to take t instead
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of l,
that is all, we have I have already illustrated,
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how to do that.
.
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This sixth problem is ever seen a seismograph
by which the ground oscillation is
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measured. So, it is a similar kind of device,
say here it is routed to the ground and the
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whole thing is
a hollow box. Inside it, you have a mass spring
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damper kind of
arrangement. So, when this ground moves, this
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body cannot move instantaneously.
So, there difference essentially measures,
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there will be a pointer of like this, which
measures the up and down motion of the ground.
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I mean, this is rather crud picturization
of what is actually inside, what is actually
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inside this firm of complicated, but this
gives
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you idea about how to do it.
Now, here you have the ground motion, ground
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is in that case not static, there is a ground
motion, say in the y direction
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and then this is K and this is the R and here
we have M and
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this fellow. Well, if you considering the
motion of this one, you do not need the need
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to
consider the mass of this, because the whole
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thing is moving up and down. So, you can
see that this point is moving with that velocity.
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So, this point and that point are moving
with their velocity in between you have to
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modulate.
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.Problem 7, the train, suppose here is
the track, here is the engine and here are
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the two
coaches. How are they connected, there always
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connected by a spring damper
arrangement. So, you have 3 masses M 1, M
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2 and M 3, there are two springs, let them
be equal, so K and K. There two dampers, let
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them be equal R and R and obviously, this
fellow is pulled with a velocity with a force
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F that is the engine force.
Force providing by the engine, so make a model
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of that, they will all individual have
their own degrees of freedom. In actual practice,
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there is a limit to which these things can
be compressed. In actual practice, there are
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two hard plates which can collide with each
other, so compression is limited, but they
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tension is not limited and this compression
be
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limited is what, it is a non-holonomic constraint.
So, if you have a non-holonomic constraint,
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then also the same set of differential
equation has to be obtained, so that is what
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you obtain those differential equations. So,
that gives you reasonably good amount of practice
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in modeling, in order to obtain first
order differential equations. Now, we promised
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to take up specifically the electrical
circuits, today where the basic law would
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be the Kirchhoff’s law.
By the way, some of you may have heard the
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name as Karchoff, it is not Karchhoff, it
is
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Kirchhoff. The guy was German and German all
the letters have pronounced as they are.
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So, you is always ooh, bus is bush. So, Karchhoff
is not the right pronunciation,
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Kirchhoff is the right pronunciation. So,
the two Kirchoff’s laws, current law and
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the
voltage law, what do they say, what do they
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say. If you have a loop in that circuit, then
you have the voltage law, it which says that
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as you traverse around the loop, the total
voltage is 0. Some of the voltages, it traverses
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around the loop is 0.
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..
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So, suppose you have
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circuit like this, then what are the loops,
well I can identify easily
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when a loops. This is a loop, this is a loop,
this is a loop, this is a loop, this is a
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loop, all
the loops and along around all these, the
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total voltage should be 0. So, that is the
point
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number 1, that the Kirchhoff’s first law.
The voltage law says that along, around all
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the
loops the total voltage should be 0.
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So, if you can identify the by names, say
if I say this is point 1, 0, 2, 3 then we
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can easily
write down the positions of the loops. For
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example, this loop would be said as 0, 1,
2
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loop, this loop would be 0, 2, 3 loop. By
the way, there is another concept that we
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to
have identified these are the nodes. This
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is the node, what is a node, nodes are essentially
the points, which can at least in theory be
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are different voltages.
So, I can see that there are 4 nodes and these
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4 nodes, I have marked as 0, 1, 2, 3 and the
individual elements are in between them clear,
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fine. So, if you write it this way, then
what are the possible KVL equations? For this
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loop, it is v 01 plus v 12 plus v 20 equal
to 0, notice I am not writing the opposite
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things, I am writing 01 not 10. The idea is
that
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you have traverse always in one direction,
which direction is choose is your business,
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you could as well as choose the opposite direction,
it will not matter. But, you always
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have to choose a particular direction and
go along, around it accordingly.
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So, similarly you can write v 02 plus v 23
plus v 30 is equal to 0, you can also write
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for
this big loop, you can write v 01 plus v 13
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plus v 32 plus v 20 equal to 0 so and so forth,
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.you can identify many loops in here; at for
all that the Kirchhoff’s voltage law would
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be
applicable.
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.
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What is the Kirchhoff’s current law? They
will be applicable to, let be draw the circuit
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once again, we are started with in that Kirchhoff’s
current law and you have to identify
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the nodes 1, 2, 3, 0. And then the currents
that either enter or exit, each of these nodes
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would sum to 0, which means that the currents
that are coming into this node, which are
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this current, this current and this current
or the ones at exiting node, this current,
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this
current and this current, they will all sum
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to 0.
So, you can write the KCL equations as say
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I am assuming the currents that are entering
a node to be 0. We have made, that is my conversion
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and how I take it is my business.
So, I consider the once that are entering
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the node as 0, so I will write, what, no I
have not
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said that yet, all I have said that is that,
there is a direction of current here, there
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is a
direction of current here, there is a direction
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of current there. All of them add to 0, it
does
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not mean that actually flowing like this,
they could flow like that.
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But, when I say i 31 plus i 21 plus i 01 equal
to 0, essentially have considered a specific
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sense of positivity of the currents. If the
currents actually flowing in the opposite
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direction, there will be negative, so what?
So, I can write similarly the KCL equations
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that is for node 1, I can also write for node
2, i 12 plus i 32 note not 2, 32 plus i 02
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is
equal to 0.
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.Here, now I can as well write the equation
as the currents that are exiting the node,
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interchangeably. So, for 3, I will write it
as if there all exiting the node, so it will
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be i 31
plus i 32 plus i 30 equal to 0, that is all
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right. So, whether you assume that they entering
direction is positive or exiting direction
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positive is up to you. You easily see that,
if they
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are all negative, they will cancel off.
But, now where had we seen that these are
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the voltage equations and these are the current
equations. Plenty clear that the number of
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equation that will have is for more than,
what
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we really need, so something is to be done
to organize this mesh. Secondly, how to write
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the differential equations, in order to write
the differential equation, there has be clues,
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something like differential has to be there.
.
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That clue comes from the fact, that if there
is an inductor in a specific line and the
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current flowing through it. Then, we can write
the d d v d t as for this L d i d t, so L
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d i d
t is equal to v, so this gives d i d t as
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v by L, so here is the voltage across the
inductor.
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Similarly, if there is a capacitor in a line,
again the current is i and the voltage is
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say v,
then it will be C d v d t is equal to i or
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d v d t is equal to i by C. So, these are;
obviously,
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differential equations.
So, the immediate conclusion is that, if in
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the circuit I notice, that there is an inductor
and then I will be able to write this equation,
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only for that inductor. And if there is a
capacitor, I will able to write this equation
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only for the capacitor. These are first order
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.differential equations, so that least to
the conclusion, that for each of the inductor
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there
will be one first order equation. For each
204
00:26:40,679 --> 00:26:44,960
of the capacitor there will be one first order
equation.
205
00:26:44,960 --> 00:26:51,620
What to do about the right hand side, this
could or could not be a chosen variable. So,
206
00:26:51,620 --> 00:26:56,160
all
we need to do express this in terms of the
207
00:26:56,160 --> 00:27:02,679
variables using the Kirchoff’s laws. Let
us
208
00:27:02,679 --> 00:27:23,880
illustrate, let us take the as yet we have
not organize this mesh too many equations.
209
00:27:23,880 --> 00:27:25,970
You
might say that why not go the same way as
210
00:27:25,970 --> 00:27:29,179
we did, in case of the Lagrangian equation
or
211
00:27:29,179 --> 00:27:34,530
the Hamiltonian equation use the mesh. That
means the open windows, in this case all
212
00:27:34,530 --> 00:27:41,660
these loops are not the independent loops,
only this and this remains the independent
213
00:27:41,660 --> 00:27:49,340
windows, yes you do that, so that will be
the Mesh current method.
214
00:27:49,340 --> 00:27:50,340
.
215
00:27:50,340 --> 00:28:02,270
Let us apply that to first a very simple circuit
and see the consequence, this one I am
216
00:28:02,270 --> 00:28:12,309
doing, because I already did for the Hamiltonian
method, so that you can compare. This
217
00:28:12,309 --> 00:28:21,700
is E L C R and we said at the time that here
is a loop, which will called q 1 and here
218
00:28:21,700 --> 00:28:25,090
is a
loop which will called q 2. But, while writing
219
00:28:25,090 --> 00:28:35,350
as the Kirchhoff’s laws equation, I will
give i 1 and i 2 currents, so i 1 is q 1 dot
220
00:28:35,350 --> 00:28:47,890
i 2 is q 2 dot fines.
Now, what are the nodes, here is a node, here
221
00:28:47,890 --> 00:29:00,030
is a node, here is a node, here is a node,
wait, they could all logically be a different
222
00:29:00,030 --> 00:29:06,120
voltages. But, this is trivial, why because
there is only one branch connected to this
223
00:29:06,120 --> 00:29:11,240
and going on to this. And therefore, the
Kirchhoff’s current law will be a trivial
224
00:29:11,240 --> 00:29:15,669
equation, so will ignore that. So, effectively
there
225
00:29:15,669 --> 00:29:21,990
.a two nodes and whenever use that we always
assume that one of them is at the ground
226
00:29:21,990 --> 00:29:26,559
potential, so that we measure the voltage
of the other, so that is the standard procedure.
227
00:29:26,559 --> 00:29:35,570
But, let us see; first let us apply the voltage
law. Mesh current method we say, in the first
228
00:29:35,570 --> 00:29:46,600
loop the voltage law will look like E, E minus
here it is, it is a voltage drop L d i d t
229
00:29:46,600 --> 00:29:55,330
minus this is the voltage rise, so instantaneous
polarity. So, this is voltage rise E minus
230
00:29:55,330 --> 00:30:07,830
L
d i 1 d t, here it is, if this is the assumed
231
00:30:07,830 --> 00:30:12,880
positive polarity. So, you have to assume,
which
232
00:30:12,880 --> 00:30:18,560
is your conversion positive polarity, which
is conversion of negative polarity, so assume
233
00:30:18,560 --> 00:30:24,020
that this is my polarity conversion.
If the upper plate is positive with respect
234
00:30:24,020 --> 00:30:26,780
to the lower plate, I will call as positive.
So, if
235
00:30:26,780 --> 00:30:45,260
that is, so then this also minus, fine and
the second equation gives v c minus i 2 R
236
00:30:45,260 --> 00:30:50,630
equal
to 0. The first equation immediately gives
237
00:30:50,630 --> 00:30:56,460
the differential equation d i 1 d t is equal
to
238
00:30:56,460 --> 00:31:10,960
take it to the other side E by L minus, so
that is first equation, no problem, done.
239
00:31:10,960 --> 00:31:14,710
But,
then this is not a differential equation,
240
00:31:14,710 --> 00:31:16,630
the second equation is not a differential
equation.
241
00:31:16,630 --> 00:31:20,980
Therefore, we have made no head way in obtaining
complete state of differential
242
00:31:20,980 --> 00:31:30,330
equation using the Mesh current method.
Then, I have to apply the Node voltage method;
243
00:31:30,330 --> 00:31:34,680
I am assuming that you have learned all
that in the first year. So, I am not redoing
244
00:31:34,680 --> 00:31:41,400
the whole concept, I am just using the method.
So, then we have to apply the mesh node voltage
245
00:31:41,400 --> 00:31:46,210
method, because the mesh current
method proved to be insufficient to obtain
246
00:31:46,210 --> 00:31:48,150
the complete set of differential equations.
So,
247
00:31:48,150 --> 00:31:53,120
if you apply the node voltage method, how
many nodes are there 1, 2, 3, but these two
248
00:31:53,120 --> 00:31:57,510
are trivial, I will apply to here.
For this one, if I say the node voltage method,
249
00:31:57,510 --> 00:32:19,730
then it will be the Kirchoff’s current law,
will be i 1 here, i 1 here minus i 2 here
250
00:32:19,730 --> 00:32:36,210
and here is minus, what is this current C.
Yes C d
251
00:32:36,210 --> 00:32:51,700
v c d t is equal to 0, which immediately gives
you the differential equation d v c d t is
252
00:32:51,700 --> 00:33:03,299
equal to 1 by C i 1, i 1 is state variable.
So, it remains minus 1 by C i 2, i 2 is not
253
00:33:03,299 --> 00:33:06,710
because
I am now obtaining the differential equation
254
00:33:06,710 --> 00:33:12,050
in terms of i 1 and v c. So, I need to do
something about it, but this one helps.
255
00:33:12,050 --> 00:33:23,049
So, I can see that this equal to i 2 is equal
to v c by R, so into that is another differential
256
00:33:23,049 --> 00:33:30,160
equation. So, the crux of the story is that,
yes it is possible to obtain the differential
257
00:33:30,160 --> 00:33:35,360
equation starting from the Kirchhoff’s laws.
But, you have to apply both the Kirchhoff’s
258
00:33:35,360 --> 00:33:40,549
.law, you cannot obtain simplify the node
voltage or simplify the mesh current method,
259
00:33:40,549 --> 00:33:44,179
you have to apply both, clear.
.
260
00:33:44,179 --> 00:33:50,419
Just let us illustrate it with another example
to get it clear, the same example that we
261
00:33:50,419 --> 00:34:02,059
have we are doing the 3 loop example, it is
a E, then an inductor, then a capacitor, then
262
00:34:02,059 --> 00:34:09,030
a
resistor, another resistor here and this visible,
263
00:34:09,030 --> 00:34:25,329
yes a capacitor, fine. In this case, we will
say here is my loop i 1, here is my loop i
264
00:34:25,329 --> 00:34:39,990
2 and here is my loop i 3. So, my E, L, C
1, C
265
00:34:39,990 --> 00:34:51,589
2, R 1, R 2 quickly write the mesh current
equations, write the mesh current equations
266
00:34:51,589 --> 00:34:55,230
quickly.
Remember, you have to go around the loops
267
00:34:55,230 --> 00:35:01,430
and you have to keep track of the positives
and negatives that is what is the most common
268
00:35:01,430 --> 00:35:06,340
mistake that you have to make. What, I
do will is in order to not to make the errors,
269
00:35:06,340 --> 00:35:23,300
I will mark them 0, 1, 2 and 3. And then say
the mesh current method will give v 12 plus
270
00:35:23,300 --> 00:35:33,380
v 20 plus v 01 is equal to 0, happy there
is
271
00:35:33,380 --> 00:35:42,530
no problem about it.
Similarly, here it is v 02 plus v 23 plus
272
00:35:42,530 --> 00:35:50,670
v 30 equal to 0, similarly here in the loop
3, it is v
273
00:35:50,670 --> 00:36:11,599
1, sorry I will write this way v 13 plus v
32 plus v 21 equal to 0, so these three are
274
00:36:11,599 --> 00:36:15,090
the
Kirchhoff’s voltage law equations. As it
275
00:36:15,090 --> 00:36:18,740
trivial, but I did write it this way, because
I
276
00:36:18,740 --> 00:36:26,460
know, the most students make a problem in
signs. So, first write it this way and then
277
00:36:26,460 --> 00:36:29,710
substitute these.
278
00:36:29,710 --> 00:36:39,750
.We have understood that the right way of
choosing the stress the variables would be
279
00:36:39,750 --> 00:36:42,841
to
say that the inductor current, the current
280
00:36:42,841 --> 00:36:48,770
through an inductor is variable and a voltage
across capacitor is variable. Why, because
281
00:36:48,770 --> 00:36:54,070
of these two, current through the inductor,
voltage across the capacitor, these are the
282
00:36:54,070 --> 00:36:57,730
minimum variables necessary. So, we will say
the current through the inductor and voltage
283
00:36:57,730 --> 00:37:13,310
across the capacitor such variables. So, all
these currents, I have to be careful about
284
00:37:13,310 --> 00:37:20,940
this particular current.
This is what will become a state variable,
285
00:37:20,940 --> 00:37:24,441
this is become a state variable, this will
become a state variable, but we have to careful
286
00:37:24,441 --> 00:37:29,410
about, what we call positive and what we
call negative. Suppose, this is call positive,
287
00:37:29,410 --> 00:37:41,810
say this direction of the current, say this
direction of voltage is positive and say this
288
00:37:41,810 --> 00:37:47,450
direction of the voltage is positive, so that
is
289
00:37:47,450 --> 00:37:51,770
my convention of positivity. Whenever, it
takes this value, it will positive, else it
290
00:37:51,770 --> 00:37:53,780
negative.
.
291
00:37:53,780 --> 00:38:12,490
Now, what is i L, i L is the current through
the inductor, I L is i 1 minus i 3, fine.
292
00:38:12,490 --> 00:38:19,240
So, that
is one thing and then the KVL equation becomes
293
00:38:19,240 --> 00:38:26,310
write the first KVL equation in terms of
that. Now, write it carefully using this convention
294
00:38:26,310 --> 00:38:40,140
of positives and negatives, it will
become E minus L d i d t. So, minus L d i
295
00:38:40,140 --> 00:38:49,420
1 minus i 3 d t, this is positive and therefore,
if
296
00:38:49,420 --> 00:38:56,580
go this way it will be negative, so minus
v c equal to 0, v c 1, yes.
297
00:38:56,580 --> 00:39:04,470
Now, this negatives and positives are important,
I am again insisting on that, because I
298
00:39:04,470 --> 00:39:10,560
know in this examination, most of you will
make a trouble in that. That is why, I when
299
00:39:10,560 --> 00:39:11,560
to
300
00:39:11,560 --> 00:39:16,020
.carefully this way, so this is one equation
this gives a differential equation, but in
301
00:39:16,020 --> 00:39:23,869
terms
of this, fine second equation becomes and
302
00:39:23,869 --> 00:39:27,420
it was this. So, we go up along the v c, so
it is
303
00:39:27,420 --> 00:39:46,281
v c 1 a minus is drop, so minus R, it will
be i 2 minus i 3, i 2 minus i 3, this 1 i
304
00:39:46,281 --> 00:39:56,790
R 2 minus
i 2 R 2 equal to 0, 1 yes.
305
00:39:56,790 --> 00:40:15,980
Third one, the one at the top, it will give,
v c 2, now here I am going anticlockwise is
306
00:40:15,980 --> 00:40:25,619
a
rise, so I have to go this way minus R 1,
307
00:40:25,619 --> 00:40:35,830
it will be i 3 minus i 2, no, i 2 minus i
3 for the
308
00:40:35,830 --> 00:40:48,750
correct sense i 3 minus i 2 minus i 3 then
minus L. Now, notice the direction this was
309
00:40:48,750 --> 00:40:53,349
the
direction of i L, am going this way, it is
310
00:40:53,349 --> 00:41:00,431
i 3 minus i 1 is flowing this way. So, either
you
311
00:41:00,431 --> 00:41:08,310
put plus of this or minus of this, so be careful
about it.
312
00:41:08,310 --> 00:41:18,510
If I say, if I read write minus here, it is
L, what you say d d t of, no, if it is not
313
00:41:18,510 --> 00:41:33,410
minus,
then i 3 minus i 1, v c 2 up going down, going
314
00:41:33,410 --> 00:41:44,609
down plus minus, minus. So, minus it will
be a drop, if the current is flowing this
315
00:41:44,609 --> 00:41:58,530
way, wait i 1 minus i 3 is the current this
way. So,
316
00:41:58,530 --> 00:42:10,570
this is this
is the equation, once you have obtained this,
317
00:42:10,570 --> 00:42:13,100
which are the differential
equations, this is the differential equation,
318
00:42:13,100 --> 00:42:18,050
this is differential equation, this is not.
So, we
319
00:42:18,050 --> 00:42:32,760
need to apply something more to that.
So, what any problem.
320
00:42:32,760 --> 00:42:38,110
.
No, here is a point, here is a point, here
321
00:42:38,110 --> 00:42:39,119
is a point, here is a point and we are going
this
322
00:42:39,119 --> 00:42:46,660
way up, this way down, this way down, this
way down. So, it is i 2 minus i 3 is the
323
00:42:46,660 --> 00:42:56,250
current this way, now that is right.
324
00:42:56,250 --> 00:42:57,250
..
325
00:42:57,250 --> 00:43:03,190
Now, so we have written down the equations,
ultimately we need to obtain the
326
00:43:03,190 --> 00:43:10,820
differential equation in terms of I L. So,
we just substitute it here, we get d i L d
327
00:43:10,820 --> 00:43:20,261
t is
equal to E by E by L minus v c 1 by L, one
328
00:43:20,261 --> 00:43:34,840
equation done, from the first. Now, in order
to proceed, we need to write the KVL equations,
329
00:43:34,840 --> 00:43:43,619
sorry KCL equations. So, KCL
equations are where is
330
00:43:43,619 --> 00:44:05,170
the equation, I L here, this current here,
this current here, fine let
331
00:44:05,170 --> 00:44:12,670
us write it.
In which node should I write, let us write
332
00:44:12,670 --> 00:44:19,660
it here at this node, if you write at this
node, it
333
00:44:19,660 --> 00:44:34,830
will be, can you see that, yes. In this node
it will be here it is I L going into it, no
334
00:44:34,830 --> 00:44:38,849
I want
to write in terms of the dividities. So, it
335
00:44:38,849 --> 00:44:43,160
is going into, this is going into and this
is going
336
00:44:43,160 --> 00:45:10,230
into, so you have to write it that way. So,
that is minus C 1 d v c d t and this is again
337
00:45:10,230 --> 00:45:29,180
minus R 1 d v c 1. Here, the potential is
v c 1 and here the potential is say v 3, this
338
00:45:29,180 --> 00:45:43,260
point
v c 1 minus v 3, so R 1 v c 1 minus v 3, no
339
00:45:43,260 --> 00:45:53,680
I want the current. So, this is the equation
in
340
00:45:53,680 --> 00:45:58,510
this node.
Now, I need v 3 in order to substitute this,
341
00:45:58,510 --> 00:46:05,500
so if can write that easily, but let extract
the
342
00:46:05,500 --> 00:46:13,380
differential equation first, d v c 1 d t is
equal to write the rest of it in the right
343
00:46:13,380 --> 00:46:27,160
hand side.
It will be 1 by C 1 i L minus 1 by R 1, C
344
00:46:27,160 --> 00:46:44,740
1, v c 1 minus 1 by, wait a minute is this
right
345
00:46:44,740 --> 00:47:09,869
plus R 1, C 1, v 3 anything else done. Now,
can have substituted this point, apply KCL
346
00:47:09,869 --> 00:47:19,030
in
node 3. What you have apply, write the KCL
347
00:47:19,030 --> 00:47:38,780
equation in node 3, it is v 2 minus v 3 by
R
348
00:47:38,780 --> 00:47:44,460
1, there is a current here, input.
349
00:47:44,460 --> 00:47:58,230
.So, it is this current is minus v 3 by R
2 and the other current is this, this and
350
00:47:58,230 --> 00:48:06,460
that, that
current would be well this is plus, this is
351
00:48:06,460 --> 00:48:16,130
minus, so it is plus C 2 d v c 2 d t equal
to 0,
352
00:48:16,130 --> 00:48:25,310
from there you can explain. So, just write
down this equation in the proper form, so
353
00:48:25,310 --> 00:48:27,080
that
you extract the differential equation out
354
00:48:27,080 --> 00:48:29,810
of it, this in the left hand side and the
rest in the
355
00:48:29,810 --> 00:49:03,130
right hand side. D v c 2 d t is equal to can
you write properly minus R 1, C 2, v c 1 plus
356
00:49:03,130 --> 00:49:15,270
1
by C 2, 1 by R 1 plus 1 by R 2, v 3.
357
00:49:15,270 --> 00:49:27,869
Now, v 3 is what, here also we substitute
v 3, here also we substitute v 3, v 3 is what,
358
00:49:27,869 --> 00:49:36,520
I
need to write down, what is v 3. So, we can
359
00:49:36,520 --> 00:49:51,380
extract it from here, v 3, v 23 which I know
and v 02, no it be easily extract from this
360
00:49:51,380 --> 00:50:00,829
equation, E minus v c 2 is this. So, you can
extract from that good, so we just substitute
361
00:50:00,829 --> 00:50:07,030
it here, it will be E minus v c 2, here also
E
362
00:50:07,030 --> 00:50:19,910
minus v c 2.
So, this is how you extract the differential
363
00:50:19,910 --> 00:50:26,580
equation, you can see that in keeping the
signs
364
00:50:26,580 --> 00:50:35,420
correct, we have to really spent a lot of
attention. In order to extract the right things,
365
00:50:35,420 --> 00:50:38,900
we
have to find out, where it that information
366
00:50:38,900 --> 00:50:48,530
available, it was spending a great deal of
attention. And moreover the problem is that
367
00:50:48,530 --> 00:50:54,800
always this assumption that the current
through the inductor and voltage across the
368
00:50:54,800 --> 00:50:58,090
capacitor will be the minimum number of
variables will not work.
369
00:50:58,090 --> 00:50:59,090
.
370
00:50:59,090 --> 00:51:05,001
Why, just imagine a circuit, which is something
like this, I am just replacing one of the
371
00:51:05,001 --> 00:51:20,710
elements in the last circuit, it was like
this, sorry it was capacitor and this is a
372
00:51:20,710 --> 00:51:21,710
capacitor,
373
00:51:21,710 --> 00:51:28,060
.this resistor and here is another capacitor.
What are you done; I have replaced this
374
00:51:28,060 --> 00:51:45,190
resistor by a capacitor. At this time, I can
identify a node, an identify a loop like this;
375
00:51:45,190 --> 00:51:47,560
it is
after a loop, it is not a window, it is not
376
00:51:47,560 --> 00:51:52,859
a mesh, but it is a loop.
If you write the case KVL in this loop, what
377
00:51:52,859 --> 00:51:59,130
you have, you have E, say this is C 2, this
is
378
00:51:59,130 --> 00:52:12,290
C 1 and this is C 3. So, E minus v c 2, because
this is a plus, this is a minus, then you
379
00:52:12,290 --> 00:52:23,740
come here it is plus v c 3, this is plus,
this is minus, so minus v c 1 equal to 0.
380
00:52:23,740 --> 00:52:27,680
You
notice, that the voltage is across the capacitors,
381
00:52:27,680 --> 00:52:31,960
these are voltage across capacitors, they
no longer independent, why because they there
382
00:52:31,960 --> 00:52:36,420
related by this equation.
All of them cannot be independent, because
383
00:52:36,420 --> 00:52:44,150
this addition, this is an externally applied
voltage, this addition is 0, therefore these
384
00:52:44,150 --> 00:52:53,290
3 cannot be independent and can you see that.
If these three cannot be the independent,
385
00:52:53,290 --> 00:52:57,300
I cannot assume all the voltages across
capacitors as independent state variables.
386
00:52:57,300 --> 00:53:02,930
So, here is the problem, notice that let us
give
387
00:53:02,930 --> 00:53:06,880
another example, say there is a circuit something
like this.
388
00:53:06,880 --> 00:53:13,609
I again make a small replacement, here is
a voltage source, here it was the inductor
389
00:53:13,609 --> 00:53:16,950
and
here it was a capacitor that remains the same.
390
00:53:16,950 --> 00:53:23,800
Here, it was a resistor that remains the
same, here only I replace it by an inductor
391
00:53:23,800 --> 00:53:33,250
and here also I replace it by an inductor,
suppose I have a circuit like this. See the
392
00:53:33,250 --> 00:53:43,030
problem, if I write this node equation, then
what we have, suppose this is now a current
393
00:53:43,030 --> 00:53:53,349
source I, imagine that the current source.
And in that case, the equation becomes I is
394
00:53:53,349 --> 00:54:08,530
equal to I, this 1, 2, 3, 0, i 12 plus i 13,
obviously or I minus i 12 minus i 13 is equal
395
00:54:08,530 --> 00:54:14,400
to 0, whatever it is, this immediately means
that these two inductor currents or not independent
396
00:54:14,400 --> 00:54:19,560
variables. So, if you have a circuit
like this, then the inductor current no longer
397
00:54:19,560 --> 00:54:21,600
remain independent variables. If you have
a
398
00:54:21,600 --> 00:54:26,660
circuit like this, the capacitor voltages
no long remain independent variable.
399
00:54:26,660 --> 00:54:34,849
So, I have identified a few main problems
with applying the Kirchoff’s laws in a row
400
00:54:34,849 --> 00:54:40,060
way. In a row way, what we have simply taking
the mesh current and the node voltage
401
00:54:40,060 --> 00:54:45,400
and proceed to write down the differential
equations. We have been able to write, but
402
00:54:45,400 --> 00:54:50,690
that will true for relatively simple circuits.
Even, for a three mesh circuit, it was
403
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becoming a bit tired some, so it is not good.
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.And secondly if you have this kind of arrangements
of the circuit, then the basic
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assumption fails, so we need to have some
kind of a systematic way of doing this. In
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the
next class will tackle with that systematic
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way of doing this.
Thank you very much.
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.