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welcome back to basic electronics in this
lecture we will look at the principle of superposition
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superposition is very useful in circuit analysis
when there are multiple sources in the circuit
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we will consider a few examples to illustrate
how superposition is applied in practice finally
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we will look at why superposition works so
let us start let us consider a circuit made
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up of the following elements resistor for
which v is r times i voltage controlled voltage
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source for which v is v is alpha times v c
v c is the controlling voltage and alpha is
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the dimensionless constant voltage controlled
current source for which i is g times v c
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v c is again a controlling voltage and g has
units of conductance current controlled voltage
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source for which v is r times i c i c is a
controlling current and r has got units of
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resistance and finally current controlled
current source for which i is beta times i
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c i c is again a controlling current and beta
is a dimensionless constant
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what is common in all of these equations they
are all linear that means a quantity on the
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left such as v or i depends on the quantity
on the right such as i here or v c here etcetera
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in the linear passion and that is why when
we have a circuit made up of these elements
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the superposition theorem can be applied in
addition to these elements we have some independent
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sources independent d c voltage sources for
which v is v zero just a constant for example
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v zero could be five volts or hundred volts
and independent d c current sources for which
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i is i zero again a constant i zero could
be five milli ampere for example or one ampere
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now if we have a circuit of this kind with
these elements and these sources such a circuit
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is a linear and we can use superposition to
obtain its response when multiple independent
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sources are involved and what do we mean by
the response of the circuit what we mean is
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the currents and voltages in that circuit
the next question is how do we use superposition
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super position enables us to consider the
independent sources one at a time with the
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other sources deactivated compute the desired
quantity of interest in each case such as
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a current or voltage and then get the net
result by adding the individual contributions
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and the advantage of this procedure is that
this procedure is generally simpler and considering
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all independent sources simultaneously all
right
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now let us address this issue of deactivating
an independent source what do we mean by deactivating
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an independent current source let us look
at the equation for a current source write
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there i equal to i zero so to deactivate this
current source all we need to do is to make
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i zero equal to zero and what does that mean
that means no current is passing through that
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element that is replace the current source
with an open circuit
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what about deactivating an independent voltage
source here is the equation for voltage source
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v equal to v zero so we need to make v zero
equal to zero that means we replace the d
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c voltage source with the short circuit because
a short circuit has no voltage drop or zero
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voltage drop across it and therefore we simply
need to replace the voltage source with the
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short circuit
all right let us now apply what we have learnt
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about superposition to this circuit here
here we have two independent sources one is
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a voltage source eighteen volts and other
is a current source three amperes and we want
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to find this current i one here so we will
take this two sources one at a time let us
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start with the voltage source so we keep the
voltage source and deactivate the current
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source and that is our case one keep v s deactivate
i s and how do we deactivate the current source
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as we saw in the last slide we need to make
the current equal to zero that is replace
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the current source with an open circuit and
if this case it is easy to calculate r one
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it is simply eighteen volts divided by two
plus four or three amperes all right so that
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is our case one
let us now look at case two in case two we
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keep the current source and deactivate the
voltage source so that is our current source
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and we have deactivated the voltage source
that means you have made the voltage here
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equal to zero that is replaced the voltage
source with a short circuit and now what is
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r one in this case r one can be obtained using
current division r one would be three amperes
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multiplied by two ohms divided by six ohms
so that is our r one three amperes times two
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by six or one ampere and note that we are
used is superscripts this two indicates case
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two and this one indicates case one and now
this is a final step we simply add these two
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quantities i one in case one and i one in
case two to obtain the net the value of i
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one and that turns to be three plus one or
four amperes
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here is our second example we have two independent
sources here one is a voltage source twelve
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volts and other is a current source six amperes
and apart from these two independent sources
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we have registers and we have this dependent
source and what kind of dependent source is
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this it's a voltage source and the voltage
here is controlled by this current if this
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current is i then the voltage is two times
i so its a current controlled voltage source
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all right
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let us now use superposition to obtain this
voltage v here and how do we go about doing
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that we have two independent sources the voltage
source and the current source we take these
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two sources one at a time compute v and then
finally add the two values of v here is our
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case one we have kept their voltage source
and deactivated the current source that means
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replaced it with an open circuit like that
and now let us find v in this case the important
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point note here is that we have left this
dependent source as it is all right and now
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how do we solve this circuit they have a loop
here and we can write the kvl equation for
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this loop and let us take voltage drop as
positive for this purpose let us go like this
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what is this voltage drop is one ohms times
i or just i up to that we come across our
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I's of twelve volts so that would be minus
twelve because this is your I's then we become
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across a voltage drop of three times i so
that is plus three times i and then another
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voltage drop of two times i and when we add
all of these terms we should get zero like
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that we can solve this equation for i that
gives us i equal to two ampere and therefore
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this voltage v is two amperes times three
ohms or six volts and note that we have used
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this superscript one here to indicate case
one
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let us now consider the second case in which
we keep the current source and deactivate
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the voltage source so here is our current
source and we have deactivated this voltage
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source that means we have replaced it with
the short circuit all right now how do you
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obtain ah i or v in this case there may be
different ways of doing this one thing that
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comes to mind is that this current through
the three ohm resistance is i plus six and
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once you figure that out you can write a kvl
equation for this loop and then obtain i and
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therefore v here is the kvl equation we are
going in this loop starting with this voltage
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drop one ohm times i or simply i the first
on there then six plus i times three that
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term there and then finally a voltage drop
of two times i across the current control
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voltage source that term there and all of
these terms must add up to two zero and that
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gives us i equal to minus three amperes this
minus sign means that the actual current is
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flowing in the opposite direction
once we know this current i we can find the
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voltage here as six plus i times three ohms
that is six minus three times three ohms or
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nine volts so that is our v with the super
script two to indicate case two and now the
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net voltage this voltage here in the original
circuit can be found simply by adding v one
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and v two and that gives us six plus nine
or fifteen volts how do we know if our solution
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is correct there are two ways of verifying
that one we can find the net quantities the
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net current here is i equal to two minus three
that is minus one ampere the net voltage we
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have already found that is fifteen volts
so therefore we know that fifteen volts divided
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by three times that is five amperes would
go like that and similarly we can find the
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voltage drops across each of these elements
make sure that k v l is actually satisfied
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by the net quantities in this loop and also
kcl is satisfied at this node the second approach
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for verifying whether our solution is correct
or not is to use circuit stimulation now in
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this ah example and in fact throughout this
course we will use this circuit simulator
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called sequel which has been developed at
iit bombay over the past few years and the
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sequel circuit file for this particular example
is given over here
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so we can run this ah simulation and verify
if our results are correct here is the u r
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l or the sequel circuit simulator website
and that url brings us to the sequel main
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page there is the description about what sequel
is about what features it occurs and this
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feature of course is other important for our
purposes it's a free circuit simulator you
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can just download and use it then this is
a link for sequel examples there is a large
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number of readymade examples and you can click
here and find out more about that there is
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a link for downloading the program and also
there are several video tutorials available
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about how to use the circuit simulator the
basics of the graphics interface tutorials
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project examples how to plot graphs and so
on and these tutorials would be particularly
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relevant if you are going to make up a new
project
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on the other hand if we are going to run an
existing project for which the circuit file
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is already available then we don't really
need to go through all of these video tutorials
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in addition there is the sequel users manual
pot one and pot two and there is some course
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material as well to run or existing ah circuit
file we note the name of that file so it is
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even one underscore superposition underscore
two and we can look for this specific project
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and then done it let us see how to do that
the circuit files are placed in various directories
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depending on which area that circuit belongs
to and so the best way to find were a given
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circuit file is to search for it
so let us do that so e one o one superposition
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two this is the file we have looking for and
this file is in a directory called ee one
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o one underscore network under sequel gui
projects all right so let us remember that
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and open this particular circuit file so this
is the executable file for the sequel gui
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and that is under sequel gui two release so
we double click on that and then the simulator
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starts
and now we look for the file of our interest
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and as we ah found earlier it is in this directory
called e e one o one network so lets open
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that and that is where that file is
so lets open that file that is what it looks
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like so this is our original circuit it has
got more the voltage source and the current
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source this circuit here is the circuit with
the voltage source deactivated and this circuit
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here is the circuit with the current source
deactivated and notice that there is this
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fictitious voltage source here called v a
zero and that is required because the current
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through that branch is used by this current
controlled voltage source we have define some
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output variables on this right hand side here
in particular we are interested in this voltage
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across r two v r two here is that voltage
in the original circuit v r two a is that's
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an voltage in this circuit with v s deactivated
and v r to be is the same voltage in this
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circuit with i s deactivated let us now run
the simulation and look at this v r two values
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so that is the table of the output variables
and the net value of v is a fifteen volts
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here as we had computed and the other two
values are nine volts and six volts
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instead of constructing these separate circuits
with v s deactivated and i is deactivated
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we could just change the parameters in the
original circuit to verify our calculations
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let us do that so let us get rid of this part
and first we will keep the voltage source
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and deactivate the current source by making
the current equal to zero and now let us run
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the simulation and now we find that v r two
is six volts as we had computed
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next let us bring this back to six amperes
and deactivate the voltage source now make
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it zero and now in the simulation again and
now we get nine volts once again as we expect
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and finally let us include both of these sources
in the circuit so make vs is equal to twelve
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volts i is equal to six amperes and see what
happens
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and now we get fifteen volts as we would expect
here is our next example we have two independent
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voltage sources v s one and v s two and we
want to find this voltage v one using superposition
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so let us take these two voltage sources one
at a time so case one v s one alone with v
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s two deactivated that is replaced with a
short circuit in this case we get v one while
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voltage division as the voltage across r two
which is r two divided by r one plus r two
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times v s bar so that is our case one and
we have marked that with this superscript
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one over here
here is a second case in which we have kept
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v s two but deactivated v s one that is replaced
it with a short circuit and in this case v
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one is the same as in the voltage across r
one and once again using voltage division
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we have v one equal to r one divided by r
one plus r two times v s two like that and
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finally our net value of v one is simply the
addition of v one in case one and v one in
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case two and that gives us this expression
over here this is our next example and it
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is essentially the same as our previous example
except we are going to treat it a little differently
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and that is by designating one of the nodes
this node as the reference node of the ground
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node and this is indeed a common practice
in electronic circuits it makes things much
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more convenient as we will see so this node
voltage is a zero volts because that is the
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reference node with respect to that node voltage
the voltage at this node would be v s one
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the voltage at this node would be v one and
the voltage at this node would be v s two
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so we can read of this circuit in which we
don't bother to show explicitly these voltage
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sources and we don't even show the ground
node it is all implicit now
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we understand when we draw a circuit diagram
like this that there is a reference node which
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is taken as zero volts and with respect to
that reference voltage we have this node voltage
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as v s one this as v one and this as v s two
and obviously this circuit is much simple
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to look at as compare to the original circuit
and that is why it is more convenient all
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right
now let us use superposition to obtain v one
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in this circuit what we will do is take v
s one first take v s two as zero and what
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is the meaning of v s equal to zero that means
this node voltage would be the same as the
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voltage at the reference node or the ground
node that means it is zero volts and that
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is shown using the ground symbol like that
and now in this case v one is given by voltage
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division essentially it is the voltage between
this node and the ground node and this v s
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one is the voltage between this node and the
ground node and therefore we can use voltage
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division so v one is the voltage across r
two which is r two divided by r one plus r
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two two time v s one and that is our v one
in case one and similarly in the second case
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in which we keep v s two but deactivate v
s one that means we connect this node to the
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reference node we get v one as r one divided
by r one plus r two times v s two and that
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is over v one incase two and finally we get
the net v one for the original circuit as
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v one in case one plus v one incase two and
that is this expression which is the same
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as the one we have seen earlier
we have seen that superposition works at least
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for the examples we have considered so far
ah now let us try to understand why it should
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be expected to work and we will do that with
the help of this specific example but our
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findings are really much more general all
right so here we have two independent sources
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voltage source and a current source and what
we will do is to take this node as the reference
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node so with respect to that this node voltage
is v s this voltage node is v one and this
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one is v two and now let us write ah the kcl
equations at node a and at node b and then
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doing that we will take a current leaving
a node as positive
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so let us look at first equation here we have
three currents at node a this current and
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this current and what is this current it is
p one minus p s divided by r one and since
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that current is leaving this node that's comes
with the plus sign that is this term over
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here what about this current that is p one
minus zero divided by r two the second term
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and this current is v one minus v two divided
by r three that is the third term and similarly
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at node b we have two currents i s and this
current now i s is entering the node here
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therefore that comes with the negative sign
like that and this current which is living
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the node is v two minus v one divided by r
three and that is the second term over here
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so these are our kcl equations and now let
us see what we can do with that let us rewrite
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the kcl equations in a matrix form and in
doing that let us use these definitions the
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conductance g one for example is one over
r one similarly g two would be one over r
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two and so on when we do that we get this
matrix equation here and let us check that
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this first equation does indeed correspond
to the first row over here what is the coefficient
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of v one in this equation it is ah g one plus
g two plus g three and that is why we have
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g one plus g two plus g three here what is
the coefficient of v two it is minus g three
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00:28:17,169 --> 00:28:25,080
and that is why we see minus g three over
here and this term minus v s times one over
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00:28:25,080 --> 00:28:32,250
r one does not depend on v one or v two and
therefore we take it to the other side and
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00:28:32,250 --> 00:28:39,029
in doing that it becomes plus g one times
v s and that is why we have g one time v s
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00:28:39,029 --> 00:28:49,240
over here and similarly you can verify that
the second equation here does correspond to
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our second kcl equation all right
now let us write this matrix as matrix a so
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we have an equation of the form a x equal
to b where x is our column vector consisting
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00:29:06,450 --> 00:29:14,350
of v one and v two and b is the right hand
side vector consisting of g one b s and five
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00:29:14,350 --> 00:29:28,980
s and from this matrix equation we can obtain
v one v two as a inverse times our r h s vector
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00:29:28,980 --> 00:29:36,429
so here is our v one v two a inverse times
our r h s vector let us now write a inverse
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00:29:36,429 --> 00:29:44,710
as this two by two matrix m one one m one
two m two one m two two and let us bring this
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00:29:44,710 --> 00:29:52,059
g one inside this matrix now and we do that
by multiplying m one one and m two one by
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00:29:52,059 --> 00:30:03,919
g one so finally our v one v two is given
by this matrix multiplied by this column vector
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00:30:03,919 --> 00:30:14,840
which consist of the two source terms v s
and i s and now this column vector can be
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00:30:14,840 --> 00:30:23,970
written as a sum of two column vectors v s
zero and zero i s and when we do that we get
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00:30:23,970 --> 00:30:30,360
this following equation and now we should
be able to see while superposition works at
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least for this circuit so this is our first
column vector coming out of this v s i s column
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00:30:39,279 --> 00:30:48,890
vector and this is the second column vector
and this first term is nothing but our v one
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00:30:48,890 --> 00:30:59,020
v two in case one denoted here by this superscript
one and this second term is nothing but our
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00:30:59,020 --> 00:31:08,720
c one v two in case two
in other words the first vector this one is
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00:31:08,720 --> 00:31:15,309
the response of the circuit due to v s alone
and i s deactivated and that is why we have
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00:31:15,309 --> 00:31:23,860
zero over here and the second vector this
one is the response due to i s alone with
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00:31:23,860 --> 00:31:35,990
v s deactivated and that is why we have zero
over here all right now turns out that all
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00:31:35,990 --> 00:31:42,919
other currents and voltages in the circuit
are linearly related to v one and v two because
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00:31:42,919 --> 00:31:48,700
this is the linear circuit and therefore any
voltage that is any node voltage or any branch
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00:31:48,700 --> 00:31:55,789
voltage or any current in the circuit can
also be computed using superposition and this
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00:31:55,789 --> 00:32:04,070
is indeed a very powerful theorem and we will
find many uses of this theorem
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00:32:04,070 --> 00:32:10,269
to summarize we have seen how superposition
can be used in circuit analysis in electronics
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these technique is very useful as we will
discover later in this course that is often
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now see you next time