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welcome back to basic electronics in this
lecture we will continue with thevenins theorem
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we will consider a circuit with a dependent
source and learn how is thevenin equivalent
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circuit can be obtained we will then see how
the thevenin form of a circuit can be converted
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into another equivalent form called the norton
form we will illustrate how this thevenin
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to norton transformation can be used in circuit
analysis at the end of the lecture we will
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look at the maximum power transfer theorem
for circuits ok let us get started
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here is another example it has a dependent
source but no independent current or voltage
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sources and let us obtain the thevenin equivalent
for this circuit as seen from a b what kind
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of dependent source is this its a voltage
source judging from this symbol and the voltage
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difference between these two nodes is two
times i one where i one is this current so
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this voltage is controlled by a current and
therefore its a current controlled voltage
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source let us find the thevenin voltage first
which is the same as v o c as shown here in
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this situation this current i one is zero
and therefore this voltage drop which is given
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by two times i one is also zero
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so therefore we can replace this c c b s with
a short circuit like that and now we find
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that v o c is going to be zero volts and therefore
our v t h the thevenin voltage is going to
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be zero volts let us now find the thevenin
resistance and to do that we deactivate the
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independent sources in the original circuit
in this case of course there are no independent
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sources so nothing needs to be done and then
we look from a b in this case because we have
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this dependent source we cannot really figure
out what the equivalent resistance as seen
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from a b is and therefore let us use another
method that is connect test source and then
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find the ratio of this voltage and that current
that gives us r t h
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so we have connected a test current source
here so what we will do is find the ratio
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v s divided by i s in this case i s and i
one happen to be the same all right where
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do we begin let us take one of the nodes as
the reference node say this one if that is
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zero volts then this voltage is two times
i one and i one is the same as i s so therefore
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this voltage is two times i s what about this
node voltage that is the same as v s so now
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we have all the node voltages and we can now
write a k c l equation to obtain v s divided
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by i s and that is our k c l equation at node
a you have three currents here i s this current
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and this current
now i s is entering this node and therefore
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we will take that as minus i s this current
v s minus zero divided by r two is leaving
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the node therefore we take that as a positive
quantity plus v s by r two what about this
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current it is v s minus two i s divided by
r one so all these three currents add up to
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zero and now we can get v s divided by i s
so after simplifying this equation we obtain
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r t h as v s divided by i s or eight over
three ohms so our thevenin equivalent circuit
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has v t h equal to zero as we saw in the previous
slide and r t h equal to eight by three ohms
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so that is what it looks like and since v
t h is zero volts we may as well replace that
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with a short circuit and therefore all we
have between a and b is this resistance r
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t h equal to eight by three ohms so this circuit
is equivalent to this original circuit
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we have seen so far that a linear circuit
can be represented by its thevenin equivalent
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circuit consisting of a voltage source v t
h in series with a resistance called r t h
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or thevenin resistance now equivalently we
can represent the same circuit with the norton
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equivalent circuit shown over here which consists
of a current source i n in parallel with a
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resistance r norton or r n for these two circuits
to be equivalent to each other we must have
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some relationship between v t h r t h and
i n r n and let us now see what those are
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let us consider the open circuit case for
the thevenin circuit the open circuit voltage
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between a and b is simply equal to v t h because
this voltage drop is zero so for the thevenin
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circuit v a b is v t h for the norton circuit
this voltage drop between a and b is i n times
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r n and that is what this equation says over
here now for these two circuits to be equivalent
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we require that this voltage must be the same
as this voltage and that gives us the condition
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that v t h must be equal to i n r n like that
next let us consider the short circuit case
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as shown here on the right what we do here
is connect a and b with a wire and look at
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this short circuit current called i s c and
we do that for the thevenin equivalent circuit
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as well as for the norton equivalent circuit
all right for the thevenin circuit i s c is
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simply v t h divided by r t h this equation
here further norton circuit the voltage drop
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between a and b is zero and therefore low
current flows through r n and therefore i
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n must flow like that and that gives us i
s c equal to i n and because the two circuits
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are equivalent these two short circuit currents
must be equal and that gives us the second
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relationship between v t h r t h i n and r
n
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let us now use this first relationship and
substitute for i n v t h divided by r t h
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because these two short circuit currents are
equal and then we get v t h equal to i n which
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is v t h by r t h like that times r n this
v t h cancels out and we get r t h equal to
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r n all right so what this means is if we
know the thevenin equivalent circuit we can
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obtain the norton equivalent circuit using
these relationships here r n and r t h are
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equal and i n is equal to v t h divided by
r t h so we can go from this circuit to that
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circuit similarly we can go from the northern
circuit to the thevenin circuit because r
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t h is equal to r n and v t h is equal to
i n times r n which follows from the equality
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of these two short circuit currents
let us consider this problem in which source
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transformation will be very useful that is
transformation from a thevenin equivalent
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to the norton equivalent and vice versa ok
so what is the problem here we want to find
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the thevenin equivalent circuit as seen from
a b as step number one what we will do is
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to convert this thevenin form consisting of
this thirty two volts in series with sixteen
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ohms into which norton equivalent what is
over i n i n is equal to v t h divided by
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r t h or thirty two divided by sixteen that
is two amperes what about r n r n is the same
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as r t h that is sixteen ohms
so with that transformation we get this equivalent
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circuit here this is our norton equivalent
circuit i n is two amperes r n is sixteen
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ohms and the rest of the circuit all of this
has been left untouched all right now these
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two current sources these two amperes and
this two amperes are actually in parallel
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they share the same nodes and therefore they
can be combined into one single source or
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for amperes as shown over here so we are four
amperes in parallel with sixteen ohms this
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resistance and then we have the rest of the
circuit here like that
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next let us convert this norton form consisting
of this four rams current source with sixteen
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ohms in parallel into a thevenin equivalent
what is v t h it is i n times r n that is
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four times sixteen or sixty four volts what
about r t h r t h is the same as r n that
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is sixteen ohms so with that transformation
we get this circuit here sixty four volts
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in series with sixteen ohms and then we have
the rest of the circuit that one
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now these two resistances r n series and that
gives us sixteen plus twenty that is thirty
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six ohms like that next we can convert this
thevenin form into its norton equivalent as
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given by this circuit here what is i n it
is sixty four divided by thirty six that turns
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out to be sixteen by nine amperes and r n
is the same as r t h that is thirty six ohms
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now this thirty six ohms is in parallel with
this twelve ohms and that gives us nine ohms
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as shown here and now we can convert this
norton form into its thevenin equivalent what
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is v t h it is i n times r n so that is sixteen
by nine times nine or sixteen volts what about
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r t h that is the same as r n that is nine
ohms
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so now we get this circuit here and now these
two can be combined to give us fifteen ohms
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so that is our final thevenin equivalent circuit
as seen from a b before leaving this example
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let us make a couple of important points one
notice that our numbers having rather friendly
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as an example take this thirty six ohms in
parallel with twelve ohms now this is a calculation
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which we can very easily do on paper without
using a calculator in real life of course
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the numbers may not be as nice for example
this could be two point seven k this could
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be two point two k and in that case we will
end up using our calculator our second point
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has to do with the method that we used in
our calculation we used source transformations
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either from thevenin to norton or from norton
to thevenin to go from this original circuit
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all the way to its final thevenin equivalent
circuit this one
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but that is not the only way to do that in
fact there may be other ways some of them
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may even be simpler for example if we want
r t h you can get that directly from the original
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circuit what do we need to do we deactivate
the independent sources that means we short
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this voltage source and open this current
source and when we do that this sixteen ohms
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comes in series with this twenty ohms so that
gives us thirty six ohms that thirty six ohms
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comes in parallel with twelve ohms and that
entire combination comes in series with six
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ohms
so the r t h calculation is fairly straightforward
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and similarly for v t h we dont need to go
all the way using source transformations we
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can stop somewhere in between for example
when we come to this stage all we need to
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do is to find the open circuit voltage between
a and b and that is very straightforward since
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there is no current in this six ohms resistance
v o c is the same as the voltage drop across
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this twelve ohm resistance and that can be
obtained simply by voltage divisions that
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would be twelve ohms divided by twelve plus
thirty six times sixty four and that will
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give us this answer
let us now talk about maximum power transfer
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which is a very useful concept in electronic
circuits here is a linear circuit consisting
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of resistors independent voltage sources and
we will consider d c voltage sources here
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independent d c current sources and dependent
sources such as current controlled voltage
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source current controlled current source etcetera
to this circuit we connect load resistance
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r l now this load resistance is going to draw
a current and that we will denote with i l
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this load resistance is going to absorb some
power and that power is given by e l equal
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to i l squared times r l
in other words there is a transfer of power
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from this circuit to the load resistance and
the question that we asked now is for a given
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circuit or for this given black box what is
the value of r l for which p l is maximum
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to answer that question let us replace the
original circuit with this thevenin equivalent
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and when we do that we get this circuit here
so this entire original circuit has been replaced
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with this thevenin equivalent circuit and
now it is straightforward to find i l i l
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it simply v t h divided by r t h plus r l
like that and once we know i l you can find
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p l as i l squared times r l and that gives
us this expression over here
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to find the value of r l for which p l is
maximum all we need to do now is to differentiate
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p l with respect to r l and put that equal
to zero that is d p l t r l should be equal
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to zero and when we do that we get this following
equation here and notice that this v t h squared
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is a constant which is independent of r l
and therefore it does not figure in this equation
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here all right
now when we simplify this further we get r
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t h plus r l equal to two r l you should go
through this algebra of course and finally
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we get r l equal to r t h so if you have a
circuit which is linear we can find its r
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t h and then if we have a load resistance
which is equal to that r t h then maximum
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power transfer will take place from the original
circuit to r l all right let us look at a
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plot of p l as a function of r l this is where
r l is equal to r t h and the function goes
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through a maximum
as an example let us consider this circuit
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here we have a load resistance between a and
b and we want to find its value for which
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p l the power transfer to r l is maximum so
step number one is to find the thevenin equivalent
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circuit for this circuit here and how do we
go about doing that let us look at r t h first
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r t h can be found by deactivating the independent
sources that means we shot this twelve volt
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source here and open this two amperes source
and then we end up with this circuit over
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here
now r t h is simply the resistance as seen
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from a b and what do we see we see that r
one and r two are in parallel and that combination
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is in series with r three so our r t h is
r one parallel r two plus r three that is
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three parallel six three parallel six plus
two ohms two ohms here and that comes to four
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ohms all right
now let us proceed with the calculation of
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v t h what is v t h v t h is the same as v
o c that means we remove r l and find the
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open circuit voltage between a and b so let
us do that now so this is our circuit for
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b t h calculation and since there are two
independent sources here we can use superposition
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it turns out to be convenient so case one
we have the twelve volts source as it is and
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two amps deactivated that means replaced with
an open circuit case two we have the two amp
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source as it is and the twelve volt source
deactivated that means shorted in the first
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case b a b is the same as the voltage across
r two because the current here is zero and
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so there is no voltage drop across r three
and v o c is then given by voltage division
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that is r two divided by r one plus r two
times twelve volts and that turns out to be
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eight volts
in the second case we can first find this
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equivalent resistance and that in fact is
the same as the resistance we found here that
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is four ohms and then the open circuit voltage
is two amperes flowing through four ohms that
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is two times four or eight eight volts like
that and the net open circuit voltage the
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thevenin voltage that we are looking for is
v o c equal to v o c in case one plus v o
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c in case two that is eight plus eight or
sixteen volts so this is our thevenin equivalent
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circuit and we can replace the original circuit
as seen by r l with this much simpler thevenin
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circuit with v t h equal to sixteen volts
and r t h equal to four ohms all right
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now we have seen earlier that the power transfer
to r l is maximum when r l is equal to r t
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h that means r l should be four ohms so we
choose r l equal to four ohms then the power
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transferred from this original circuit to
r l is maximum it is now a simple matter to
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calculate this maximum power let us find i
l first what is i l it is v t h divided by
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r t h plus r l and since r l is equal to r
t h this i l is given by v t h divided by
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two times r t h like that and that turns out
to be two amperes once we know i l we can
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find the maximum power as i l squared that
is two squared times r l that is four ohms
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so that comes to sixteen watts
let us now look at the simulation results
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for the example we just discussed and the
circuit file is given over here so you can
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also try out the simulation and look at the
results here is our circuit and what we are
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going to do is plot the power absorbed by
r l that is the power transferred by this
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circuit to r l as a function of r l we will
also plot the current through r l and the
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voltage across r l as a function of r l
first let us look at i r l that is this current
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as a function of r l and that is given by
this red curve over here when r l is zero
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what is the current through r l we can look
at this circuit and find that out when r l
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zero the current reach sixteen volts divided
by four ohms that is four amperes and that
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is what is seen over here as r l is increased
the resistance in the circuit increases and
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therefore the current goes down like that
eventually of course it will reach zero
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next let us look at this voltage across r
l denoted by v l over here as a function of
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r l and that is given by this blue graph here
what is v l it is r l divided by r l plus
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00:24:23,010 --> 00:24:29,460
r t h times sixteen volts if r l is zero v
l is going to be zero and that is our starting
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00:24:29,460 --> 00:24:38,110
point here as r l is increased v l also increases
and eventually if r l becomes very large then
193
00:24:38,110 --> 00:24:44,910
this entire voltage is going to appear over
here so this blue graph is going to approach
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00:24:44,910 --> 00:24:52,980
sixteen volts as r l tends to infinity
now the power transfer to r l is given by
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00:24:52,980 --> 00:25:02,310
the product of v r l and i r l v r l increases
as r l increases whereas i r l decreases as
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00:25:02,310 --> 00:25:07,960
r l increases so clearly the product of these
two is going to go through a maximum at some
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00:25:07,960 --> 00:25:14,679
point and that is what is happening over here
and that happens at r l equal to four ohms
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00:25:14,679 --> 00:25:18,890
and as we have seen in the last slide that
condition is the same as r l equal to r t
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00:25:18,890 --> 00:25:25,770
h that is four ohms and the maximum power
is sixteen watts and that agrees with the
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00:25:25,770 --> 00:25:32,030
calculation that we looked at in the last
slide
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00:25:32,030 --> 00:25:36,660
to summarize we have learnt how to convert
the thevenin form into the norton form and
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00:25:36,660 --> 00:25:41,900
vice versa we have also seen the maximum power
transfer theorem for circuits and we have
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00:25:41,900 --> 00:25:46,840
seen that maximum power is transferred when
the load resistance is equal to the thevenin
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00:25:46,840 --> 00:25:49,999
resistance all right
see you next time