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welcome back to basic electronics in this
lecture we will look at a very useful theorem
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called the thevenins theorem we will first
try to understand why it is possible to represent
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ah circuit with this thevenin equivalent form
we will then look at how to extract the social
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credit parameters namely the thevenin resistance
and the thevenin voltage after that we will
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consider ah couple of examples to illustrate
how was circuit can be represented by is thevenin
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equivalent we will also see how the thevenin
parameters can be extracted graphically so
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let us get started
thevenins theorem is a very useful and important
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theorem and particularly in electronics circuits
we will find it ah very useful but before
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will look at the statement of the thevenins
theorem let us see briefly where it comes
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from and we will do that with the help of
this ah example here is the circuit to which
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we have connected ah load register r l and
we are interested in this voltage across r
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l and in particular we want to ask this question
how is v related to the circuit parameters
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we begin by ah assigning node voltages to
this various nodes in the circuits with respect
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to a reference node so let us first choose
ah reference node we have taken that as the
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reference node so the node voltage there is
zero volts and with expect to that reference
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node this voltage at node a a is v one the
node voltage at b is b two and the voltage
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at node c is v three now what we will do is
write k c l equation in terms of v one v two
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and v three at this nodes nodes a b and c
and ah in doing that let us define g one a
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conduction as one over r one g two equal to
one over r two etcetera
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these are our k c l equation let us look at
k c l at node a that is our node a we have
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three currents entering or living node a these
one current going like that and that is v
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one minus v three divided by r one so that
is g one times v one minus v three and since
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that current is living this nodes we will
take it as positive then we have this current
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v one minus v two divided by r two that means
g two times v one minus v two and then we
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have this current i zero entering the nodes
sense this current is entering the node we
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take that as negative and thats why it appears
as minus i zero in this equation and all these
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currents must add up to zero and that gives
us k c l at node a
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in a similar manner we can write k c l at
node b and k c l at node c in terms of the
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node voltages v one v two and v three and
we can now write these equations in a matrix
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form like that let us look at one of these
equations lets say the first one this first
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equation comes from k c l at node a and let
us verify that what is the coefficient of
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v one in this equation it is g one plus g
two so that goes over there thats going to
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multiply v one what about v two the coefficient
is minus g two so that goes there and that
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multiplies v two and the coefficient of v
three is minus g one so that goes over there
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in addition we have minus i zero over here
and since that does not depend on v one v
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two or v three we take that to the right hand
side so it becomes plus i zero and that is
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how it appears here
so this matrix equation is of the form g times
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v equal to i s where g is this matrix here
three by three v is this column vector consisting
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of v one v two v three the node voltages in
our circuit and i s is the source current
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vector this one all right now if you recall
our objective is to find the relationship
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between v and the rest of the circuit and
in order to do that we want to solve this
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matrix equation and obtain v two since v two
is the same as b here because this node voltage
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is zero
and v two can be obtained using cramers rule
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and what does cramers rule say it says that
v two is given by d one divided by d two where
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d one is the determinant of this matrix with
the second column replaced with this r h s
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vector and d two is the determinant of this
matrix as it is so let us do that and simplify
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things after that
ok so this is what we have for v two the ratio
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of two determinants determinant of this original
g matrix with the second column replaced with
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the r h s vector like we mentioned in the
last slide and the determinant of the g matrix
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itself all right now let us denote this determinant
by delta one and what we should note about
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delta one is that it does not have r l or
g l in it so this value is independent of
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the load resistance value let us look at the
denominator now that is determinant of g this
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is our g matrix and this determinant does
depend on g l we have g l over here and what
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we will do now is write this column as the
sum of two columns column one minus g two
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g two zero column two zero g l zero and if
you do that we can write this determinant
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as the sum of two determinants this determinant
as no g l term in it whereas this one does
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have g l in it right there
let us denote this first determinant by delta
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and let us write this second determinant as
g l times delta two where delta two is the
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determinant of this matrix here and notice
that we dont have g l here anymore because
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that has been taken outside all right so with
these substitutions we can write v two as
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delta one divided by determinant of g that
is delta one divided by delta plus g l delta
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two that term where delta delta one delta
two are all independent of g l so the only
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dependence of v two on g l is right there
so here is v two once again delta one divided
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by delta plus g l times delta two and let
us emphasize once again that delta one delta
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and delta two have no dependence on g l or
r l and they will only depend on the circuit
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parameters here all right let us now look
at the open circuit value of ah v two that
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means the value of v two when r l is infinite
and when r l is infinite g l which is one
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over r l becomes zero and therefore we have
v two equal to delta one divided by delta
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so that is our v two in the open circuit condition
we can now rewrite v two by dividing both
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the numerator and denominator by delta so
in the numerator we get delta one divided
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by delta in the denominator we get one here
so one plus g l times delta two over delta
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now delta one by delta is nothing but v two
o c v two open circuit so we replace that
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with v two o c and that gets divided by one
plus delta two over r l times delta here we
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have used the fact that g l is one over r
l and now let us ah multiply both numerator
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and denominator by r l to get r l divided
by r l plus delta two over delta times v two
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o c
now it turns out that this quantity delta
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two divided by delta has units of resistance
and you should really go back one slide and
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check that out so let us define r thevenin
that is thevenin resistance as delta two divided
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by delta and with that we get v two equal
to r l divided by r l plus r thevenin times
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v two o c like that
oh the stop sign has come and why is it there
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that means there is some unfinished business
and that is this delta two by delta you need
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to really go back and check that it has units
of resistance so do that stop the video here
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and when you complete this assignment start
again with the video
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so this is our expression for v two and it
is looking very much like ah voltage division
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formula if we have a source which is equal
to v two o c like that and r thevenin and
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r l in series then the voltage across r l
which is v two here would be r l divided by
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r l plus r t h times v t h so this circuit
and this expression are compatible with each
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other if you have v t h equal to v two open
circuit
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in other words this expression allows us to
replace the original circuit with an equivalent
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simpler circuit and that simpler circuit is
the thevenin equivalent circuit so here is
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our original circuit and this entire circuit
can now be replaced with this much simpler
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circuit consisting of only one voltage source
in series with one resistance the voltage
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is called v t h or thevenin voltage and the
resistance is called r thevenin or r t h and
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later of course we will look at how we can
obtain v t h and r t h from the original network
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and that brings us to thevenins theorem if
we have a circuit consisting of registers
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independent voltage sources independent current
sources and dependent sources such as current
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controlled voltage source current controlled
current source etcetera then thevenin theorem
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says that this circuit can be represented
by its thevenin equivalent circuit consisting
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of a voltage source v t h in series with a
resistance r t h
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and now imagine that we have a load resistance
r l connected between a and b to the original
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circuit as well as to the thevenin equivalent
circuit then thevenin theorem says that the
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load resistance does not see any difference
between the original circuit and its thevenin
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equivalent that means the voltage across the
load resistance or the current through the
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load resistance would be identical in these
two cases all right
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now the next question that comes up is how
do we figure out v t h and r t h from the
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original network so let us look at that first
let us look at how to get v t h from the original
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circuit since the two circuits are equivalent
the open circuit voltage must be the same
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in both cases now let v o c be the open circuit
voltage for the left circuit here that means
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we dont connect anything between a and b and
measure this voltage v o c between a and b
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for the thevenin equivalent circuit what is
the open circuit voltage if we dont connect
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anything between a and b there is no current
and therefore there is no voltage drop across
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r t h and therefore this open circuit voltage
would be the same as this v t h
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and therefore we conclude that this v t h
is the same as the open circuit voltage seen
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between a and b in the original circuit and
that is what this equation tells us
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what about r t h the thevenin resistance here
that can be found by different methods and
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let us see what those methods are method one
here is our original circuit and that is the
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thevenin equivalent representation what we
do is deactivate all independent sources in
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the original circuit and we do the same thing
in the thevenin equivalent circuit as well
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and in the thevenin equivalent circuit the
only independent source is this voltage source
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here and when we deactivate that we replace
that with a short circuit like that and what
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we see then between a and b is simply the
thevenin resistance r t h and now if we look
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into the port a b of the original circuit
with the voltage sources and current sources
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deactivated then whatever resistance we see
must be the same as r t h
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often r t h can be found by inspection of
the original circuit with the independent
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voltage sources and independent current sources
deactivated and if that does not work what
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we can do is to use a test source we can use
either or test voltage source like that or
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we can use a test current source like this
and what we do then is for example if we use
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a voltage source then we find this current
i s and then r t h is given by v s divided
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by i s in this case where we have a test current
source then we find this voltage drop between
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a and b and then r t h is given by v s by
i s
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let us ah look at our second method now this
is our original circuit and that is the thevenin
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equivalent we look at two things one the open
circuit voltage between a and b in both cases
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second the short circuit current which flows
from a to b when a and b are shorted and we
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do that again in both cases and because the
original circuit is equivalent to this thevenin
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representation v o c here v o c here must
be the same and similarly i s c here and i
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s c here must be the same so we use this fact
to obtain r t h for the thevenin representation
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let us see how
let us look at this circuit the open circuit
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voltage between a and b is the same as v t
h here because there is no voltage drop across
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r t h the current being zero and therefore
we get v o c equal to v t h this equation
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here all right let us now look at the short
circuit current i s c is simply v t h divided
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by r t h because there is no other resistance
in the circuit so that gives us i s c equal
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to v t h divided by r t h and v t h is the
same as v o c therefore i s c is equal to
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v o c divided by r t h and that gives us a
formula to obtain r t h that is r t h equal
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to v o c divided by i s c
now since this v o c and this v o c are identical
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because the two circuits are equivalent and
also because this i s c and this i s c are
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the same what we can do is in the original
circuit we find the open circuit voltage and
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the short circuit current and then r t h is
given by that open circuit voltage divided
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by the short circuit current and note here
that we do not deactivate any sources in this
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case so the independent sources are left as
they are in the original circuit
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let us find the thevenin equivalent of this
circuit here as seen from r l so we need to
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find v t h and r t h let us find v t s first
and how do we do that v t h is the same as
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v o c the open circuit voltage between a and
b so we remove r l and then find this voltage
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and that is the same as v t h that is the
straightforward calculation there is no current
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through this two ohms resistance so no voltage
drop there and therefore v o c is the same
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as this voltage drop and that can be found
by voltage division that is three ohms divided
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by nine ohms times nine volts so that comes
to three volts so our v t h which is the same
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as v o c is three volts
next let us find r t h and how do we do that
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in the original circuit we deactivate the
independent sources in this case there is
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only one independent source thats a voltage
source so we shot it and then we look from
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this port a b and this is what we see from
a b this six ohms and three ohms come in parallel
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and that combination comes in series with
these two works and therefore our r t h terms
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out to be three parallel six plus two or four
ohms so over thevenin equivalent circuit as
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v t h equal to three volts and r t h is equal
to four ohms
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another example this is a little more complicated
than the previous example we have two independent
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sources now our current source six amperes
and a voltage source forty eight volts and
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we want to find the thevenin equivalent circuit
as seen from a b let us look at r t h first
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how do we find r t h we deactivate the independent
sources that is this current source and this
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voltage source the current source is replaced
with an open circuit and the voltage source
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is replaced with a short circuit and now we
look from a b and find the resistance that
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resistance is over forty eight and that is
fairly straightforward to do this for ohms
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and two ohms are in series so therefore we
have six ohms and that combination comes in
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parallel with twelve ohms and that entire
combination is four ohms
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on this side we have four ohms in parallel
with twelve ohms and that turns out to be
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three ohms so altogether between a and b we
have four ohms plus three ohms or seven ohms
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and that is our r thevenin let us now look
at v thevenin v thevenin is the same as v
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o c this voltage here and how do we go about
finding that voltage drop we note that this
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current i must be zero because there is no
return path for this current there is an open
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circuit here and if this current were not
zero then that would lead to violation of
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the charge conservation principle all right
once we figure that out the rest of the calculation
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is straightforward what we do now is find
this voltage drop between a and c this voltage
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drop between c and b and then v a b is simply
v a c plus v c b so let us do that
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since this current is zero we have two independent
circuits here one circuit is here and the
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other circuit is here and that makes things
very easy all right let us look at v a c first
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what is this resistance as we have already
found out that is four ohms and therefore
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six amperes going through four ohms gives
us twenty four volts so that is v a c what
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about v c b v c b is given by voltage division
of this forty eight volts between twelve ohms
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and four ohms so it would be twelve divided
by sixteen times forty eight and that terms
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out to be thirty six words so altogether between
a and b we have twenty four volts plus thirty
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six volts and that comes to sixty volts so
that is our open circuit voltage and that
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is the same as v t h so we have v t h equal
to sixty volts or t h equal to seven ohms
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and our thevenin equivalent circuit therefore
looks like this
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let us now look at ah graphical method for
finding v t h and r t h this is ah thevenin
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equivalent circuit of some original circuit
and what we do is connect a voltage source
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here and plot this current as a function of
this voltage and the plot that we are going
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to get is shown over here and let us now try
to understand why the plot looks like this
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what is this current it is v t h minus v divided
by r t h this equation and note that it has
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a negative slope the slope is minus one over
r t h and that has reflected in this plot
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here
what about the intercept of this ah line on
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the v axis to find that we put i equal to
zero and we get v equal to v t h so that gives
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us the x intercept and v t h remember is the
same as v o c what about the y intercept to
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find that what we do is put v equal to zero
that gives us i equal to v t h divided by
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r t h this quantity here and if you recall
that is the same as the short circuit current
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that means it is the current that would flow
if we connect these two nodes by a wire just
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a short circuit all right so now from a plot
like this we immediately find v t h as the
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intercept on the v axis and i s c which is
equal to v t h by r t h as the intercept on
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the i axis and from these two numbers we can
find our thevenin voltage and thevenin resistance
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and as a remark let us mention that instead
of a voltage source here we could have also
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connected a load resistor say r varied that
resistance and then plotted i versus v again
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we would get the same plot once again
let us apply the graphical method that we
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just discussed for this circuit which we have
already considered earlier and here is the
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sequel circuit file so you can run this simulation
and obtain the plot that we are going to look
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at very soon ok so how do we go about it we
connect a voltage source between a and b like
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that and then plot this current as a function
of this voltage and when we do that v o c
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is given by the intercept on the v axis and
i s c the short circuit current is given by
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the intercept on the i or the current axis
and here is the result i as a function of
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v the intercept on the v axis is sixty volts
and that gives us v o c and the intercept
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on the current axis is eight point five seven
amperes and that gives us i s c and v t h
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is the same as v o c sixty volts and r t h
is the same as v o c divided by i s c and
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that turns out to be sixty divided by eight
point five seven or seven ohms and these numbers
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of course are the same as what we got earlier
when we analyzed ah this circuit using different
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methods
to summarize we have become familiar with
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thevenins theorem we have also seen how do
i represent a given circuit with this thevenins
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equivalent form this background will be useful
to understand r c circuits amplifiers etcetera
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that is all for now
see you next time