Hi friends in this video we are going to
see the new technique to simplify and
solve a problem and it is called as
Norton's theorem so lets see what the
statement of Norton's theorem is saying
its states that it is possible to
simplify any complex linear circuit to
an equivalent circuit with just a single
current source and a parallel resistance
connected to a load where current is
called as Norton's current which is
obtained between the terminals where
load is connected after replacing load
by a short-circuit link and a value of
resistance which is also called as
norton's resistance is obtained by
removing load resistance and replacing
all the power sources by their internal
resistances so what is the important
thing here you have to understand is how
to get this Norton's current the
procedure to get a Norton's current is
we have to replace load by a short
circuit link that means we have to
remove RL and we have to place a wire
between the terminals where load
resistance was connected and norton's
resistance is obtain  first of all we have
to remove load and then all the power
sources we need to replace with their
internal resistances so if you see
properly the second part of norton's
theorem is same as second part of
thevenin's theorem so there is absolute
similarity between how to find out
norton's resistance
and thevenin's resistance all the difference
is how to get this norton current to
elaborate this concept lets take a
simple problem and solve it by norton's
theorem so I will consider a circuit
with voltage sources and three
resistances connected like this I will
consider this is V 1 this is V 2 R 1 R 2
and lets consider this as load
resistance RL through which we are
supposed to find out the current using
Norton's theorem so objective is to get
IL  using Norton's theorem so it is
saying four steps first step is a
calculation of I N which we call as
Norton's current calculation of I N  for
this as per the theorem you will get
Norton's current if you replace load by
a short circuit link see the
modification that we are supposed to
make while calculating I N instead of
this RL I will have a short circuit link
like this and the rest of the part of
the circuit will be same so I will get
this circuit with V1 V2 acting R1 R2
will remain as it is only thing is that
RL we replaced by short circuit so that
I will get part between A and B shorted
having current I N
to get this iron you can use any of the
technique either a circuit reduction
technique by source transformation or
nodal analysis or mesh analysis my
advice and we will see this through
certain problems whenever you solve a
problem of a Norton's theorem while
calculating I N preferably go for mesh
analysis I will elaborate this concept
at the time of solving numericals also
so find I N by using mesh analysis so
once we got iron we can move ahead with
the second step and that is calculation
of RN so how to get RN which is a
norton's resistance we need to remove
load remove means we have to open
circuited and all the power sources
voltage source and current source need
to be replaced by their internal
resistances so for calculation of iron
three things you have to remember what
are those first we have to remove RL lets
 not confuse ourselves here while
calculating I N what I have done RL I
have replaced with a short circuit so I
should write over here RL is short
circuited and for calculation of RN I am
saying remove RL that means I will open
circuit RL what about the sources here
we are having two voltage sources being
a voltage source is internal resistance
is zero so voltage sources should be
replaced by zero resistance which is
short circuit so voltage sources
are short circuited though here we are
not having a current source but still in
general if current source is present
that we need to replace by open
circuited because internal resistance of
ideal current sources in finite and
infinite resistance means open circuit
so if I apply these modifications to the
original circuit I will get circuit like
this having resistances R1 R2 remember
we have open RL leaving behind two
terminals A and B through which we have
to find out
R TH or you can say RN so sometimes I may
say it's R TH but the procedure of R TH
and RN finding is same so I will write
here R TH is same as RN in this case I
will get RN as R 1 parallel R 2 which is
multiplication of two resistances
divided by addition of them so RN we
have found out third step is drawing
Norton's equivalent circuit lets go
back to the theorem once again what it's
saying you can replace the circuit to an
equivalent circuit with just a single
current source and a parallel resistance
so the norton's second circuit will be
like this a current source with which we
are having a parallel resistance current
source is IN and parallel resistance is
RN
and these are the points A and B how we
are going to use Norton cyclone
circuit to find out IL that we will see
in next step for that I need to connect
load resistance to the norton's
equivalent circuit and our job is to get
current flowing through it its a
standard problem of a current divider
rule so while calculation of IL we need
to use current divider rule only so by
current divider rule I will get il which
is nothing but the resistance which is
connected parallel to RL is RN
divided by addition of these two
resistances multiplying total current
supply which is IN so every time while
calculating IL we are going to use this
formula which is current divider rule so
here we have seen how norton's theorem can
be applied to a problem which is a
simple problem we have taken in order to
get IL  now one more thing we want to
develop and that is relationship of
thevinin's and Norton's equivalent circuit
so actually what I am doing over here
now I will consider thevenin's equivalent
circuit
as per the theorem I will have circuit
like this so here at thevenin's voltage VT
thevenin's resistance RTH A and B point as it
is now if you see properly it is nothing
but a voltage source since it is having
a resistance in parallel with it which
we consider as a internal resistance of
this voltage source so it is practical
voltage source let's say its counterpart
which is norton's equivalent circuit  norton's equivalent
circuit will have a current
source and parallel to it we will have a
resistance so current value is IN which
we call as an  norton's  current and this is
norton's resistance RN A and B point as
it is now if you see properly
I can relate this to a current source this
current source is having a resistance
across it which I can consider as
internal resistance so ultimately it is
practical current source the thing is
that these two circuits are linked to
one another how I can convert this
practical voltage source into a
practical current source so if I have a
practical voltage source which is a
Thevenin equivalent circuit I can get
its equivalent current source which will
be Norton's equivalent circuit and the
relationship is IN equal to V TH 
divided by R TH and
RN will be same as R TH so what I
have done over here I have convert
voltage source to current source voltage
to current source conversion now next
suppose I know practical current source
I can get our equivalent voltage source
which is a Thevenin equivalent circuit
and for that
V TH  is calculated In multiplied by RN and R TH 
will be same as R N what is this if I
know not on the current circuit I can
get equivalent Thevenin circuit so here
it is current to voltage source
conversion so this is a relationship
between thevenin's and Norton's
equivalent circuit in subsequent videos
we will solve numerical based on
Norton's theorem thank you
