Hi friends in this video we are going to
see the application of Norton's theorem
in order to solve a problem so let us
have a problem
so in this circuit I am having a voltage
source of 10 volt one more voltage
source of 5 volt and resistances are of
5 to 1 and 4 OHM and these are the
terminals given A and B across which we
have to replace the entire circuit as a
Norton's equivalent circuit so the
heading is find equivalent Norton's
circuit between terminals A and B if you
see properly what I need to do to get IN
 I need to short-circuit these two
points and get a current but it is very
difficult here
because it may give rise to a
three-dimensional structure which is
very difficult to analyze so what I will
do I will redraw the circuit keeping all
the concepts as it is so my first step
over here before applying Norton's theorem
redrawing the circuit
see how I am going to redraw the same
circuit but with a different orientation
I will keep this part of the circuit as
it is
so here it is 10 volt phi OHM 2 ohm with
terminal A now what I am going to do
from this 2 ohm will consider this branch
and I will draw like this from this
point I have 1 OHM  resistance connected
to B terminal so it is like this 1 OHM
connected to B terminal from B 4 OHM
resistance branch will start so this is
4 ohm and this end of a resistance is
connected to positive terminal of 5 volt
battery so it is like this positive is
connected to 4 OHM and this negative
will go same circuit I have redraw like
this why I have done this because it is
very easy to analyze now so the step
number 1 is calculation of IN what we
need to do for that we need to connect A
and B points by a straight line or short
circuit link so here only I will replace
it with a short circuit link like this
and I need to find out current flowing
through this so for this purpose I will
use mesh analysis because mesh analysis
is the simplest technique while solving
Norton's theorem in subsequent videos
will solve more numericals where this
concept will be clear so right now there
are three loops so 3 currents I will get
this is I 1
I 2 and I 3 as per the direction of
currents I will get a voltage drops now
all the elements are having the voltage
is associated with it so lets apply
KVL to loop number 1 I will start from
this point so from here to here there is
no element here plus minus so its
minus 2 I1 minus plus plus 2 I 2 plus
minus minus 5 I1 minus plus plus 10
equal to 0 if I simplify this I will get
minus 7 I 1 plus 2 I 2 equal to minus 10
equation number 1 similarly I can apply
KVL to loop number 2 so for this I will
start from this point no element is
granted here plus minus so it's minus I
2 minus plus is plus I 3 no element over
here plus minus is minus 2 I 2 once
again and minus plus plus 2 I 1 since
there is no battery in this loop
constant is equal to 0 if I simplify I
will get 2 I 1 minus 3 I 2 plus I 3
equal to 0 equation number 2 similarly I
can apply KVL to loop 3
loop three I will start from this point
so in this direction minus plus is the
voltage rise plus five ohm plus minus
minus 4i three plus minus minus I 3
minus plus plus I 2 equal to zero if I
simplify I will get I 2 minus 5 I 3
equal to minus five equation number
three if I solve all these three
equations I will get I 1 as 1.9231 
 ampere I 2 as 1.7308
ampere and I 3 equal to 1.3462
ampere now how to decide I N go back to
the circuit if you see over here I want
this current from B to A or A to B so I 2
is satisfying that and it is in this
direction so I can say IN  is nothing
but I 2 which is equal to 1.7308
ampere  and I am
considering the positive value the
positive values from B to A so here I
will write from B to A
step number one is over lets go to the
second step that is calculation of RN
second step is a calculation of RN for
this we need to do three things one is
we need to open circuit RL second
current sources need to be replaced by
open circuit and voltage sources need to
be replaced by short circuited in R
case there is no load resistance given
so it's already open so no need to
mention it we are having voltage sources
and 10 volt and 5 volt  those voltage
sources need to be replaced by short
circuited so if you do this modification
the circuit will be like this so we
replace 10 volt by short similarly 5
volt is also short circuited so I will
get circuit like this with terminals A
and B across which we have to find out
RN and resistances are 5 to 1 and 4 ohm 
this 2 and 5 are in parallel because
those resistances are forming a loop 2
into 5 divided by 2 plus 5 that will
give you answer 10 divided by 7 which is
1.4286 ohm similarly this 1 and 4
are also in parallel so the answer is 1
into 4 divided by 1 plus 4 so it is 4
divided by 5 0.8 OHM so if I
redraw the circuit I will get two
resistances connected in series which
finally comes between points A and B so
its a 1.4286 and this is 0.8
 comes in series
so I can directly say RN which is nothing
but RAB series combination of 0.8
 and 1.4286
which will give you RN as 2.2286 
 Ohm
so second step is also over now third
step is Norton's
equivalent circuit so as per the theorem
it will have a current source parallel
with a resistance current source is
called as Norton's current which is IN
and in our case it is 1.7308 ampere
resistance is RN which is norton's
resistance the value we got is 2.2286
 OHM now current we got
is from B to A so in order to get a
current from B to A here
I should mark this point as B and this
point as A since load resistance is not
given so no need to perform the step
number four so this will be our final
answer where we have replaced entire
circuit by a simple current source with
a resistance across it in subsequent
video we will see more numericals
based on Norton's theorem thank you
