Hello friends lets take a problem
where we can use Thevenin's theorem to
solve a problem so let's have a problem
I'm having a voltage source of 100 volt
and resistances are of 20 80 this is 40
and 50 and here its a load resistance
RL this load resistance can have 3
values 5 10 ohm and 20 ohm  so in all the
3 cases
I need to find out what will be the load
current using Thevenin's theorem let's
follow the procedure that we have
discussed first calculation of V th
you in order to get a V th we need to
remove RL circuit will be like this
once we have removed  RL we will get
two terminals will mark that as A and B
our job is to get voltage across A and B
so I will consider A positive with
respect to B this voltage we will denote
as V th in order to get a V th we
need to trace a path so either you can
select this path or this path as we have
discussed we need to select a simplest
path so lets select this path or you
can go for this path also let's select
this path where I am having 20 and 40
ohm resistances connected right so my
job is to get voltage across 20 ohm and
voltage across 40 ohm if I know the
resistances I can get voltages provided
current should be known so in this my
objective is get this two currents now
it is again a very simple circuit where
100 volt is connected between these two
points and these two points where I can
directly say current flowing through 20
ohm resistance will be this and given by
100 divided by 20 plus 80 which is 1
ampere same way current flowing through
40 ohm resistance will be 100 divided by
40 plus 50 if I solve I will get 1.1111
ampere so I got the parameters
now let's select this path
this is point A and B we consider A
positive with respect to B and the
voltage across them is V th this is 20
ohm 40 ohm current flowing through this
is I 20 and current flowing through this
is I 40 since we have assumed a current
direction same as the current direction
we will get voltage drops so apply KVL
to this loop I can start from point B
and will come to the same point so from
B to A I am having minus plus and the
voltage is V th minus plus switch once
more plus 20 multiply by I 20 plus minus minus
40 multiply by  I 40 equal to 0 so V th will be 40 multiply by 
I 40 I 40 we calculated as 1.1111
ampere 20 I 20 we got as 1 equal to 0
if I solve this I will get V th as
24.4444
volt again a positive answer I am
getting so assumption is once again
correct so A positive with respect to B
step number one is over lets go to the
second step
calculation of R th for this we need to
remove RL and whatever sources are
present we need to replace with their
internal resistances so here we are
having only one voltage source so that
voltage source is 100 volt so 100 volt
is short-circuit because it is a voltage
source internal resistance is 0 so if
I do the modification here I will get a
circuit like this
I am short-circuiting 100 volt this is
20 80 40 and 50 Ohm lets not forget to
mark these points A and B and here we
have to find out R th across A and B
again same concept lets have this
point of C so this entire thing is C
only so lets redraw the circuit
I will get from A to C there is a 20 ohm
again from A to C I have 80 ohm from C
to B that is 40 ohm resistance and from
C to B that is 50 ohm so it is very
simple now because this 20 and 80 are in
parallel the equivalent is 20 multiplied
by 80 divided by 20 plus 80 resultant I
will get as 16 ohm 40 and 50 are in
parallel resultant is 40 multiplied by
50 divided by 40 plus 50 here it is 22.2222
ohm and now these two
resistances will come in series 16 come
in series with 22.2222 ohm so
R ab nothing but R th will be 16 plus 22.2222
so finally R th I will get as 38.2222
ohm
step number 2 is over lets go to step
number 3 which is Thevenin
equivalent circuit
it will have V th in series with R th so
V th is 24.4444
volt and R th we are getting 38.2222
ohm now here
A positive with respect to B so i
consider this point is A this point is B
and the last step
calculation of IL in R case we have
consider three values of RL so lets
connect the RL to this circuit now first
case if RL is 5 ohm the current flowing
through RL which we denoted as IL will
be V th divided by R th plus RL so if i
substitute the RL 5 ohm  V th and R th value I
will get IL as 0.5656
 ampere lets decide the direction
how we are going to decide the direction
lets consider the circuit so in
original circuit RL is connected like
this and if I remove that I will get
terminals A and B right so in R answer
I am getting a current flowing from A to
B so here also it is A to B but for this
horizontal orientation of the load
resistance the current Direction is like
this which is from A to B so this will
be the value of IL if I consider RL
5 ohm second  if RL is 10 ohm
simply we have to replace RL as 10 ohm
considering V th R th  same if we solve
we will get IL at 0.5069 ampere
current Direction is same and in third
case RL equal to 20 ohm if I replace RL
as 20 I will get IL 0.4198
ampere direction is A to B which is same
as the previous two cases so this will
be end of another problem that we use
Thevenin's theorem in subsequent videos we
will solve more numericals based on this
thank you
