Hello friends lets apply Thevenin's
theorem to a problem so lets take a
simple problem having a one source
so in this problem I have just consider
one source of 80 volt and remaining
elements are just resistances 10 50 this
is again 10 50 and 60 Ohm objective is to
get current flowing through 60 ohm using
Thevenin's theorem now we are going to
solve the problems in 4 steps step
number one calculation of V th
as per the theorem whenever we have to
find out the V th we need to find out
voltage between the terminals where load
is connected so I will first mark those
terminals A and B and for V th we need to
remove that load resistance so in this
case load resistance is 60
ohm so remove RL which is 60 ohm in R
case so I need to redraw the circuit so
circuit will become like this
80 volt as it is and resistances
accept RL we'll be there in a circuit
and our job is to get V th across
terminals A and B so whenever you see
the circuit properly how to get V th you
need to trace a simple loop which will
cover A and B points meaning I will
consider this loop 50 ohm 10 ohm which
is covering this A and B points so my
objective is we need to find out voltages 
 these two branches in order to
get V th so finally I will come to know I
should know current flowing through 50
ohm and current flowing through 10 ohm
in order to get V th so once you are sure
what exactly we need to find out then we
will redraw the circuit and we will just
find out those parameters so in R case
our parameters are current flowing
through 50 ohm and current flowing
through 10 ohm you can select any other
part but I am saying i am suggesting you
take a shortest path having minimum
number of elements so this will be the
shortest path for the problem so to get
our objectives lets redraw the
circuit one more time here it is very
simple these are the two terminals where
elements are connected so 80 volt then
to this 80 volt I am having two
resistances connected like this
and two more resistances connected like
this
this is 10 ohm 50 50 and 10 ohm point A
is between this 10 and this 15 so this
is point A point B between 50 and 10 50
and 10 point B so it is a very simple
circuit now the current flowing through
this branch which I will denote as I
50 because I am interested in knowing
this current and here I will denote this
as I 10 so since both these branches are
connected to 80 volts source current
flowing through 50 ohm resistance will
be 80 divided by 10 plus 50 which is 60
so 80 by 60 so through 50 ohm current
flowing is 1.3333 ampere now
for this I will denote it as current
flowing through I 10 ohm it was same
because this branch is connected to 80
volts source only should 80 divided by
50 plus 10 equals 80 divided by 60 equal to
1.3333 ampere now we got our
objectives current flowing through 10
ohm current flowing through 50 ohm now
what you do we have to consider that
loop which we have considered for
calculation of V th so here we consider
like this plus minus v th so I will
redraw that loop so it will be like this
50 and 10 these are the two points A and
B
so here I consider A positive with
respect to B and the voltage is V th for
this 50 ohm I got current like this I 50
because of the direction downward it
will develop the voltage polarity in the
current direction same way for this 10
ohm I calculated current in downward
direction so voltage developed across it
is also in the downward direction
meaning in a direction of current only
so for simple loop I will apply KVL so I
will start from B point and come back to
the same point so minus plus it is plus
V th plus minus it is minus 50 is the
resistance and I 50 is the current minus
plus plus 10 is the resistance I 10 is
the current equal to 0 V th is unknown I
will keep this on left hand side minus
50 into I 50 will go so its a 50
multiplied by I 50 which is 
1.3333 ampere this will go other side minus
10 multiplied by again same 
1.3333 so if I solve I will get V th as 53.3333
 volt since I am getting the
positive answer that mean I assume A
positive with respect to B is correct so
right here A positive with respect to B
step number one is over lets go to
step number two calculation of R th
for calculation of R th we have to remove
RL in this case it is 60 ohm and all
voltage sources we need to short-circuit
and all current sources we need to open
circuit so in this case we are having
only one voltage source and that is 80
volt so 80 volt is short circuited so I
will consider this circuit so in this
circuit
I will short circuit 80 volt so I will
get circuit like this 80 volt is short
circuited this is 10 ohm 50 50 and 10
and these are the points A and B never
forget to mention these points because
this will change your circuit
orientation altogether meaning if I
forget here A and B points you simply
say 10 and 50 in series 50 and 10 in
series which is wrong because I should
have A and B points intact
so I cannot say 50 and 10 are in series
okay so lets consider this point here
so this point I will consider as say C
if you see this point is also C this
point is also C and this point is also C
so lets redraw the circuit so after
reading the circuit I will get from A to
C there is 10 ohm so A to C that is 10
ohm similarly A to C there is 50 ohm
C to B one more 50 ohm and B to C there is
10 ohm
so I will get this 10 and 50 are in
parallel same way this 50 and 10 are in
parallel so 10 parallel 50 I will get 10
multiplied by 50 divided by 10 plus 50
if I solve I will get answer 
8.3333 
Ohm same way this will give me 50
parallel 10 again 1 on the same so it is
8.3333
Ohm so if I redraw the circuit
considering this modification it will be
like this from A and from B 2
resistances I will get of the same value
8.3333 and 8.3333 so
if I solve I will get R ab which is
nothing but R th as 8.3333 plus
8.3333 nothing but 16.6667
ohm
so now second step is over lets go to
the third step third step is you have to
draw Thevenin's equivalent circuit so
terminal equivalent circuit consists of
V th in series with R th now before
marking points A and B we need to check
V th so V th I am getting 53.3333
volt A
positive with respect to B so positive I
have mark over here so obviously this
point will be A this point will be B
this is V th of the value 
53.3333 volt this
is R th of the value 16.6667
 ohm so this is a Thevenin's
equivalent circuit now the last part is
to get value of IL so for this I will
use Thevenin's equivalent circuit and
here I will connect a load resistance
between points A and B so load
resistance I am connecting which is of a
60 ohm and getting a current IL is
now very simple so IL will be V th
divided by R th plus RL if I substitute
53.3333 divided by 16.6667
 RL is 60 I will get IL
equal to 0.6957 ampere 
 and the orientation I will get is
vertical and I am getting a positive
answer for
IL so direction that I have considered
it is correct which is downward so this
is an end of problem number one where we
have used Thevenin's theorem in subsequent
videos we will solve more numericals
based on this thank you
