Hi friends in this video we are going to
take a numerical which will give you a
application of KVL and KCl together so
lets have a numerical so let's mark
all the values to the elements so I will
consider this as a battery of 10 volt 20
and 30 volt let's give values to the
resistance in increasing order starting
from 1 so I could have a random value so
lets start 1 2 3 4 5 and 6 now I will
apply KCl to node A B C and D such a way
that I will get all branch currents so
let's take this as I 1 out of this I 1
which is an incoming current for node A
I will consider some of the current is
outgoing in this 2 ohm resistance so
obviously KCl at A I will get I 1 minus
 I 2 same will be the current flowing
through the 3 ohm so here I will mention
I 1 minus I 2 now I consider this
current as I 3 and KCl at point B will
give me current flowing through this
branch as I 1 minus  I 2 minus I 3 same will
be the current through this branch so I
will mention here it is I 1 minus I
2 Minus  I 3 this current will be I 2 and
this current will be same I 3 and
KCL at C I will get current
through 6 ohm as I 2 plus  I 3 and finally if
I apply KCl at D I am getting this
branch current is I 1 which is same as
the current flowing through the 1 ohm
resistance so I have done all current
markings based on KCl at different nodes
so I will write here by applying KCl at
nodes A B C and D all branch currents
are marked which will be our objective
meaning I may ask the question find out
the current flowing through the 1 ohm 2
3 4 5 or 6 so we will take only one of
the outcomes so I will consider let's
calculate what is the current flowing
through the 2 ohm and I have to apply KCl
and KVL to get an answer so I apply a
KCl and I have marked all branch
currents based on this direction of
currents I will have the voltage drop
across resistances so in the direction
of current I will have voltage drops of
the polarities which I have shown like
this
so my circuit is ready remember these
three are the voltage sources so it will
not have the direction also the polarity
based on direction of current it will be
fixed longer arm plus shorter arm minus now
my circuit is ready for application of
KVL so lets apply KVL to loop 1 loop
2
and loop 3 everytime I am tracing a
path in clockwise direction
so keeping this in mind let's write the
equation so first lets apply KVL to
loop 1 so I will get minus 1 I 1 minus 2
I 2 minus 6 current flowing through 6
ohm is I 2 plus I 3 and potential
difference for a battery is plus 10
equal to 0 now I will simplify I will
get minus I 1 minus 2 I 2 minus 6 I 2
minus 6 I 3 I am taking this constant to
other sides so that will become minus 10
so finally I will get expression minus I
1 minus 8 I 2 minus 6 I 3 equal to minus
10 this will be equation number 1
similarly I can apply KVL to loop number
2 so if I apply KVL to loop number 2 I
will get minus 20 minus 3 I 1 minus I 2
current flowing through the 3 ohm minus 4 
I 3 plus 2 I 2 equal to 0 let's simplify
so I will consider minus 20 as it is
minus 3 I 1 plus 3 I 2 minus 4 I 3 plus 2
I 2 equal to 0
so let's Club the like terms so 3 I 2 2 I2
 will become 5 I 2
and lets take this constant  minus 20 on
another side of equation so my equation
will become minus 3 I 1 plus 5 I 2
minus 4 I 3 equal to 20 equation number
2 lets apply KVL to loop number 3 in a
similar manner
so apply KVL to loop 3
so I will have 6 into I 2 plus I 3
plus 4 I 3 minus 5 I 1 minus I 2 minus I 3
and plus 30 equal to 0 see all this plus
minus we have considered as per the
direction of current and whether it is
having a voltage drop or voltage rise in
case of voltage drop we consider the
negative in case of voltage rise we
consider positive so let's simplify I
will take this plus 30 on another side
of equation which will become minus 30
let's club all light terms so here minus
5 I 1 will be alone then this 6 I 2 and
this 5 I 2 will become 11 I 2 6 I 3
for I 3 and 5 I 3 will give you 15 I 3
equal to minus 30 this will be equation
number 3 so I have 3 equations and 3
unknowns solving equation 1 2 & 3 I will
get answer I 1 equals 3.3333
ampere I 2 is same 3.3333 ampere and I 3
I will get minus 3.3333 ampere
minus means whatever direction of I 3 we
have assumed that need to be reversed
that
the only thing otherwise the magnitude
will remain same now our objective was
to get a current flowing through the
2 ohm so current flowing to the 2 ohm is
I 2 and we are getting answer for I 2 is
3.3333
ohm so finally I can write current
flowing through the 2 ohm will be 3.3333
ampere this way you can get current
flowing through any of the branches by
using a combination of I1 I2 I3 meaning
suppose they are asking what is the
current flowing through the 5 ohm
so simply I 1 minus I 2 minus I 3 you
have to put the values of I1 I2 and I3
whatever we got and get an answer so
similarly we can have a current flowing
through all the branches so this way KVL
and KCl together can be applied to a
problem to get a current flowing through
any of the branches given we will see
more numericals based on this KVL and
KCl it is a modification we call that as
a nodal and mesh in subsequent videos
thank you
