Hello friends in this video we are going
to see a modification to nodal analysis
technique and that is called as super
node analysis so I will consider a
circuit having resistances connected in
parallel and voltage source connected
between 2 nodes and then I am
considering a current source with a
resistance parallel to it I will give
the values R1 R2 R3 lets take this
as voltage source of V and current
source of I ampere now we have to
analyze this circuit using nodal
analysis
let's define how many nodes are there in
this circuit node number 1
so I'll write VX as a potential of node
X VY a potential of node Y and
obviously I will consider this node
where four branches are connected as a
reference
now here problem is I am not having any
resistance connected in this branch so
if I write a equation I make it a infinite
to avoid this instead of writing the
equation for this branch current I will
simply consider this as IV current
flowing through this branch I mentioned
as IV is nothing but current flowing
through this voltage source lets apply
KCL at node X
so I am considering random current
directions only except for the current
source which I will give current
duration same as whatever current value
and it's direction it is having applying
KCl at node X gives me 0 minus VX divided
R1 equal to VX minus 0 divided by R2
plus IV see here I am NOT going to apply
any equation like VX minus VY minus V
divided by R because R is not known or
it's not given so if I write that
equation it will give you infinite value
so there is a problem so instead of
writing an equation for this branch current
I will simply mention IV and I will
consider this has equation number 1
similarly I can apply KCl at node Y so I
at node Y this current is incoming
along with this I and only outgoing
current this current flowing through R 3
so I will write equation IV plus I equal
to VY minus 0 divided by R3 this is
equation number 2 remember we have solved
numericals where I am getting the
equations in terms of node voltages only
so here along with our node voltages I
am having this IV term extra I is a
constant so that is not a problem so I
need to eliminate this IV to see how I
am going to eliminate IV so from
Equation 1
I can say IV aequal to minus VX by R1
minus VX by R2
so just rearrange the terms so that IV
will be on one side of equation and
remaining terms will be on another side
of equation similarly from Equation 2
I can get IV as VY divided by R3 minus
I another equation I will get equation
number 4 I can equate these 2
equations
and I can say minus VX by R1 minus VX by
R2 equal to VY by R3 minus I lets rearrange
 the terms so after rearranging the
terms I am getting VX in a bracket 1 by
R1 plus 1 by R2 plus VY 1 by R3 I'll
take this minus I this side I equation
number 5 so now what I have done I have
written 2 equations in terms of IV as
a third variable since I am having two
nodes so I should only have 2
equations and 2 variables but this
particular circuit having voltage source
between 2 nodes given because of that
I am getting third variable IV also so I
eliminated the third variable IV like
this so ultimately this equation I will
get second equation I need to form so
let's go back to the circuit once again
so if you see circuit closely
I will come to know V is a voltage
source given between two points X and Y
so obviously it is a potential
difference between these two nodes so I
can write VX minus VY equal to V
remember voltage source V and current
source I are the constants so I will get
equation number 6 so equation 5 and
equation 6 are in terms of node voltages
and solving them
I will get node voltages VX and VY so
our work is done but we need to apply
such a long process in order to avoid
that let's generalize the thing now
let's go back to the equation 5 once
again it is VX bracket 1 by R1 plus 1
by R2 plus VY 1 by R3 equal to I so
let's generate this equation directly ok
so I will consider a circuit once again
so better I will draw the circuit
removing V voltage source so the circuit
will be like this
so I remove the voltage source means I'll
consider that as a short circuit now
these are R1 R2 R3 current source having
a value I this potential is VX this
potential is VY and current directions
I will keep same as the original circuit
and now I will apply KCL to this node
which we consider as arbitrarily same
I'll name it as super node it is a
combination of 2 nodes so lets apply
KCL at super node
so for super node I am having these two
incoming currents so i will write a equations
so for this current it is starting from
this reference and ending on VX divided
by R1 this current it's I because it is
a current source equal to outgoing
currents two outgoing currents one is at
this node VX minus 0 divided by R2 and
for this node it is VY minus 0 divided
by R3 after rearranging the terms I will
get VX in a bracket 1 upon R1 plus 1
upon R2
VY 1 upon R3 equal to I i will
consider this are the equation number 7
and if I look at equation number 5 and
equation number 7 equation number 5 is
like this VX 1 by R1 plus 1 by R2 plus
py 1 by R3 equal to I and equation 7 is
also same so instead of considering two
equations and then eliminating third
variable getting one more equation
better we can apply this concept which
we call as a super node so you have to
apply KCL as if considering this is a
one node but having 2 potentials and
write the terms get only one equation
and other equation you will get from
this voltage source so here voltage
source is counted between 2 nodes VX
minus VY  will be V so I will
consider this as equation of voltage
source so here we are seeing a concept
of super node in subsequent videos we
will solve numerically based on this
thank you
