Examples; Numericals : Delta Star transformation - Current Electricity, Physics, Class 12 Video Lecture

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ahead with this one this is also a very
useful method which helps us to
calculate equivalent resistance of
complicated circuits so before I look at
the problem in the before I look at the
circuit in the problem let me tell you
what is Delta star transformation Delta
star transformation tells that if you
have any circuit of this form let us
suppose if you have a circuit in this
pattern this is in the shape of Delta
right so you can transform a delta into
a star let us suppose this circuit has
resistances let us say there is a delta
circuit which has resistances r1 r2 r3
so you can convert this Delta circuit
into a star circuit how you can
transform this circuit into this pattern
this circuit can be transformed to this
pattern the three resistances can be
transformed in this pattern such that
this resistance that means the
resistance here will be equal to the
product of these two resistances that is
r1 r2 divided by sum of all the three
resistances
similarly the resistance here that is
the resistance here so the resistance
which comes in between the two
resistances those two resistances will
be the product for example here this
resistance comes between r2 and r3 right
so it will be this will be equal to I 2
RT divided by r1 plus r2 plus r3
similarly the value of this resistance
will be what it comes between what art 1
and R 3 so this will be equal to r1 r3
divided by r1 plus r2 plus r3 right so
these three values if you you can
convert this Delta into a star such that
the values of these three resistance
be the values which I indicated here so
this is known as Delta star
transformation because sometimes it
happens that in complicated circuits you
you just cannot solve the equivalent
resistance with the help of current
distribution because the circuit looks
so very complicated that we are not even
able to determine how the current will
get distributed in different arms in the
circuit so in those cases these kind of
rules help us a lot so now when I look
at a problem where we will use this
Delta star transformation let us suppose
we have this circuit and we have to
calculate the equivalent resistance
between these two points let us say this
is point a and this is point B now you
must be thinking like this looks like a
weave Stoneridge so why not apply
wheatstonebridge but wait I haven't
mentioned the values of the resistance
here let us suppose this is of 2 ohms
this is of 1 ohm this is of 2 own
business of 1 ohm and this is of 2 off
by looking at this figure the first
thing that will strike your mind is the
Wheatstone bridge now if you quickly
apply the rule of wheatstonebridge that
is r1 by r2 that is 2 and r3 by r4 that
is 1 by 2 that means R 1 by I 2 is not
equal to r3 by r4 therefore the weight
is not there for the bridge is not
balanced here now if the bridge is not
balanced then we cannot apply the rule
of the Wheatstone bridge right so this
is basically I mean this Delta star
transformation is also an alternative
method to solve problems where the beat
whether wheatstonebridge is not balanced
okay so for scenarios where the bridge
is not balanced you can take help of
this Delta star transformation if you
see here you have 2 Delta Z this is 1
Delta this is another Delta so we what
we can do is we can transform 1 Delta
into a star and see if that helps us so
if we transform this Delta into a star
that means we can transform this like
this instead of this day
we transform it into a star like this in
that case how will my circuit look like
the circuit will become somewhat like
this now we will not have these three
instead of these three we will have
these so now your circuit will become
like this side so instead of this dead
and now I have this star so let us let
us put the values so what will be the
value of this one this arm that is this
and this arm comes between this and this
so this will be 1 into 2 divided by 1
plus 2 plus 2 right this is 1 this is 2
this is 2 so what will be the value of
this arm this arm comes between 1 & 2 so
this will also be 1 into 2 divided by 1
plus 2 plus 2 what will be the value of
this on this arm comes between 2 & 2 so
this will be 2 into 2 divided by 1 plus
2 plus 2 so that means we got a circuit
like this so this is 2 this is 1 this is
2 by 5 this is again 2 by 5 and this is
4 by 5 so now do you think it is easy to
calculate the equivalent resistance
between points a and B yes it is
relatively simpler now because this 2
and 2 by 5 are in series similarly 1 & 4
by 5 are in series so what we can do
that means these 2 are in series
similarly these 2 are also in series so
we can further reduce the circuit like
this instead of 2 & 2 by 5 we replace
them with time resistance so what will
be the resistance of the
this will be 2 plus 2 by 5 so this will
become what it will be 2 plus 2 by 5 so
2 plus 2 by 5 becomes 12 by 5 so this
will be 12 by 5 similarly what will be
this resistance it will be 1 plus 4 by 5
so 1 back plus 4 by 5 becomes 9 by 5
then you have this 2 by 5 and this is
your point B so now you have 12 by 5 and
9 by 5 in parallel so what will be the
equivalent resistance of that so we you
replace 12 by 5 and 9 by 5 and calculate
the equivalent resistance let us call
that resistances r1 let us call this as
r1 r1 is the equivalent resistance of 12
by 5 and 9 by 5 which will be in series
with 2 by 5 right so let us calculate R
1 here R 1 is nothing but 12 by 5 in
parallel with 9 by 5 that is 1 by r1
this is 1 - 5 by 12 plus 5 by 9 so that
comes out to be
so this comes out to be therefore the
equivalent resistance becomes the
parallel combination of these two as I
mentioned in the previous slide there
was 12 by 5 and 9 by 5 this was in
series with 2 by 5 right this was my
scenario so now what happens this
parallel combination you can replace it
with their equivalent resistance let us
say this is 2 by 5 and this is R 1 where
R 1 is the parallel combination of 12 by
5 and 9 by 5 so we can say 1 by r 1 is
equal to 5 by 12 plus 5 by 9 so this
comes out to be 35 by 36 therefore the
value of R 1 comes to be 36 by 35 ohms
so therefore I equivalent becomes series
combination of R 1 and 2 by 5 so R 1 is
36 by 35 plus 2 by 5 so this comes out
to be 50 by 35 which is equal to 10 by
seven ohms so thus 10 by seven ohms is
the equivalent resistance of this
circuit so this is how we use the Delta
star transformation method in this
problem so Delta star transformation
helps us to transform a delta into a
star but it is very much important that
you put correct values of the resistance
cells on the arms of the star right and
this method can be used as an
alternative method to a Wheatstone it
can be used as a method to unbalanced
wheatstonebridge thank you please visit
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