Examples; Numericals Wheatstone Bridge - Current Electricity, Physics, Class 12 Video Lecture

hello friends this video on current
electricity part 25 is brought to you by
exam feel calm no more fear from exam
please make sure that you have watched
all the videos to the part 24 before
going ahead with this one we have to
calculate that we will in resistance
between the points E and P so since the
circuit looks quite complicated so it is
little difficult to know the to guess
the current distribution here I mean
even if we try to say that okay this
much current goes here we will get
confused to find the to keeper track of
what current flows in which an so what
we do the approach which we follow for
SEC circuit says divide this circuit
into smaller parts and solve the smaller
parts and reduce the circuit to simpler
circuits so the first part which we
consider here is the part in C D so we
look at only this part ACD so when we
consider only this part later suppose
from here current comes out so how much
current will go here let us say some
current i1 goes here and I 2 goes here
so this I 1 is the same current which
will flow through this 3 ohms as well
and this I 2 will flow through this 6
ohms and they both will joined together
here and form come back the same current
I right so if we consider only this
circuit what do we see this 3 ohm and
this 3 ohm these two are in series
because same current flows to both of
them and the their series combination is
in parallel with this 6 ohm because they
have their starting and ending point as
the same so if you draw this it will
look somewhat like this let us say this
is point a and this is point D so you
have one resistance here and other
resistance here this is 3 ohm this is 3
ohm and this is 6 ohm this is wow it
looks like so these three this two are
in series so that means 3 plus 3 becomes
6 ohms so now you can reduce this to a
this is six and this also is six so you
have two resistances in parallel so for
the part a CD what is your equivalent
resistance let us denote it by r1 so
that is the parallel combination of 6 &
6 so 1 by r1 is equal to 1 by 6 plus 1
by 6 that is equal to 1 by 3 therefore
r1 is equal to 3 ohms so instead of this
entire a CD we can just replace a CD
with this resistance 3 ohms so now my
circuit becomes little simpler so let us
draw the circuit again let us say this
is point a this is point B so from point
B you have one resistance till point F
from here you again have another
resistance from F you again have a
resistance till point E here you again
have another resistance so from EU again
have a resistance till point D and now
instead of drawing this ADC I just
replace it with value resistance and
that resistance is R 1 that is 3 ohms
now all leather resistance values are
given in the problem so goodness values
as it is so now if you look at this
figure you have reduced the figure
little bit so now your diagram is little
simpler so now let us concentrate on the
next part that is a d e so part by part
we will solve the problem so if you look
at ad e again if some current i1 flows
through this and some current I 2 flows
through this the same current i1 flows
through this so here also this 3 and
this 3 are in series and the series
combination is in parallel with 6 ohm so
it is exactly similar to the triangle
ACD so that means the equivalent
resistance here so equivalent resistance
of a CD let us denote it by r2 that will
also be equal to 3 ohms because the
values are also same so you can now
reduce your circuit this way say this is
a
in essence being so from me you have one
resistant still F here also you have
another resistance from F you have
another resistance still e so now
instead of drawing AED you just draw one
resistance and that resistances 3 ohms
right so now your circuit became even
more simpler so now it is time to look
at a F II so if you look at a Fe they're
also the same thing this geometry or
modern series and the combination is in
parallel with 6 ohms so here let us say
the equivalent resistance of a Fe is r3
which will again come out to be 3 ohms
so here you can again further simplify
the set it like this this is 3 ohms from
here you have 3 ohms to the point F now
instead of drawing a if e you just draw
R 3 that is 3 ohms so now you have this
final triangle so in this also you can
say the same thing if you look at it say
I 1 goes to this I 2 goes to this the
same I 1 will flow through this because
from here current is entering say I
current is entering it is dividing into
I 1 and I 2 and from this the current
will terminate right because we have to
find out the equivalent resistance
between these two points K and P so I 1
goes here I 2 goes here so the same I 1
will flow through this also so that
means this 3 and this 3 are in series so
he can draw it like this let us say this
is a this is B this is C ohms and the
other part is this 3 plus 3 that is 6
ohms right so this is how it will look
like therefore our equivalent of the
circuit that is equivalent resistance of
the circuit will be parallel combination
of 6 ohms and 3 ohms so that is 1 by our
equivalent is equal to 1 by 6 plus 1 by
3 so this is equal to 1 by 2 or we can
say our equivalent is equal to 2 ohms so
this is the equivalent
resistance of the circuit so this is how
we break down complex circuits into
simpler circuits and make our solution
easier okay so let us go and look at the
next problem that we will talk about
wheatstonebridge
I have already discussed about
wheatstonebridge when we were talking
about capacitors in circuits so the same
wheatstonebridge also has its
application in case of resistors in
circuits so before I solve this problem
when we have to calculate the equivalent
resistance between point A and B let me
tell you we tell you what a Wheatstone
bridge is Wheatstone bridge tells that
if you have a circuit which looks
somewhat like this if the circuit is of
this pattern if there is a circuit with
four resistors which are arranged in
this pattern and if r1 by r2 is equal to
r3 by r4 then we say that the Wheatstone
bridge is balanced so under this
condition the bridge is safe to be
balanced and when the bridge is balanced
in that case we say that no current will
flow through the resistor R 5 so that
means there will be no current which
will flow through this resistor that
means R 5 will not contribute in the
equivalent resistance of the circuit
right so this is the principle of
wheatstonebridge
it says that if you have five resistors
arranged in this fashion and if r1 by r2
is equal to r3 by r4 in that case we say
that the bridge is balanced and there
will be no current flow through the
middle and that is through the resistor
in the middle so the middle resistor
that is r5 here will not contribute
anything to the equivalent resistance of
the circuit so now we will solve this
problem where we will where we will see
an application of deep stone bridge if
you look at this circuit and the
question do you think it resembles which
stone bridge you don't but now
we will do something which will make you
to resemble the wheatstonebridge what is
it
just imagine it this way if you point a
and B apart from each other let us
suppose this wires are nothing but they
are friends so if you hold point a with
your one hand and point B with the other
hand and pull them apart from each other
then what will happen this B point B
will come somewhere here then you pull
it apart similarly this point a will
come somewhere here and now along with a
this entire thing this three arms will
go out so it will look somewhat like
this right this is how it will look like
if you pull a and B apart from each
other so how will your circuit look like
the circuit will look somewhat like this
so now if you look at this circuit does
the circuit resemble with stone bridge
yes it does it is just that you will
have to see it in the correct
orientation instead of this a and B here
we have this a and B so if you want you
can arrange the circuit in the correct
pattern to get a better look of the same
so this is your circuit right so always
try to observe the circuits very
carefully okay so in this case if you
look at r1 r2 r3 and r4 they are all
same therefore in this case also r1 by
r2 is equal to r3 by r4
therefore the bridge is balanced that
means there will be no current flow
across this capacitor across this
resistor I so your circuit will become
somewhat like this right now you have to
calculate the equivalent resistance of
this circuit which is very simple
because these two resistors are
we can see very clearly are in series
they are joined end to end similarly
these two resistors are again in series
so from this we can again draw this
circuit so R and R in series will form
two I similarly R and R in series will
form two I therefore the equivalent
resistance of the circuit will be given
by parallel combination of 2r and - I
therefore 1 by R equivalent is equal to
1 by 2 I plus 1 by 2 R which is equal to
1 by I therefore equivalent resistance
of the circuit is equal to I here so did
you ever imagine that this circuit which
we saw previously will ultimately become
this circuit no so that's what is what
will come in you with practice so try to
practice more and more problems and try
to look at problems in all possible ways
sometimes just in this problem if you
see just pulling out a and B make things
so simple right so now we will look at
few other methods and rules which helps
you to calculate equivalent resistance
of complicated circuits so this was the
first one bit study which we studied not
the datas wheatstonebridge
so the next method which we will see is
the Delta star transformation thank you
please visit examview dot-com to watch
free educational videos try3 online
tests get the best quality study
materials study from the best tutors and
mentors and much more thank you once
again