Wheatstone Bridge Video Lecture | Physics for JEE Main & Advanced

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and now we will learn about the
wheatstonebridge
wheatstonebridge is nothing new to you
because we have already spoken about
wheatstonebridge then we studied about
capacitors and we also talked about
wheatstonebridge when we were talking
about resistors right because this
proves to be a an excellent method to
solve some specific kind of circuits to
calculate equivalent resistance in case
of some special circuits right so we
have talked about wheatstonebridge but
now we will talk about wheat stone
bridge in bit more detail we will also
prove the truth of the wheat stone
bridge so first let us quickly review
what is a wheat stone bridge it is a
special arrangement of this kind as
shown in the figure if there are
resistances which are arranged in this
pattern r1 r2 r3 r4 and such that there
is a galvanometer which is placed in
this pattern that means there is a
galvanometer which is placed between the
diagonally opposite points right and
this circuit is connected to the battery
across the other pair of diagonally
opposite points that means this if this
is where the battery is connected so
what do we see let us name this circuit
let us suppose this is a b c d so this
kind of circuit is known as a Wheatstone
bridge which has four resistances
arranged in this pattern that is one the
four sides of a square and on the
diagonal of the square you have a
galvanometer and the other two
diagonally opposite points are connected
to the battery so in this case we say
that VD is known as the galvanometer arm
D D is known as galvanometer arm because
it contains the Kelvin
and AC is known as the battery um
because the battery is connected across
this arm so AC is called battery arm and
BD is known as galvanometer arm so
please remember that the galvanometer
arm and the battery arm should not be
the same I mean it cannot be the same
arm right so that is a very important
thing to recognize the boot stonebridge
right okay so now the boot Stoneridge
principle states that if I went by I 2
is equal to r3 by r4 then we say that
the bridge is balanced so if this
divided by this is equal to this divided
by this we see that it is a balanced
wheatstonebridge
we call it a balanced wheatstonebridge
so what happens in a balanced
wheatstonebridge it is seen that if the
bridge is balanced there is no current
which flows through this and so no
current flows through the galvanometer
arm no current flow through the
galvanometer arm so there will be no
current flow so when you look at the
galvanometer there is no deflection at
all if the galvanometer shows no
deflection that itself means that this
bridge is balanced that is r1 by r2 is
equal to r3 by r4 right so now in so
this is what the principle of a
Wheatstone bridge is so in our next
slide that is in this slide we will
prove the truth of them which stone
bridge principle it so what is the
principle it sees that if r1 by r2 is
equal to r3 by r4 then there is no
current which flows through this circuit
so what do we do we will try to prove it
other way around we will assume we
assume that the current through the
galvanometer arm let us suppose the
current through the galvanometer
is IG so we will assume that IG is equal
to zero and we have to prove that r1 by
r2 is equal to r3 by r4 so this is how
we will prove the boot Stonebridge
principle we are proving it the other
way round we are assuming that the
current through the galvanometer arm is
zero so if it is zero then this
condition should be true that is r1 by
r2 should be equal to r3 by r4 so let us
try to do this so first of all we will
mark the current distribution which will
make our task easy let us suppose this
has EMF V this is a B C and D so let us
try to distribute the current so let us
suppose some current I comes out of the
positive end it reaches here so it gets
divided into two parts let us suppose
some current i1 goes here so how much
current will go this side i minus i1
this I 1 reaches here and then some
current IG flows to the galvanometer arm
so the remaining amount of current is i1
minus IG what happens here this I minus
i1 reaches here and IG comes from this
end so this current this side it will be
I minus i1 plus IG so this will come
here and from this side this will come
here both together then again form I so
this I will go back to the match right
so this is my current distribution so
now I will apply the loop law to which
loop to a be da that is this loop so let
us assume one direction suppose I
assumed this direction for the loop a B
D a so in this what is the EMF there is
no EMF F so we can take it as zero what
are the resistances involved are
add 3 ad1 add 3 and the resistance of
the galvanometer the galvanometer also
all these has a resistance of its own
let us call this resistances RG what is
the current flowing through r1 it is i1
how much current is flowing through this
galvanometer it is IG and how much
current is flowing through R 3 it is I
minus i1 now let us consider the signs
assuming considering our assumed
direction assume direction and I 1 are
in the same direction so this will be
positive IG and assume Direction is also
in the same direction so this will also
be positive but I assume direction and I
- I - in the opposite direction so this
will be negative so this is one equation
which we get now let us apply the loop
lot to the load BC DB that is BC DB this
side so here again I assume a direction
let us suppose I assume this direction
so here also there is no EMF at all so
this will be 0 is equal to what are the
resistances involved r2 r4 and IG so
what is the current flowing through I -
it is i 1 minus i g current through an
for it is i 1 it is I minus I 1 plus I G
what is the current through I G it is IG
so consider the science now this is my
assumed direction so i1 minus i2 is in
my assumed direction IG is also in my
assumed direction but I minus i1 plus i2
the opposite direction so this will have
a negative sign right so these are the
two equations which I get now we assume
that I G is equal to zero so we will put
I T is equal to zero in both the
equations so first equation it becomes I
1 R 1 minus I minus I 1 into R 3 is
equal to zero so this becomes I 1 R 1 is
equal to I minus I 1 into R 3 right so
we call this equation 1 so from this
equation we got this similarly we will
put I T is equal to 0 in this expression
so we get I 1 R 2 minus I minus I 1 into
ad 4 is equal to 0 so from this what do
we get we get I 1 R 2 is equal to I
minus I 1 into R 4 so this is my
equation 2 so this was my equation 1 and
this is my equation 2 now if you divide
equation by by equation 2 what do we get
divide equation 1 by equation 2 this is
your equation 1 this is equation 1 and
this is equation 2 so if you divide
equation 1 by equation to you get r1 by
r2 is equal to r3 by r4 and this is what
you wanted to prove right if you look at
this point you wanted to prove that r1
by r2 is equal to r3 by r4 therefore
your route Stonebridge principle is
proved and say
that if r1 by r2 is equal to r3 by r4 we
say that the Wheatstone bridge is
balanced and there is no current flowing
through the galvanometer arm thank you
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