# Wheatstone bridge equation | Wheatstone bridge applications

We stone bridges like any other an electric
circuit Invented by some well hunter Christy in 1833
and improved and popularized by such Charles Weston in 1843 This circuit network is an arrangement of
full resistors Which are arranged as you see we have resistor
r1 r2? r3 and r4 Those are the four resistors that make up
the Weston bridge r1 is an unknown resistor Which we are definitely supposed to find R2 is a standard open Resistor whose resistance is known then r3
and r4 Resistors that are variable with variable
resistance but whose resistances still can also be read So when we vary the potential differences
using this variable resistors when we value the potential differences? using these variable resistors The potential difference or the potential
at B, and the potential at D will keep varying If the potential at D is greater than the
potential at D. It means that current will flow From D to B, and it will cause that galvanometer
to deflect if the potential at B is Greater than the potential at D. It means
that current coming in will flow also in through the galvanometer from B to D and it will cause this galvanometer to deflect
in the opposite direction so It means that if we vary these resistances
such that the potential at B And the potential at D is the same Then it means that the galvanometer will not
deflect When the governor does not deflect it means
that there is no current going through the galvanometer This condition whereby there is no current
that is going through this galvanometer is what we call the balance condition so in deriving the balance condition for Western
bridge we are going to Do it in two ways? You can either derive it using kirchoff’s
laws of electrical networks Or shall delay on the basis of the physical
conditions that are supposed to exist for balance condition to be called soul So we shall begin off with catch of slows in this circuit diagram This is a Western bridge Just a pushin of the twisting bridge We have this first loop and the second loop Inside a whoo stone bridge that is having
resistances r1 r2 and two variable resistors r3 and r4 plus our galvanometer G previously we had just said that for the
wisdom bridge could be considered to be at balanced condition. It means that No current is supposed to be flowing through
the galvanometer So this is I naught current coming into the
bridge when it gets to that Junction it splits into
i1 and i2 and i1 will go all the way and I to go all the way to Aqaba combine here
back to IO So it begin with loop 1 we know that according to the voltage law Of catch off all the potential differences
all the potential what I mean all the voltages within this loop Must be equal to 0. They are jab rate sum of all the voltages
in this loop. I equal to 0 So we begin by first allocating this The potentials all the resistance this is
current is flowing internet comes in it is moving through this resistor When it moves into this resistor? Weight getting in from is a higher potential
where it is getting out from is a lower potential Like why is this kind where it is getting
in from it is a higher potential. It’s getting out from is a lower potential Same-store here current comes in here at the
higher potential gets out here. This is a plus or a lower potential We do so because current moves from a position
of higher potential towards a place of lower
potential So now we look at our loops loop 1 Now firstly we see that this We are moving in the clockwise direction remember. We said that you choose one direction either
clockwise or anti-clockwise so as we are moving From here we move we come up here Through this resistor we see that we are me
getting in from negative to positive so it means across this resistor We are having a potential gain so if you are having a potential gain it means that they pertain the voltage across
this resistor is a positive so we are having a positive I R3 this is I 2 so it’s I 2 times r3 How then we come it comes like that this is
from positive to negative so meaning there is a potential drop across this resistor So it becomes minus Current i1 times there is this the R 1 and
that is going to be equal to 0 so this eventually becomes I 2 R 3 Is 0 to I 1 R 1 and that is our equation 1? We go to the other loop the second loop. We call it loop 2 In our second loop we are still moving in
the clockwise direction So as we are moving the up here we go through
this resistor. It is from high potential to low potential So there is a put potential drop so it means
the voltage across this resistor is Negative so it’s going to be negative. I 1 times R 2 And then when we come here We are moving from negative to positive as
we are coming like this From negative to positive, so there is a potential
gain across this resistor R 4 so the potential the voltage across this
is going to be a positive so it’s going to be plus I to Our 4 and that is also going to be equal to
0 and this will give us I To r4 will be equal to i1 I2 that’s why fission – like that So we have equation 1 and equation 2 and for
us to get our balance equation balance condition. We simply are going to divide the equation 2 by equation 1 so dividing 2 by 1 Divide the question 2 by equation 1 if equation
2 is going to be I 2 R 4 is going to be equal to I 1 So for balance condition Resistor are for the resistance of our folks
over r3 Will be equal to R. 2 over R 4 and this is
our balance condition for a With Stonebridge, and it is that is how we
derive it when we are using Kirchhoff’s laws Now the other where we can derive it is by
using the physical quantities or the physical conditions for balance if you look at our with Stonebridge We said that when current is moving in here
if the potential at D is greater than the potential at C It means that current will move from the higher
potential towards G to that potential and our galvanometer will deflect if the potential at C is Greater than the potential at D. It means
that our current will move like that some of it will move there, but some will
move like that and this galvanometer will deflect and That won’t be a balanced condition so for
balanced condition to happen. It should be that the potential at C Should be equal to the potential at D. And
if the potential at C is equal to the potential at D It means that there is no current going through
this galvanometer And so the galvanometer will definitely show
zero deflection if that is so it means that the potential between a C and a is The same as the potential between a and B Likewise the potential between C. And B is
the same as the potential between D. And B So for us to derive our balance equation using
the physical properties It means that we are going to find the potential
difference across, SC and equate that to the PD across FD We make equation 1, then we go ahead and get
the potential across seed be Equated to the potential across dB make another
equation call it equation 2, then we shall divide the two equations, and we shall get
our balanced condition For this first equation is going to be V.
They put a PD across this Across, SC. Would be the current. I times R. One is going to be equal to the current i2 times R 3 That is equation 1, then you go to version
2 the PBR cross this should be equal to the PD across
that so the PD across This is going to be i1 times R. 2 is equal
to i2 times R 4 and you’ll find that we shall be coming
up still with equations that are similar to that and Then we do just like here we divide them,
and we still get the same balanced condition for the histone bridge This brings us to the end of this video. Thank you for watching Somebody out they need to watch this video
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Cara Mia