Meter bridge experiment  – Kisembo Academy
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amita, bridge also known As the slidewire bridge is based on the principle
of the westone bridge and it is used to find either the resistance of an unknown Conductor or to compare two unknown resistances It has a resistance where whose cross sectional
area is uniform This, where is mounted on a meter roll and? As you can see here this way from a to b Is 1 meters long or call it 100 centimeters so if the length of this portion, from here
to here is l then it, means that the length of the remaining portion will be 100 minus l This is in centimeters? Emitter bre has got two resistances, we have
the left-hand gap, we have the right-hand gap resistance r1 And r2. Occupying those gaps then, we have
a galvanometer this galvanometer here is Supposed to deflect is supposed to It’s having a galvanometer and the sliding
jockey The sliding jockey is placed at different places, along the slidewire a b Until this galvanometer here Gives no different deflection or until this
galvanometer indicates zero deflection Now the point at, which this galvanometer
guy is not Deflecting at all is what we are calling the
balanced condition of this meter bridge just, like, the way, we? We did it with a. Histone bridge So it means that at balance the resistance
r1 divide that by the resistance of r3 Which is the resistance of the slide the web
from here to there it should be equal to the resistance r2 Divide that by the resistance r 4 which is the resistance of the wire between
that point and that point and this condition should only be satisfied at balance point at balance
point is when this galvanometer is indicating zero deflection So how, do we get the resistance between these,
two, hours are three first of all we know the length of the wire so What, we need to find Also, is the resistance per unit length of
this wire which we shall call the resistance per centimeter
of that well so, we get the resistance per centimeter
of the wire and multiplied By the length of the wire to get the resistance
of the, well for both this and that that is how, we get our? R3. Resistance per centimeter multiplied that,
by the balanced length l, and also our for this is supposed to be r4? And also our for we get the resistance per
centimeter and we multiply that by the balanced length Which is 100 minus l? So in essence it means that it’s going to
be r1. Divided, by the resistance per centimeter times l is equal to r2 Divided that by the resistance force mu times
hundred minus l When you multiply both sides you’re, going
to get this Multiply this on both sides this will disappear
and you remain, with r, over l is equal to r2 Over 100 minus l now resistance this is r1
as We had seen in our diagram is the resistor
the resistance of the resistor in the left-hand gap and R2. Is the resistance of the resistor in the
right-hand gap l is the balanced length That is after moving the sliding jockey and
the? Government shows zero deflection and this
100 minus l will be the balanced length in the in the right-hand gap This, is right hand gap So this Is the balanced condition for the meter bridge
the question Goes the left hand gap and the right hand
gap of a meter bridge across bike waves having resistances of five ohms and three ohms respectively When the five ohm is shunted, by the length
of wire or by A length of wire the balanced length is going
to be zero point five eight meters from the left hand end of the bridge And now they’re Asking us to find the resistance of the shunt now this is the summary question the summary
of the course the question in terms in diagrammatic form They’re, telling us that the right hand the
left hand gap has 5 ohms the right hand gap has 3 ohms these are resistors but Then this, gap here is shunted, by a certain
length of wire which, we shall call rs another telling us that when this side, is
shunted, by a length of wire there is the balanced length is 0.58 meters from the left-hand side of the
wire So it means since this is or is 1 meter long
if this is your point 5 8 meter when, we subtract? 1? This, from 1, we remain is 0.4. 8 there and
now we’re being, asked to find the Resistor of that where that has been shunted,
when there is no current going Through, that is at balanced condition, when
there is no current, going. Through gee? Previously, we had already explored that at
balanced condition the resistance of this gap of the left-hand gap Over the resistance of the wire which is the
resistance of the zero point 5 Which is the balanced length, which is 0.5.
Actually 0.58 should be giving us the resistance of
the right-hand gap Divide that, by the balanced length, which
is 0.42 meters Now, we have the resistance of the right-hand
gap which is going to be that 3 ohms the resistance of the left-hand gap meanwhile Is having 2 resistance these resistors are
in parallel so it means for us to get the resistance of the left-hand gap we need, to find the effective resistance
of this 2 So to get the effective resistance of this
just r. To come and later substitute it in there the resistance of the left-hand gap will be equal to Let’s do it like, this it’s going to be 1
over the resistance of the left-hand gap Is going to be equal to 1 over the resistance
of the shunt where plus 1 over the resistance of the 5
ohms When, we make our left-hand gap the subject
of the formula, we find that the resistance of the left-hand gap Will end up being equal to 5 rs Divide all that, by, five plus rs So it is going to be the product Divide that, by the sum of those resistances
so We get this this the resistance of the left
hand gap and we substituted up there? So it means that our equation this is now
going to become Five rs Divide that, by, five plus Rs. This is the effective resistance That is divide that by the balanced length,
which is 0.58? Should, give us the resistance of the right
hand gap and the resistance on the right hand gap is three ohms so this is going to be to be divide that, by the balanced length which
is going to be 0.42 so we’ll end up with five rs 5 + Rs. Multiply that by 1 or open 5:8 giving
us 3 over 3.4 – We end up with our value of rs as 24.1. Ohms? And that will be the resistance of that shunt Where this brings us to the end of this video
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