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A meter bridge is considered as an electrical apparatus which is conditioned to govern the resistance of a particular conductor. To calculate the effective series and parallel equivalent of resistances, the meter bridge method is used.

The meter bridge contains a long meter bridge wire length of 1metre, and the meter bridge has two dissimilar segments which are left and right segments. In the left segment, we attach the known resistance, whereas in the right segment that resistance is to be connected, whose charge is to be measured.

A jockey is present which senses the balance point. The length is calculated from the left end which is similarly named zero points to the balance point.

Also, a meter bridge entitled a slide wire bridge is a gadget that operates on the Wheatstone bridge principle. A meter bridge is operated for concluding the unknown resistance of a conductor unlike that in Wheatstone bridge.

An apparatus like the meter bridge is employed for measuring the unknown resistance of a coil. The figure given below is the diagram of a beneficial meter bridge tool.

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In this figure,

R is named as the Resistance, P is the resistance approaching from AB, S is the Unknown Resistance, and Q is the resistance between the two joints of B and D.

The long wire is named ‘AC’ 1 m in length, and it is prepared from constantan or manganin possessing a uniform cross-sectional area.

As we know, L1 + L2 = 100 cm

Let’s assume that, L1 = L

We get, L2 = 100 – L

The unknown resistance ‘X’ is obtained from a relation in the given wire can be written as:

\[X = R(\frac{L_{2}}{L_{1}}) = R \frac{(100 - L)}{L}\]

And the material’s specific resistance for a certain wire can be illustrated by this relation; R = (3.14) r2X / l

Where, l = length of the wire, and r = the radius of the cable.

The devices that are necessary for measuring the unknown resistance of a conductor using a meter bridge are:

A resistance box

Galvanometer

Meter bridge

Connecting Wires

Unknown resistance of a length 1 meter

Screw gauge

One way key

Jockey

In the meter bridge, the resistances as of the lateral types are replaced by a wire around 1m of the distance of the uniform cross-section. Another pair consists of one known and an unknown pair of resistances.

The galvanometer’s one part linked next to both resistances. At that time, the other fragment of the wire results in the null point where there is no deflection shown by the galvanometer. In this situation, the bridge is considered balanced.

Generally, there are two types of circuits, known as parallel and series. Both series and parallel circuits contain above single load.

Resistors can be linked as series, or parallel, or both the series and parallel combination.

In series combinations, two or more resistors can be possible for the connection; only when the same current is travelling through each one of them during some potential difference is supplied transversely to the combination.

The combination’s equivalent resistance available in that series combination can be calculated as the arithmetic summation of all the resistances

As examined, in series circuit electrons can pass through in one path only. At this time, as given in the figure below; the current will be identical traveling through every single resistor. Besides, the voltage will be different across the resistor in the series connection.

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In the parallel circuit, the situation is different here. Electrons can pass through many branches in the particular circuit. Here, the voltage stays normal across each resistor without any voltage rise or drop.

A parallel circuit can be designed in many styles, which indicates that resistors can be organized in various forms. It can be utilized as a current separator.

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We have conducted these experiments to calculate & authenticate the laws of combination of resistances using a meter bridge.

To verify the laws of combination series of resistances using a metre bridge

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The resistance available in resistance wire or a coil is given as:

\[r = \frac{(100 - l)}{l} \times R\]

Here,

R = the resistance from the resistance box present at the left gap.

l = the length of the meter bridge wire from zero ends equal to the balance point.

As soon as two resistors r1 and r2 are linked with series connection, the relation for the combined resistance becomes:

Rs = r1 + r2

2. To verify the law of parallel combination of resistance using a metre bridge

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Let’s take two resistances as r1 and r2 are linked with the series connection. The series combination resistance denoted as RS.

Here, Rs = r1 + r2

When they are linked as parallel, the relation for the combined resistance is denoted as Rp

\[R_{p} = \frac{r_{1}r_{2}}{r_{1} + r_{2}}\]

FAQ (Frequently Asked Questions)

Q1. In a Meter Bridge Experiment, it is Suggested to have Identified and Unidentified Resistances of the Equivalent Standards. What Does this Satisfy?

Ans: It makes the null point to lie in the middle of the wire. At this condition, a minor alteration in the jockey’s position produces a huge deflection in the galvanometer.

Thus, the null point can determine the value more accurately. We acclaim you to prove this analytically.

Q2. In a Meter Bridge Experiment, two Students have Identical Setup and Follow a Similar Procedure. But one of the Students uses very Low Source Voltage V. How does this Student get more Errors?

Ans: This increases the error in the position of the null point. The voltage difference per unit length of the wire is very small in this case.

Thus, the minimum displacement of the jockey that produces a perceptible deflection of the galvanometer (sensitivity to galvanometer) is more. This increases the error in finding the null point.

Q3. In an Improved Meter Bridge, if the Length of the Bridge Wire is Condensed to 50 cm From 100 cm. What does this Phenomenon Result?

Ans: If the length of the bridge wire is condensed to 50 cm from 100 cm, this incident increases the error in finding the null point.

Q4. Find the Total Resistance, RT of the Following Resistors Connected in a Parallel Network.

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Ans: As we know the formula for total resistance is

1/R_{T} = 1/R_{1 }+ 1/R_{2 }+ 1/R_{3} = 1/200 + 1/470 + 1/200 = 0.012

R_{T} = 1/0.012 = 83.33 Ω