Meter Bridge Notes

Meter Bridge: A meter bridge, also known as a slide wire bridge, is a device that works on the Wheatstone bridge idea. A metre bridge is used to find the unknown resistance of a conductor. In Physics, while theory forms the basis of our knowledge, practicals form our understanding. What is this device? This is a meter bridge! It consists of a wire of one meter, which is why it is called a ‘meter bridge’. It is used to measure the resistance of wires, coils or any other material. Please read on to learn about the meter bridge formula, meter bridge diagram, and more.

In this article, we will provide you with all information about the meter Bridge, the principle of the meter bridge, meter bridge experiment class 12 etc, continue to learn the concept thoroughly and make no mistakes while answering questions on the meter bridge.

What is Meter Bridge?

A meter bridge is an electrical apparatus using which we can measure the value of unknown resistance. It is made using a metre long wire of uniform cross-section. This wire is either nichrome or manganin or constantan because they offer high resistance and low-temperature coefficient of resistance.

A meter bridge or Slide wire bridge is designed from a Wheatstone bridge. It is the most basic and functional application of a Wheatstone bridge.

The Principle of Meter Bridge

A meter bridge works on the principle of a Wheatstone bridge. A Wheatstone bridge is based on the principle of null deflection, i.e. when the ratio of resistances in the two arms is equal, no current will flow through the middle arm of the circuit. Consider the diagram of the Wheatstone bridge as shown below. It consists of four resistances \(P,\,Q,\,R\) and \(S\) with a battery of EMF \(E\).

In the balanced condition, no current flows through the galvanometer, and terminals, \(B\) and \(D\) are at the same potential. This condition arises when, \(\frac{P}{Q} = \frac{R}{S}\)

Construction of a Meter Bridge

1. A meter bridge has a \(1\,\rm{m}\) long wire of uniform cross-section area, which is stretched tight.
2. Between two metal strips that are bent at right angles, this wire is then carefully clamped, as shown in the diagram below:
3. Within the gap between the metal stripes, resistances are connected. In the first gap, \(R\), a resistance box, and in the second gap, a small resistor wire \(S\) is connected.
4. The endpoints within which the wire is clamped are connected to a key through the cell.
5. A galvanometer is connected to the metallic right in the middle of the two gaps.
6. A jockey is connected at the other end of the galvanometer (Here, a jockey is a metal rod with a knife-like edge at one end that slides over the potentiometer wire to make an electrical connection). The jockey is slid over the meter bridge wire till the galvanometer shows zero deflection.

Meter Bridge Working

1. To begin with, move the jockey to the endpoints of the wire, i.e., \(A\) and \(C\). The deflection of the galvanometer should be opposite on both ends.
2. From side \(A\), start sliding the jockey slowly over the wire and carefully observe where the deflection of the galvanometer comes out to be zero.
3. If such a point is not obtained, try varying the resistance across the bridge by changing the resistance on the variable resistance.
4. Slide the jockey over the wire and carefully observe the point on the wire where the deflection of the galvanometer comes out to be zero. This is the null point as represented by the point ‘\(B\)’ in the diagram.
5. Obtain the length of the null point using the meter scale attached along the wire. This is the ‘balancing length’ of the meter bridge.
6. Let the distance between points \(A\) and \(B\) be ‘\(l_1\)’.
7. Let the distance between points \(B\) and \(C\) be ‘\(l_2\)’, where \(l_2 = 100 – l_1.\)

When the galvanometer shows null deflection, the meter bridge behaves like a Wheatstone bridge and can be represented as:

Find the Resistance of the Given Wire by Metre Bridge

If \(S\) is the unknown resistance in the above circuit, we can calculate its value using the meter bridge. In the balanced condition,
\(\frac{R}{{\text{Resistance}}\;{\text{across}}\;{\text{length}}\;AB} = \frac{S}{{\text{Resistance}}\;{\text{across}}\;{\text{length}}\;BC}\)
We know that the resistance \(r\) of a wire of length \(l\), area of cross-section \(A\) and resistivity \(ρ\) is given as, \(r = \frac{{\rho l}}{A}.\)
Using this relation, if \(ρ\) be the resistivity and \(A\) be the area of cross-section of the given meter bridge wire, then the resistance across length \(AB = \frac{{\rho l_1}}{A}.\)
The resistance across length \(BC = \frac{{\rho l_2}}{A}.\)
Substituting these values in the above relation, we get:
\(\frac{R}{{\frac{{\rho {l_1}}}{A}}} = \frac{S}{{\frac{{\rho {l_2}}}{A}}}\)
or, \(\frac{R}{{{l_1}}} = \frac{S}{{{l_2}}}\)
\(\frac{R}{{{l_1}}} = \frac{S}{{100 – {l_1}}}\)
Thus, the unknown resistance, \(S = \left( {100 – {l_1}} \right)\frac{R}{{{l_1}}}\)
We can calculate the specific resistivity of the unknown resistance by using the formula,
\(\rho = \frac{{\pi {d^2}S}}{{4L}}\)
Where \(d\) is the diameter of the wire, \(S\) is the unknown resistance (of the wire), and \(L\) is the length of the wire.

Meter Bridge Experiment Class 12

Equipment Required

1. Meter Bridge
2. Galvanometer
3. Connecting wires
4. Unknown resistance
5. Resistance Box
6. Jockey
7. One-way key
8. Screw Gauge
9. Lechlanche cell

Procedure

1. Collect all the required instruments and make all the necessary connections, as demonstrated in the above figure.
2. Take some appropriate kind of resistance out from the resistance box ‘\(R\)’.
3. Now, place the jockey at point \(A\); look that there is a deflection within the galvanometer. When the jockey is moved from point \(A\) to Point \(C\), the deflection of the galvanometer must go from one side to the other side. If it is not observed, adjust the known resistance value.
4. Start sliding the jockey from \(A\) towards \(C\) and obtain the point where the deflection of the galvanometer is zero.
5. Proceed with the above strategy for various values of the ‘\(R\)’. Note probably around \(5 -10\) readings.
6. The point where the galvanometer gives null deflection is the balance point of the meter bridge for the given unknown resistance.
7. Measure the distance between point \(A\) and the balance point of a given wire using an ordinary meter scale and the radius of the wire using a screw gauge (Take at least five readings for both the quantities).
8. Compute the mean value of the unknown resistances obtained above. It will be equal to the sum of all the values of resistance divided by the total number of readings taken.

Errors in the Meter Bridge

The most common error that can affect the measurement accuracy of a meter bridge is the end error. The end error can come up due to the following reasons:

1. We know that along the length of the bridge wire, a scale is provided. If the zero of the scale does not coincide with the starting point of the bridge wire, the \(100\,\rm{cm}\) mark on the scale will not coincide with the endpoint of the wire. This will lead to incorrect measurements of the balancing length.
2. The non-uniformity of the metal wire might lead to the generation of stray resistance, and it will create an end error.

We can minimize end error by taking multiple readings of the experiment by interchanging the unknown and known resistance in the circuit and by calculating the final value of resistance by taking the mean of all the observations.

Solved Examples of Meter Bridge

Q.1. In a meter bridge, there are two unknown resistance \(R\) and \(S\). Find the ratio of \(R\) and \(S\) if the galvanometer shows a null deflection at \(20\,\rm{cm}\) from one end?
Ans: The null deflection in the galvanometer is obtained at \(20\,\rm{cm}\) from one end.
Let, \(L_1 = 20\,\rm{cm}\)
So, \(L_2 = 100 – 20 = 80\,\rm{cm}\)
Thus, the ratio of unknown resistance will be: \(\frac{R}{{{L_1}}} = \frac{S}{{{L_2}}}\)
Thus, \(\frac{R}{{{S}}} = \frac{1}{{{4}}}\)

Q.2. A \(20\,\rm{Ω}\) resistor is connected in the left gap, and an unknown resistance is joined in the right gap of the meter bridge. Also, the null deflection point is shifted by \(40\,\rm{cm}\) when the resistors are interchanged. Find the value of unknown resistance?
Ans: In the first case, let the deflection point is taken as \(L\). Let the balance point gets shifted to \(l\) by \(40\,\rm{cm}\) when the resistors are interchanged.
Thus, \(L – l = 40\,\rm{cm}\)
Also, \(L + l = 100\,\rm{cm}\)
Solving the above equations, we get:
\(l = 30\,\rm{cm}\)
\(L = 70\,\rm{cm}\)
Let \(R = 20\,\rm{Ω}\)
And unknown resistance be \(S\), thus,
\(\frac{R}{S} = \frac{L}{l}\)
\(\frac{R}{S} = \frac{70}{30}\)
\(\frac{20}{S} = \frac{7}{3}\)
\(∴ S = 8.57\,\rm{Ω}\)

Summary

A meter bridge is an electrical apparatus using which we can measure the value of unknown resistance. It is made using a metre long wire of uniform cross-section. This wire is either nichrome or manganin or constantan. The principle of working of a meter bridge is the same as the principle of a Wheatstone bridge. A Wheatstone bridge is based on the principle of null deflection. Thus, the unknown resistance, \(S = \left( {100 – {l_1}} \right)\frac{R}{{{l_1}}}.\)

FAQs on Meter Bridge

Q.1: What is the principle of meter bridge?
Ans: Meter bridge is based on the principle of the Wheatstone bridge.

Q.2: Why do we use constantan or manganin wire in a meter bridge?
Ans: Constantan, manganin, or nichrome wires provide a low-temperature coefficient of resistance, so they are used in a meter bridge.

Q.3: What is a meter bridge, and what is it used for?
Ans: A meter bridge is an electrical apparatus that is used to measure the unknown resistance of a conductor. It consists of a wire of length of one meter. Hence it is called a meter bridge.

Q.4: What is the end error in a meter bridge?
Ans: End error occurs when the zero scales of the meter scale do not coincide with the starting of the wire. It is caused due to the shifting of zero scale or the stray resistance in the wire.

Q.5: Give the formula to measure the unknown resistance for a meter bridge.
Ans: The unknown resistance, \(S = \left( {100 – {l_1}} \right)\frac{R}{{{l_1}}}\), where \(R\) is known resistance and \(l_1\) is the balancing length of the wire.

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