In a meter bridge experiment, the unknown resistance X is balanced against a known resistance R. If the connections to the galvanometer and the battery are interchanged, will the balance point change?

Interchanging the galvanometer and battery connections does not affect the balance point, as the balance condition relies on the ratio of resistances and lengths, which stays unchanged. This is due to the principle of proportional resistances.

How does using a thick wire in a meter bridge affect the results?

A thicker wire has a lower resistance per unit length, which improves the sensitivity of the meter bridge. This results in more noticeable shifts in the balance point for smaller changes in resistance, allowing for more accurate measurements of low resistances.

What methods can be used to improve the exactness of a meter bridge ?

Accuracy can be improved by using a uniform wire, ensuring good connections, finding the null point near the wire's center, averaging multiple readings, using a sensitive galvanometer, and minimizing end resistance errors with a null experiment and correction.

In a meter bridge, the jockey is moved from one side of the balance point to the other. Does the current in the galvanometer change direction?

As the jockey moves across the balance point, the potential difference across the galvanometer reverses, causing the current to reverse direction. The null point is detected when the galvanometer shows zero deflection, and it deflects when the jockey moves away.

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Meter Bridge

A meter bridge is a simple and essential laboratory instrument used to measure the resistance of electrical conductors. It operates on the principle of the Wheatstone bridge, consisting of a one-meter-long wire mounted on a base. The device allows the measurement of unknown resistances by balancing the bridge with the help of a jockey and a galvanometer. The meter bridge is widely used in educational setups to demonstrate the concepts of resistance and electrical circuits.

1.0Definition Of Meter Bridge

It operates on the principle of the Wheatstone Bridge, which is used to determine the value of an unknown resistance in a wire.

2.0Diagram Of Meter Bridge

3.0Construction Of Meter Bridge

It typically consists of a one-meter-long manganin wire of uniform cross-section, stretched along a meter scale mounted on a wooden board, with its ends soldered to two L-shaped thick copper strips, A and C.
A third copper strip is fixed between the two copper strips, creating two gaps: ab and cd .
A resistance box R.B is connected in the gap ab and the unknown resistance X is connected between the gap cd.
A movable jockey and Galvanometer are connected across BD.

4.0Working Of Meter Bridge

After adjusting the resistance box, the jockey is moved along wire AC until the galvanometer shows no deflection, signaling the Wheatstone bridge's balance.
If P and Q represent the resistances of the AB and BC segments of the wire, then the balanced condition of the bridge is given by, $\frac{P}{Q} = \frac{R}{X}$

If AB=l cm ,then BC=(100-l) cm

Resistance P of the wire between A and B will be proportional to l i.e., $P \propto l$
Specific resistance and cross-sectional area A are the same for the whole of the wire.
Similarly, if Q is resistance of the wire between B and C, then Q ∝ 100-l

$\frac{P}{Q} = \frac{l}{100 - l}$

Applying the condition for balanced Wheatstone bridge, we get

$RQ = PX \Rightarrow X = R \frac{Q}{P}$ $or X = \frac{( 100 - l )}{l} R$

Since R and l are known, therefore, the value of X can be calculated.

Note: For better accuracy, R is so adjusted that l lies close to 50 cm.

5.0End Corrections Of Meter Bridge

In meter Bridge circuit, some extra length comes (is found under metallic strips) at end point A and C . So, some additional length ( and ) should be included at ends for accurate results.
Hence in place of l we use $l + α$ and in place of $100- l$ , we use $100- l + β$ (where and are called end correction).
To estimate $α$ and $β$ , we use known resistance $R_{1}$ and $R_{2}$ at the place of R and X in meter Bridge.
Suppose we get null point at l1 distance then

$\frac{R _{1}}{R _{2}} = \frac{l _{1} + α}{100- l _{1} + β} \dots\dots\dots\dots\dots (1)$

Now we interchange the position of R1 and R2, and get null point at l2 distance then

$\frac{R _{2}}{R _{1}} = \frac{l _{2} + α}{100- l _{2} + β} \dots\dots\dots\dots\dots . (2)$

Solving equations (1) and (2)

$α = \frac{R _{2} l _{1} - R _{1} l _{2}}{R _{1} - R _{2}}$ and $β = \frac{R _{1} l _{1} - R _{2} l _{2}}{R _{1} - R _{2}} - 100$

These end corrections ( and )are used to modify the observations.

Example-1.In meter bridge experiment, x $Ω$ is connected on the left side and $2Ω (< x)$ is connected on the right side. Now the position of these two resistances are interchanged and balancing length is shifted by 20 cm. Find the value of x.

Solution:

$\frac{x}{l _{1}} = \frac{2}{100 - l _{1}} \Rightarrow x = \frac{2 l _{1}}{100 - l _{1}} \dots\dots\dots . (1)$

On inter changing

$\frac{2}{l _{1}} = \frac{x}{100 - l _{2}} \Rightarrow x = \frac{2 ( 100 - l _{2} )}{l _{2}} \dots\dots . (2)$

$∴ l_{1} - l_{2} = 20$

using(1) and (2)

$l_{1} + l_{2} = 100$

$l_{1} - l_{2} = 20$

$2 l_{1} = 120 \Rightarrow l_{1} = 60$

$x = \frac{2 l _{1}}{100 - l _{1}} = \frac{2 ( 60 )}{100 - 60} = \frac{120}{40} = 3Ω$

Example-2.The meter bridge shown is in balanced position with $\frac{P}{Q} = \frac{l _{1}}{l _{2}}$ . If we now interchange the positions of galvanometer and cell, will the bridge work? If yes, what will be the balance condition?

Solution:

$\frac{P}{l _{1}} = \frac{Q}{l _{2}} \Rightarrow \frac{P}{Q} = l_{1} l_{2}$

Yes, the bridge will work and the balance condition will , $\frac{P}{Q} = \frac{l _{1}}{l _{2}}$

6.0Key Points On Meter Bridge

The meter bridge is also known as the slide-wire bridge.
In the meter bridge experiment, avoid sliding the jockey. Instead, lift it and gently touch the constantan wire at specific points to find the zero deflection. Pressing may alter the wire's cross-section or introduce stray resistance, affecting the results for others.

7.0Sample Questions On Meter Bridge

Q-1.When using 100Ω and 200Ω resistances for R and X, the null deflection occurs at l₁ = 33.0 cm. After interchanging the resistances, the null deflection shifts to l₂ = 67.0 cm. The end corrections α and β should be adjusted accordingly.

Solution:

$α = \frac{R _{2} l _{1} - R _{1} l _{2}}{R _{1} - R _{2}}$ $= \frac{( 200 ) ( 33 ) - ( 100 ) ( 67 )}{100 - 200} = 1 c m$

$β = \frac{R _{1} l _{1} - R _{2} l _{2}}{R _{1} - R _{2}} - 100$ $= \frac{( 33 ) ( 100 ) - ( 200 ) ( 67 )}{100 - 200} - 100 = 1 c m$

Meter Bridge

1.0Definition Of Meter Bridge

2.0Diagram Of Meter Bridge

3.0Construction Of Meter Bridge

4.0Working Of Meter Bridge

5.0End Corrections Of Meter Bridge

6.0Key Points On Meter Bridge

7.0Sample Questions On Meter Bridge

Table of Contents

Frequently Asked Questions

In a meter bridge experiment, the unknown resistance X is balanced against a known resistance R. If the connections to the galvanometer and the battery are interchanged, will the balance point change?

How does using a thick wire in a meter bridge affect the results?

What methods can be used to improve the exactness of a meter bridge ?

In a meter bridge, the jockey is moved from one side of the balance point to the other. Does the current in the galvanometer change direction?

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