Meter Bridge

The meter bridge, a practical application of the Wheatstone bridge, is a fundamental device used in physics labs to measure unknown electrical resistances. It consists of a one-meter-long wire of uniform cross-section, typically mounted on a wooden base, with a sliding contact that helps in determining the point of balance. This point of balance is key to calculating unknown resistances with high precision. In real life, the concept of a meter bridge is analogous to balancing weights on a seesaw. Just as you adjust positions on a seesaw to achieve equilibrium, the meter bridge requires adjusting the sliding contact to balance the circuit. In this article, we will discuss the concept of meter bridge. A meter bridge is a critical instrument in current electricity research and is typically employed for the accurate measurement of unknown resistances.

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What is a Meter Bridge?

It is used to find the resistance of a given wire using a meter bridge and hence determine the specific resistance of its materials. It works on the principle of Wheatstone's bridge. The meter bridge arrangement is shown in the above figure. The wire connected between A and C is 1 meter in length and has a uniform cross-section. A constant current is passed through the wire AC so that the potential of the wire is proportional to the length of the wire. R is the known resistance which is selected from the resistance box. S is the unknown resistance whose value can be measured. The arrangement also has a galvanometer with a jokey. We will slide the galvanometer through wire AC so as to obtain null deflection in the galvanometer. Let B be the point on AC where null deflection is obtained and length AB =l . Pand Q resistance of the portion AB and BC respectively then by principle of Wheatstone's bridge

$\frac{P}{Q}=\frac{R}{S}\Rightarrow S=\frac{(100-l)}{l}R$

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Solved Examples Based on Meter Bridge

Example 1: The resistance of meter bridge AB in a given figure is $4\mathrm{\Omega }$. With the cell of emf $\epsilon =0.5\mathrm{V}$ and rheostat resistance ${\mathrm{R}}_{\mathrm{h}}=2\mathrm{\Omega }$ the null point is obtained at some point
$\mathrm{J}$. When the cell is replaced by another one of emf $\epsilon ={\epsilon }_2$ the same null point $\mathrm{J}$ is found for $R_h=6\mathrm{\Omega }$. The emf ${\epsilon }_2$ (in $\mathrm{V}$ ) is:

1) 0.3

2) 0.4

3) 0.6

4) 0.5

Solution:

Meter bridge

$\frac{P}{Q}=\frac{R}{S}\Rightarrow S=\frac{(100-l)}{l}R$
wherein
$\begin{array}{rl}& AB=l\\ & BC=(100-l)\end{array}$

For the question

When $R_h=2\mathrm{\Omega }$
$\frac{dV}{dL}=\left(\frac{6}{4+2}\right)\times \frac{4}{L}$
where $L=100\mathrm{cm}$
Let the null point be at $l\mathrm{cm}$Let the null point be at cm

${\epsilon }_1=0.5\mathrm{V}=\left(\frac{6}{2+4}\right)\frac{4}{L}l......(1)$
for $R_h=6\mathrm{\Omega }$

${\epsilon }_2=\left(\frac{6}{4+6}\right)\frac{4}{L}\times l\dots \dots \dots (2)$

From equation (1) and (2)
$\begin{array}{rl}& \frac{0.5}{{\epsilon }_2}=\frac{10}{6}\\ & \Rightarrow {\epsilon }_2=0.3\end{array}$
Hence, the answer is option (1).

Example 2: In a meter bridge experimental S is a standard resistance, and R is the resistance wire. It is found that the balancing length is l=25 cm. If R is replaced by a wire of half length and half diameter that of R of the same material, then the balancing distance (in cm) will now be _____.

1) 400

2) 40

3) 20

4) 30

Solution:

$$\begin{array}{rl}& \frac{X}{R}=\frac{75}{25}=3\\ & R=\frac{P^l}{A}=\frac{4P^l}{\pi d^2}\\ & R^{\prime }=\frac{4\rho \left(\frac{l}{2}\right)}{\pi {\left(\frac{d}{2}\right)}^2}=2R\\ & \text{then,}\frac{x}{R^{\prime }}=\frac{X}{2R}=\frac{3}{2}\\ & l=40.00\mathrm{cm}\end{array}$$
Hence, the answer is option (2).

Example 3: Consider a 72cm long AB as shown in the figure. The galvanometer jockey is placed at P on AB at a distance x cm from A. The galvanometer shows zero deflection.

The value of x, to the nearest integer, is ____

1) 48

2) 96

3) 24

4) 72

Solution:

In Balanced conditions
$\begin{array}{rl}& \frac{12}{6}=\frac{x}{72-x}\\ & x=48\mathrm{cm}\end{array}$

Hence, the answer is option (1).

Example 4: On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 kΩ. How much was the resistance (in ) on the left slot before interchanging the resistances?

1) 550

2) 990

3) 505

4) 910

Solution:

Meter bridge

To find the resistance of a given wire using a meter bridge and hence determine the specific resistance of its materials

wherein

Let's say resistances are R and 1000-R

Case 1: $\frac{R}{l}=\frac{1000-R}{100-l}$ eq 1
Case II: $\frac{1000-R}{l-10}=\frac{R}{110-l}$ eq 2
Multiply both equations
$\begin{array}{rl}& \frac{R(1000-R)}{l(l-10)}=\frac{R(1000-R)}{(100-l)(110-l)}\Rightarrow l^2-10l=11000+l^2-210l\\ & \Rightarrow 200l=11000\\ & \text{or}l=55\mathrm{cm}\\ & \Rightarrow \frac{R}{55}=\frac{1000-R}{45}\end{array}$
or $45R=55000-55R$
or $R=550\mathrm{\Omega }$
Hence, the answer is option (1).

Example 5: In the shown arrangement of the experiment of a meter bridge if AC corresponding to the null deflection of the galvanometer is x, what would be its value if the radius of the wire AB is doubled.?

1) $x$
2) $\frac{x}{4}$
3) $4x$
4) $2x$

Solution:

At null point

$\frac{R_1}{R_2}=\frac{R_3}{R_4}=\frac{x}{100-x}$

if the radius of the wire is doubled then the resistance of AC will change and the resistance of CB will also change.

But since $\frac{R_1}{R_2}$ does not change so $\frac{R_3}{R_4}$ should also not change at a null point. Therefore point C does not change.

Hence, the answer is option (1).

Summary

A slide wire bridge, which is also known as a meter bridge, is equipment that compares unknown resistance to a known resistance level. The principle behind it is the Wheatstone Bridge Theory which consists of one meter-long uniform cross-section wire stretched on a wooden board and calibrated. Should one move the contact point on this metallic conductor, one gets the balanced point where there is no deflection on the galvanometer.