|Aim||To find the resistance of a given wire using a meter bridge and hence determine the specific resistance of its material,|
|Materials Required||Meter bridge|
Battery or power supply
A wire whose resistance is to be measured
|Theory||The meter bridge is a simple and efficient device used to measure the resistance of a wire, and hence determine the specific resistance of its material. It is based on the principle of Wheatstone’s bridge, where the ratio of resistances in a balanced circuit is constant.|
The working of the meter bridge is simple. When a wire of unknown resistance is connected to one side of the meter bridge, and a known resistance is connected to the other side, the bridge is unbalanced, and the galvanometer shows a deflection. By adjusting the resistance of the known resistance box, the bridge can be balanced, and the ratio of resistances can be determined. Using this ratio, the resistance of the unknown wire can be calculated.
|Procedure||Firstly, set up the meter bridge circuit by connecting the resistance box and the galvanometer to the meter bridge. Connect the wire whose resistance is to be measured to the left gap of the meter bridge, and the known resistance to the right gap of the meter bridge.|
Now, adjust the resistance box to get a balanced condition in the meter bridge circuit. This is achieved when there is no deflection in the galvanometer.
Once the balanced condition is achieved, note down the length of the wire on either side of the meter bridge, as well as the resistance of the resistance box. Let these be denoted by L1 and L2, and R respectively.
Calculate the resistance of the wire using the formula:
R1/R2 = L1/L2
where R1 is the resistance of the wire and R2 is the resistance of the resistance box.
Solving for R1, we get:
R1 = (L1/L2) * R
Now, using the known length and cross-sectional area of the wire, calculate the specific resistance (resistivity) of the wire material using the formula:
Specific resistance (p) = (R * A) / l
where A is the cross-sectional area of the wire, and l is its length.
You have now determined the resistance of the wire and the specific resistance of its material.
|Observation Results:||The following are the observation results for the given experiment:|
Resistance of resistance box (R): 10 Ω
Length of wire on left side (L1): 40 cm
Length of wire on right side (L2): 60 cm
Deflection in galvanometer: Zero (balanced condition)
|Calculations||Using the formula for the resistance of the wire,|
R1 = (L1/L2) * R
Substituting the values, we get:
R1 = (40/60) * 10 R1 = 6.67 Ω
Using the formula for specific resistance (p),
p = (R * A) / l
Where the length of the wire (l) is the sum of L1 and L2, and the cross-sectional area (A) is known.
Assuming the cross-sectional area of the wire to be 1 mm², we get:
l = L1 + L2 l = 40 + 60 l = 100 cm = 1 m
p = (R1 * A) / l p = (6.67 * 1) / 1 p = 6.67 Ωm
Therefore, the resistance of the wire is 6.67 Ω and the specific resistance of its material is 6.67 Ωm.