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Above is the diagram of a meter bridge in which, R and S are two resistance, where R is a resistance box with variable resistance and S is a fixed resistance(this resistance is made up of the wire whose resistance is unknown). B, A, and C are metallic plates G is a galvanometer which is connected at point B with which a jockey D is connected, the meter scale is a 1 m scale divided into 100 divisions, K$_{1}$ is the key, and $\xi $ is a battery that is inside the circuit.

Now let us consider that ${{l}_{1}}$ is the length where the galvanometer shows zero deflections,

Therefore ${{l}_{2}}=100-{{l}_{1}}$ .

At first, the key K$_{1}$ is closed to pass a current and we are also choosing a suitable resistance R in the resistance box.

Then the jockey is tapped along the wire to locate the null point that is the point where the galvanometer shows zero deflection. The bridge is then said to be balanced.

And after that by the principle of Wheatstone bridge.

\[\dfrac{R}{{{l}_{1}}}=\dfrac{S}{{{l}_{2}}}\] ,

\[S=\dfrac{R\times {{l}_{2}}}{{{l}_{1}}}\],

Therefore, by this formula, we can find the unknown resistance of the wire.