\(I=1 \,\mathrm{A}\)        .......\((i)\)

When the key between the terminals \(1\) and \(2\) is plugged in, then

Potential difference across \(R=I R=k l_{1}\)       ......\((ii)\)

where \(k\) is the potential gradient across the potentiometer wire

When the key between the terminals \(1\) and \(3\) is plugged in, then

Potential difference across \((R+X)=I(R+X)=k l_{2}\)      ....\((iii)\)

From equation \((ii),\) we get

\(R=\frac{k l_{1}}{I}=\frac{k l_{1}}{1}=k l_{1} \Omega\)      .......\((iv)\)

From equation \((iii),\) we get

\(R + X = \frac{{k{l_2}}}{I} = \frac{{k{l_2}}}{1} = k{l_2}\,\Omega \quad {\rm{ (Using }}({\rm{i}}))\)

\(X = k{l_2} - R\)

\( = k{l_2} - k{l_1}{\rm{ }}\,\,\,\,\,\,\,\,\,\,\,{\rm{(Using}}\left( {iv} \right){\rm{)}}\)

\(=k\left(l_{2}-l_{1}\right) \,\Omega\)