When the two known resistance ' \(R\) ' and ' \(S\) ' are connected in the left and right gaps of a meter bridge respectively, the null point is found at a distance ' \(l_1\) ' from the zero end of a meter bridge wire. An unknown resistance ' X ' is now connected in parallel with ' S ' and null point is found at a distance ' \(l_2\) ' form zero end of meter bridge wire. The unknown resistance ' X ' is

As per the first condition,
\(\frac{\mathrm{R}}{l_1}=\frac{\mathrm{S}}{100-l_1}\)
As per the second condition,
\(\begin{array}{ll}
& \frac{\mathrm{R}}{l_2}=\frac{\mathrm{XS}}{(\mathrm{X}+\mathrm{S})\left(100-l_2\right)} \\
\therefore & \frac{l_{\mathrm{S}} \mathrm{~S}}{100-l_1}=\frac{l_2 \mathrm{XS}}{(\mathrm{X}+\mathrm{S})\left(100-l_2\right)} \\
\therefore \quad & \frac{100-l_1}{l_1}=\left(1+\frac{\mathrm{S}}{\mathrm{X}}\right) \frac{100-l_2}{l_2} \\
\therefore \quad & 1+\frac{\mathrm{S}}{\mathrm{X}}=\frac{l_2\left(100-l_1\right)}{l_1\left(100-l_2\right)} \\
\therefore \quad & \frac{\mathrm{S}}{\mathrm{X}}=\frac{l_2\left(100-l_1\right)-l_1\left(100-l_2\right)}{l_1\left(100-l_2\right)}=\frac{100\left(l_2-l_1\right)}{l_1\left(100-l_2\right)} \\
\therefore \quad & \mathrm{X}=\frac{\mathrm{Sl}_1\left(100-l_2\right)}{100\left(l_2-l_1\right)}
\end{array}\)