When the two known resistance ' $R$ ' and ' $S$ ' are connected in the left and right gaps of a meter bridge…

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
  1. $\frac{\mathrm{S} l_1\left(100-l_2\right)}{100\left(l_2-l_1\right)}$
  2. $\frac{\mathrm{S} l_2\left(100-l_1\right)}{100\left(l_1-l_2\right)}$
  3. $\frac{100\left(l_2-l_1\right)}{\mathrm{Sl}_1\left(100-l_2\right)}$
  4. $\frac{100\left(l_2-l_1\right)}{\mathrm{S} l_2\left(100-l_1\right)}$

Solution

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

Asked in: MHT CET 2024 (09 May Shift 1)

Practice more Current Electricity questions on Aicharya