Question

When two known resistances R and S are connected in the left and right gaps of a metre bridge, the balance point is found at a distance l1from zero end of the metre bridge wire. An unknown resistance X is now connected in parallel to the resistance S and the balance point is now found at distance l2 from the zero end of the metre bridge wire. Obtain a formula for X in terms of l1 ,l2 and S

Answer 1

Dear Student,

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{When}\mathrm{resistance}\mathrm{R}\mathrm{and}\mathrm{S}\mathrm{are}\mathrm{connected}: \\ \mathrm{Since}\mathrm{balance}\mathrm{point}\mathrm{is}\mathrm{found}\mathrm{at}\mathrm{a}\mathrm{distance}{\mathrm{l}}_1\mathrm{from}\mathrm{the}\mathrm{zero}\mathrm{end}, \\ \mathrm{R}\mathrm{\alpha }{\mathrm{l}}_1 \\ \mathrm{and},\mathrm{S}\mathrm{\alpha }\left(100-{\mathrm{l}}_1\right) \\ \mathrm{Therefore},\frac{\mathrm{R}}{\mathrm{S}}=\frac{{\mathrm{l}}_1}{\left(100-{\mathrm{l}}_1\right)}..............\left(1\right) \\ \left(\mathrm{b}\right)\mathrm{When}\mathrm{unknown}\mathrm{resistance}\mathrm{X}\mathrm{is}\mathrm{connected}\mathrm{in}\mathrm{parallel}\mathrm{to}\mathrm{S}: \\ \mathrm{The}\mathrm{effective}\mathrm{resistance}\mathrm{in}\mathrm{the}\mathrm{right}\mathrm{gap}, \\ \mathrm{S}\text{'}=\frac{\mathrm{SX}}{\mathrm{S}+\mathrm{X}}.................\left(2\right) \\ \mathrm{Since}\mathrm{balance}\mathrm{point}\mathrm{is}\mathrm{obtained}\mathrm{at}\mathrm{a}\mathrm{distance}{\mathrm{l}}_2\mathrm{from}\mathrm{the}\mathrm{zero}\mathrm{end}, \\ \mathrm{R}\mathrm{\alpha }{\mathrm{l}}_2 \\ \mathrm{S}\text{'}\mathrm{\alpha }\left(100-{\mathrm{l}}_2\right) \\ \mathrm{Therefore},\frac{\mathrm{R}}{\mathrm{S}\text{'}}=\frac{{\mathrm{l}}_2}{\left(100-{\mathrm{l}}_2\right)}.....................\left(3\right) \\ \mathrm{substituting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{S}\text{'}\mathrm{from}\left(2\right), \\ \frac{\mathrm{R}\left(\mathrm{S}+\mathrm{X}\right)}{\mathrm{SX}}=\frac{{\mathrm{l}}_2}{\left(100-{\mathrm{l}}_2\right)}.....................\left(4\right) \\ \mathrm{Dividing}\mathrm{equation}\left(4\right)\mathrm{by}\left(1\right), \\ \frac{\mathrm{S}+\mathrm{X}}{\mathrm{X}}=\frac{{\mathrm{l}}_2}{\left(100-{\mathrm{l}}_2\right)}\times \frac{\left(100-{\mathrm{l}}_1\right)}{{\mathrm{l}}_1} \\ \frac{\mathrm{S}}{\mathrm{X}}+1=\frac{{\mathrm{l}}_2\left(100-{\mathrm{l}}_1\right)}{{\mathrm{l}}_1\left(100-{\mathrm{l}}_2\right)} \\ \frac{\mathrm{S}}{\mathrm{X}}=\frac{{\mathrm{l}}_2\left(100-{\mathrm{l}}_1\right)}{{\mathrm{l}}_1\left(100-{\mathrm{l}}_2\right)}-1 \\ \frac{\mathrm{S}}{\mathrm{X}}=\frac{100{\mathrm{l}}_2-{\mathrm{l}}_1{\mathrm{l}}_2-100{\mathrm{l}}_1+{\mathrm{l}}_1{\mathrm{l}}_2}{{\mathrm{l}}_1\left(100-{\mathrm{l}}_2\right)} \\ \mathrm{X}=\frac{{\mathrm{l}}_1\left(100-{\mathrm{l}}_2\right)}{100\left({\mathrm{l}}_2-{\mathrm{l}}_1\right)}\mathrm{S} \end{gathered}$$