question_answer

In a meter bridge balance point is found at a distance \[{{1}_{1}},\] with resistance R and S shown in the figure. When an unknown resistance X is connected in parallel with the resistance S, the balance point shifts to a distance \[{{1}_{2}}.\] Find the expression for X in terms of \[{{1}_{1}},{{1}_{2}}\] and S.

Answer:

In a meter bridge, the balance condition is given by \[\frac{R}{S}=\frac{{{l}_{2}}}{100-{{l}_{1}}}\]or\[R=\frac{S{{l}_{1}}}{100-{{l}_{1}}}\] When X is connected in parallel with S, equivalent resistance, in that arm             \[{{S}_{eq}}=\frac{SX}{S+X}\] Now, the new balance condition is given by             \[\frac{R}{\left( \frac{SX}{S+X} \right)}=\frac{{{l}_{2}}}{(100-{{l}_{2}})}\] Substituting \[R=\frac{S\,\,{{l}_{1}}}{(100-{{l}_{1}})}\]             \[\left( \frac{S\,\,{{l}_{1}}}{(100-{{l}_{1}})} \right)\times \left( \frac{S+X}{SX} \right)=\frac{{{l}_{2}}}{100-{{l}_{2}}}\]             \[{{l}_{1}}(100-{{l}_{2}})(S+X)={{l}_{2}}(100-{{l}_{1}})X\] \[{{l}_{1}}(100S-{{l}_{2}}S-{{l}_{2}}X+100X)={{l}_{2}}(100-{{l}_{2}})X\] \[100S\,\,{{l}_{1}}-{{l}_{1}}\,\,{{l}_{2}}S-{{l}_{1}}\,\,{{l}_{2}}X+100X{{l}_{1}}=100X{{l}_{2}}-{{l}_{1}}\,\,{{l}_{2}}X\]             \[1000\,\,S{{l}_{1}}-{{l}_{1}}\,\,{{l}_{2}}S=100X({{l}_{2}}-{{l}_{1}})\]                         \[X=\frac{{{l}_{1}}S\,\,(100-{{l}_{2}})}{100\,\,({{l}_{2}}-{{l}_{1}})}\]