question_answer

When a resistance \[{{\operatorname{R}}_{1}}\]is connected across the terminal of a cell, the current is \[{{\operatorname{I}}_{1}}\], when the resistance is changed to\[{{\operatorname{R}}_{2}}\], the current is \[{{\operatorname{I}}_{2}}\].The internal resistance of the cell is:

A) \[\frac{{{I}_{1}}{{R}_{1}}+{{I}_{2}}{{R}_{2}}}{{{I}_{1}}+{{I}_{2}}}\]         

B) \[\frac{{{I}_{1}}{{R}_{2}}+{{I}_{2}}{{R}_{1}}}{{{I}_{1}}+{{I}_{2}}}\]   

C) \[\frac{{{I}_{1}}{{R}_{2}}+{{I}_{2}}{{R}_{1}}}{{{I}_{1}}-{{I}_{2}}}\]                  

D) \[\frac{{{I}_{2}}{{R}_{2}}+{{I}_{1}}{{R}_{1}}}{{{I}_{1}}-{{I}_{2}}}\]

Correct Answer: D

Solution :

\[{{\operatorname{I}}_{1}}=\frac{E}{r+{{R}_{1}}}{{\operatorname{I}}_{2}}=\frac{E}{r+{{R}_{2}}}\] \[\frac{{{\operatorname{I}}_{1}}}{{{\operatorname{I}}_{2}}}=\frac{r+{{R}_{2}}}{r+{{R}_{1}}}\] \[{{\operatorname{I}}_{1}}r+{{I}_{1}}{{R}_{1}}={{\operatorname{I}}_{2}}r+{{I}_{2}}{{R}_{2}}\] \[r\left( {{\operatorname{I}}_{1}}-{{I}_{2}} \right)={{I}_{2}}{{R}_{2}}-{{I}_{1}}{{R}_{1}}\] \[r=\frac{{{I}_{2}}{{R}_{2}}-{{I}_{1}}{{R}_{1}}}{{{\operatorname{I}}_{1}}-{{\operatorname{I}}_{2}}}\]