question_answer

Two cells, having the same emf, are connected in series through an external      resistance R. Cells have internal resistances \[{{r}_{1}}\]and \[{{r}_{2}}\]\[({{r}_{1}}>{{r}_{2}}),\] respectively. When the circuit is              closed, the potential difference across the first cell is zero. The value of R is

A)  \[{{r}_{1}}-{{r}_{2}}\]                                    

B)  \[\frac{{{r}_{1}}+{{r}_{2}}}{2}\]

C) \[\frac{{{r}_{1}}+{{r}_{2}}}{2}\]                                 

D) \[{{r}_{1}}+{{r}_{2}}\]

Correct Answer: A

Solution :

                Net resistance of the circuit \[={{r}_{1}}+{{r}_{2}}+R\] Net emf in series \[=E+E=2E\]                 Therefore, from Ohms law, current in the circuit                 \[i=\frac{Net\,emf}{Net\,\text{reistance}}\] \[\Rightarrow \]               \[i=\frac{2E}{{{r}_{1}}+{{r}_{2}}+R}\]                       ??(i)     It is given that, as circuit is closed, potential difference across the first cell is zero. That is,                 \[V=E-i{{r}_{1}}=0\] \[\Rightarrow \]               \[i=\frac{E}{{{r}_{1}}}\] Equating Eqs. (i) and (ii), we get                 \[\frac{E}{{{r}_{1}}}=\frac{2E}{{{r}_{1}}+{{r}_{2}}+R}\] \[\Rightarrow \]               \[2{{r}_{1}}={{r}_{1}}+{{r}_{2}}+R\] \[\therefore \]  R = external resistance \[={{r}_{1}}-{{r}_{2}}\]