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 \[={{r}_{1}}-{{r}_{2}}\]You need to login to perform this action.
You will be redirected in
3 sec