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 :
| [a] Key Idea: Current in the circuit is given by Ohm's law. |
| Net resistance of the circuit \[={{r}_{1}}+{{r}_{2}}+R\] |
| Net emf in series \[=E+E=2\,E\] |
 |
| Therefore, from Ohm's law, current in the circuit |
| \[i=\frac{\text{Net}\,\text{emf}}{\text{Net}\,\text{resistance}}\] |
| \[\Rightarrow \] \[i=\frac{2E}{{{r}_{1}}+{{r}_{2}}+R}\] (i) |
| It is given that, as circuit is closed, potential difference across the first cell is zero. That is, |
| \[V=E-i{{r}_{1}}=0\] |
| \[\Rightarrow \] \[i=\frac{E}{{{r}_{1}}}\] (ii) |
| Equating Eqs. (i) and (ii), we get |
| \[\frac{E}{{{r}_{1}}}=\frac{2\,E}{{{r}_{1}}+{{r}_{2}}+R}\] |
| \[\Rightarrow \] \[2{{r}_{1}}={{r}_{1}}+{{r}_{2}}+R\] |
| \[\therefore \] \[R=\] external resistance \[={{r}_{1}}-{{r}_{2}}\] |
| Note: The question is wrong as the statement is when the circuit is closed, the potential difference across the first cell is zero which implies that in a series circuit, one part cannot conduct current which is wrong, Kirchhoffs law is violated. The question must have been modified. |
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