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:                                                                    [AIPMT (S) 2006]

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, Kirchhoff’s law is violated. The question must have been modified.