question_answer

Two cells of equal e.m.f. and of internal resistances \[{{r}_{1}}\] and \[{{r}_{2}}({{r}_{1}}>{{r}_{2}})\] are connected in series. On connecting this combination to an external resistance R, it is observed that the potential difference across the first cell becomes zero. The value of R will be [MP PET 1985; KCET 2005; Kerala PMT 2005]

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

B)            \[{{r}_{1}}-{{r}_{2}}\]

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

D)            \[\frac{{{r}_{1}}-{{r}_{2}}}{2}\]

Correct Answer: B

Solution :

                   Let the voltage across any one cell is V, then \[V=E-ir=E-{{r}_{1}}\,\left( \frac{2E}{{{r}_{1}}+{{r}_{2}}+R} \right)\] But V = 0 Þ \[E-\frac{2E{{r}_{1}}}{{{r}_{1}}+{{r}_{2}}+R}=0\] Þ \[{{r}_{1}}+{{r}_{2}}+R=2{{r}_{1}}\] Þ \[R={{r}_{1}}-{{r}_{2}}\]