question_answer

Two cells with the same emf E and different internal resistances \[{{r}_{1}}\] and \[{{r}_{2}}\] are connected in series to an external resistance R. The value of R so that the potential difference across the first cell be zero, is

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

B)  \[{{r}_{1}}+{{r}_{2}}\]

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

D)  \[\frac{{{r}_{1}}+{{r}_{2}}}{2}\]

Correct Answer: C

Solution :

                 There are two batteries with emf E each and the internal resistances \[{{r}_{1}}\]and \[{{r}_{2}}\]respectively. Hence we have \[I(R+{{r}_{1}}+{{r}_{2}})=2E,\] thus,                      \[I=\frac{2E}{2+{{r}_{1}}+{{r}_{2}}}\] Now the potential difference across the first cell would be equal to \[V=E-I{{r}_{1}}.\]From the question, V = 0, hence, \[E=I{{r}_{1}}=\frac{2E{{r}_{1}}}{R+{{r}_{1}}+{{r}_{2}}},\] thus, \[R+{{r}_{1}}+{{r}_{2}}=2{{r}_{1}},\] hence \[R={{r}_{1}}-{{r}_{2}}.\]