question_answer

Two cells with a same emf E and different internal resitances \[{{r}_{1}}\text{ }and\text{ }{{r}_{2}}\] 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 :

                 The figure is drawn accordingly. Potential across \[{{r}_{1}}\] is given by as \[{{V}_{1}}=E-i{{r}_{1}}\]                                              ?(i) Similarly, potential across \[{{r}_{2}}\] is given by as \[{{V}_{2}}=E-i{{r}_{2}}\]                                              ?(ii) For \[{{V}_{1}}\]to be zero. \[E-i{{r}_{1}}=0\,\]or \[E=i{{r}_{1}}\] Now, Eq. (ii) on by putting values \[E=i\,{{r}_{1}},\]becomes \[{{V}_{2}}=i{{r}_{1}}-i{{r}_{2}}=i({{r}_{1}}-{{r}_{2}})\] Total potential, \[V={{V}_{1}}+{{V}_{2}}={{V}_{2}}\] Hence, \[i\,R=i({{r}_{1}}-{{r}_{2}})\] So,          \[R={{r}_{1}}-{{r}_{2}}\]