Question 183 of 380

Two cells of the same EMF $E$ but different internal resistances ${r}_{1}$ and ${r}_{2}$ are connected in series with an external resistance $R$ as shown in the figure. The terminal potential difference across, the second cell is found to be zero. The external resistance $R$ must then be:

A${r}_{1}{r}_{2}$B${r}_{1}-{r}_{2}$C${r}_{2}-{r}_{1}$D${r}_{1}+{r}_{2}$

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Answer: (C) ${r}_{2}-{r}_{1}$

In this question we have to find the Terminal potential difference of second cell, we are given the internal resistances of both the cells and an external resistance. Applying the following expression:-
$V=E-i{r}_{2}$
$\Rightarrow 0=E-i{r}_{2}$
Applying the expression of current
$0=E-\left(\dfrac{2E}{{r}_{1}+{r}_{2}+R}\right)\cdot {r}_{2}$
$\Rightarrow R={r}_{2}-{r}_{1}$