Question 181 of 380

Two sources of equal emfs are connected in series. This combination is connected to an external resistance $R$. The internal resistances of the two sources are ${r}_{1}$ and ${r}_{2}\left({r}_{1}\gt {r}_{2}\right)$. If the potential difference across the source of internal resistance ${r}_{1}$ is zero then the value of $R$ will be

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

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

Current through the wire will be,
$I=\dfrac{2\epsilon }{R+{r}_{1}+{r}_{2}}$.
Now potential difference across cell $1$ will be, $\epsilon -I{r}_{1}=0$
$\Rightarrow \epsilon =\dfrac{2\epsilon }{R+{r}_{1}+{r}_{2}}\times {r}_{1} \Rightarrow R+{r}_{1}+{r}_{2}=2{r}_{1} \Rightarrow R={r}_{1}-{r}_{2}$