Question 188 of 380

Two cells of emf $2E$ and $E$ with internal resistance ${r}_{1}$and ${r}_{2}$ respectively are connected in series to an external resistor $R$ (see figure). The value of $R,$ at which the potential difference across the terminals of the first cell becomes zero is

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

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Answer: (B) $\dfrac{{r}_{1}}{2}-{r}_{2}$

$i=\dfrac{3E}{R+{r}_{1}+{r}_{2}}$
$T.P.D.=2E-i{r}_{1}=0$
$2E=i{r}_{1}$
$\Rightarrow 2E=\dfrac{3E\times {r}_{1}}{R+{r}_{1}+{r}_{2}}$
$2R+2{r}_{1}+2{r}_{2}=3{r}_{1}$
$R=\dfrac{{r}_{1}}{2}-{r}_{2}$