Question

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. $\frac{r_1+r_2}{2}$

Answer 1

Answer: (B)

The two cells have potential $E$ each and respective internal resistance $r_1$ and $r_2$.

Let the current flowing in the circuit be $i$.

Using Kirchhoff's voltage law:

$\Rightarrow -E+i{r}_1+iR-E+i{r}_2=0$

$\Rightarrow i=\frac{2E}{(R+r_1+r_2)}$

Potential across first cell, $V_{AB}=i{r}_1-E=\frac{2E{r}_1}{(R+r_1+r_2)}-E$

For $V_{AB}=0$, we get: $\frac{2E{r}_1}{(R+r_1+r_2)}-E=0$

$\Rightarrow$ $\frac{2r_1}{(R+r_1+r_2)}=1$

$\Rightarrow 2r_1=R+r_1+r_2$

$\Rightarrow R=r_1-r_2$