Question

Two cells with the same e.m.f. $E$ 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: (C)

$r_{1}-r_{2}$

Answer 2

There are two batteries with emf $E$ each and the internal resistances $r_{1}$ and $r_{2}$ respectively.
Hence we have $I\left(R+r_{1}+r_{2}\right)=2 E$ thus, $I=\frac{2 E}{R+r_{1}+r_{2}}$
Now the potential difference across the first cell would be equal to $V=E-I r_{1} .$
From the question, $V=0$, hence, $E=I r_{1}=\frac{2 E r_{1}}{R+r_{1}+r_{2}}$,
thus, $R+r_{1}+r_{2}=2 r_{1}$,
hence $R=r_{1}-r_{2}$.
So, the correct option is (C) : $r_1-r_2$.