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

The Correct Option is C

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) : \(r_1-r_2\).

Answer 2

Given :
Current flowing through the circuit :
\(I=\frac{E_{eq}}{R+r_1+r_1}=\frac{E+E}{R+r_1+r_2}\)
Potential difference across the first cell,
\(V_1=E-lr_1=E-\frac{2E}{r_1+r_2+R}r_1=0\)
\(E=\frac{2Er_1}{r_1+r_2+R}\)
\(1=\frac{2r_1}{r_1+r_2+R}\)
\(r_1+r_2+R=2r_1\)
\(R=r_1-r_2\)
So, the correct option is (C) : \(r_1+r_2\)