Question

Two cells, having the same emf, are connected in series through an external resistance $R$. Cells have internal resistances $ r_{1} $ and $ r_{2} $ $ (r_{1} > r_{2}), $ respectively. When the circuit is closed, the potential difference across the first cell is zero. The value of $R$ is

• A. $ r_{1}-r_{2} $
• B. $ \frac{r_{1}+r_{2}}{2} $
• C. $ \frac{r_{1}+r_{2}}{2} $
• D. $ r_{1}+r_{2} $

Answer 1

The Correct Option is A

Net resistance of the circuit $=r_{1}+r_{2}+R$ Net emf in series $=E+E=2 E$ Therefore, from Ohms law, current in the circuit $i=\frac{\text { Net emf }}{\text { Net reistance }} $ $\Rightarrow i=\frac{2 E}{r_{1}+r_{2}+R}$ ...(i) It is given that, as circuit is closed, potential difference across the first cell is zero. That is, $V=E-i r_{1}=0 $ $\Rightarrow i=\frac{E}{r_{1}}$ Equating Eqs. (i) and (ii), we get $\frac{E}{r_{1}}=\frac{2 E}{r_{1}+r_{2}+R}$ $ \Rightarrow 2 r_{1}=r_{1}+r_{2}+R $ $\therefore R =$ external resistance $=r_{1}-r_{2}$