Question

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



$\left[\text{Mains-2021, March 21st, Shift 2}\right]$

• A. $\frac{r_1}{2}-r_2$
  
• B. $r_1-r_2$
• C. $r_1+r_2$
• D. $\frac{r_1}{2}+r_2$
  

Answer 1

The correct option is A $\frac{r_1}{2}-r_2$



Current in the circuit, $i=\frac{3E}{R+r_1+r_2}$

If potential difference across terminals of first cell is zero.

Then, $V_A-V_B=0$

$\Rightarrow 2E-i{r}_1=0$

$\Rightarrow 2E=\frac{3E}{R+r_1+r_2}{r}_1$

$\Rightarrow 2R+2r_1+2r_2=3r_1$

$\therefore R=\frac{r_1}{2}-r_2$