Two cells with same emf $\mathrm{E}$ but different inteınal resistances, $r_1$, and $r_2$ are connected in…

Two cells with same emf $\mathrm{E}$ but different inteınal resistances, $r_1$, and $r_2$ are connected in series to an external resistance $\mathrm{R}$. If the potential difference across the first cell is zero then the value of $\mathrm{R}$ is
  1. $\frac{r_1-r_2}{2}$
  2. $\frac{r_1+r_2}{2}$
  3. $r_1-r_2$
  4. $\left(r_1+r_2\right)$

Solution


Equivalent emf, $\mathrm{E}_{\mathrm{eq}}=\mathrm{E}+\mathrm{E}=2 \mathrm{E}$ Equivalent resistance, $R_{e q}=r_1+r_2+R$ Current flowing through the circuit, $ i=\frac{2 E}{r_1+r_2+R} $ Potential drop across the first cell, $\mathrm{V}_1=\mathrm{E}-\mathrm{ir}_1$ $ \begin{aligned} & 0=E-i_1 \\ & 0=E-\left(\frac{2 E}{r_1+r_2+R}\right) r_1 \\ & 0=r_1+r_2+R-2 r_1 \\ & 0=r_2-r_1+R \\ & R=r_1-r_2 \end{aligned} $

Asked in: AP EAMCET 2023 (18 May Shift 2)

Practice more Current Electricity questions on Aicharya