When a resistance $R_1$ is connected across a cell, the current is $I_1$ and if the resistance $R_1$ is…

When a resistance $R_1$ is connected across a cell, the current is $I_1$ and if the resistance $R_1$ is replaced by $R_2$, the current is $\mathrm{I}_2$. Then the internal resistance of the cell is
  1. $\frac{\mathrm{I}_1 \mathrm{R}_1+\mathrm{I}_2 \mathrm{R}_2}{\mathrm{I}_1+\mathrm{I}_2}$
  2. $\frac{\mathrm{I}_1 \mathrm{R}_2-\mathrm{I}_2 \mathrm{R}_1}{\mathrm{I}_1-\mathrm{I}_2}$
  3. $\frac{\mathrm{I}_1 \mathrm{R}_2-\mathrm{I}_2 \mathrm{R}_1}{\mathrm{I}_1-\mathrm{I}_2}$
  4. $\frac{\mathrm{I}_2 \mathrm{R}_2-\mathrm{I}_1 \mathrm{R}_1}{\mathrm{I}_1-\mathrm{I}_2}$

Solution

Current $I_1=\frac{E}{R_1+r}$ and $R_1$ is replaced by $R_2$ then, $ \begin{aligned} & I_2=\frac{E}{R_2+r} \\ & \therefore \frac{I_1}{I_2}=\frac{R_2+r}{R_1+r} \Rightarrow I_1 r+I_1 R_1=I_2 r+I_2 R_2 \\ & \Rightarrow r\left(I_1-I_2\right)=I_2 R_2-I_1 R_1 \Rightarrow r=\frac{I_2 R_2-I_1 R_1}{\left(I_1-I_2\right)} \end{aligned} $

Asked in: AP EAMCET 2023 (15 May Shift 1)

Practice more Current Electricity questions on Aicharya