In the experiment to determine the internal resistance of a cell (E1) using potentiometer, the resistance…

In the experiment to determine the internal resistance of a cell (E1) using potentiometer, the resistance drawn from the resistance box is ${ }^{\prime} \mathrm{R}^{\prime}$. The potential difference across the balancing length of the wire is equal to the terminal potential difference (V) of the cell. The value of internal resistance (r) of the cell is
  1. $\mathrm{R}\left(\frac{\mathrm{E}_{1}}{\mathrm{~V}}+1\right)$
  2. $\mathrm{R}\left(\frac{\mathrm{V}}{\mathrm{E}_{1}}-1\right)$
  3. $\mathrm{R}\left(\frac{\mathrm{V}}{\mathrm{E}_{1}}+1\right)$
  4. $\mathrm{R}\left(\frac{\mathrm{E}_{1}}{\mathrm{~V}}-1\right)$

Solution

In the experiment to determine the internal resistance of a cell $\left(\mathrm{E}_1\right)$ using a potentiometer, the resistance drawn from the resistance box is 'R'. The potential difference across the balancing length of the wire is equal to the terminal potential difference $(\mathrm{V})$ of the cell. The value of internal resistance $(\mathrm{r})$ of the cell is $R\left(\frac{E_1}{V}-1\right)$.

Asked in: MHT CET 2020 (13 Oct Shift 2)

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