MHT CET · Physics · Current Electricity
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
- A \(\mathrm{R}\left(\frac{\mathrm{E}_{1}}{\mathrm{~V}}+1\right)\)
- B \(\mathrm{R}\left(\frac{\mathrm{V}}{\mathrm{E}_{1}}-1\right)\)
- C \(\mathrm{R}\left(\frac{\mathrm{V}}{\mathrm{E}_{1}}+1\right)\)
- D \(\mathrm{R}\left(\frac{\mathrm{E}_{1}}{\mathrm{~V}}-1\right)\)
Answer & Solution
Correct Answer
(D) \(\mathrm{R}\left(\frac{\mathrm{E}_{1}}{\mathrm{~V}}-1\right)\)
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