The formula for internal resistance of a cell will be, (Here $ l_{1} \& l_{2} $ are the balancing lengths of cell in open and closed circuit respectively.)

Question

The formula for internal resistance of a cell will be, (Here $ l_{1} \& l_{2} $ are the balancing lengths of cell in open and closed circuit respectively.), $ \mathrm{r}=\left(\frac{\mathrm{l}_{1}-\mathrm{l}_{2}}{\mathrm{l}_{2}}\right) \mathrm{R} $, $ \mathrm{r}=\left(\frac{\mathrm{l}_{2}-\mathrm{l}_{1}}{\mathrm{l}_{2}}\right) \mathrm{R} $, $ \mathrm{r}=\left(\frac{\mathrm{l}_{1}-\mathrm{l}_{2}}{\mathrm{l}_{1}}\right) \mathrm{R} $, $ \mathrm{r}=\left(\frac{\mathrm{l}_{2}-\mathrm{l}_{1}}{\mathrm{l}_{1}}\right) \mathrm{R} $

MARKS App · Accepted Answer

Given that balancing lengths of potentiometer in open and closed circuit as l 1 , l 2 and If R is the resistance of resistance box and x is the potential gradient then Internal resistance of the cell will be, r = E – V V R = l 1 x – l 2 x l 2 x R = l 1 – l 2 l 2 R

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