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In potentiometer experiment, the balancing length with a cell $\mathrm{E}_{1}$ of unknown e.m.f.
is $\mathcal{\ell}_{1}{ }^{\prime} \mathrm{cm}$. By shunting the cell with resistance $\mathrm{R} \Omega$, the balancing length becomes $\frac{\ell_{1}}{2} \mathrm{~cm}$, the internal resistance (r) of a cell is
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is $\mathcal{\ell}_{1}{ }^{\prime} \mathrm{cm}$. By shunting the cell with resistance $\mathrm{R} \Omega$, the balancing length becomes $\frac{\ell_{1}}{2} \mathrm{~cm}$, the internal resistance (r) of a cell is
Solution:
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Verified Answer
The correct answer is:
$\mathrm{r}=\mathrm{R}$
$\mathrm{r}=\mathrm{R}\left(\frac{\ell_{1}}{\ell_{2}}-1\right) \quad \ell_{2}=\frac{\ell_{1}}{2}$
Putting the values and solving we get $r=R$
Putting the values and solving we get $r=R$
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