The length of a wire of a potentiometer is $100\, cm$, and the $emf$ of its standard cell is $E\,volt$. It is employed to measure the $e.m.f$ of a battery whose internal resistance is $0.5 \,\Omega$. If the balance point is obtained at $l = 30\, cm$ from the positive end, the $e.m.f.$ of the battery is
where $i$ is the current in the potentiometer
AIEEE 2003, Medium
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(a) From the principle of potentiometer $V \propto l$
$ \Rightarrow $ $\frac{V}{E} = \frac{l}{L}$; where $V$ = $emf$ of battery, $ E$ = $emf$ of standard cell, $L$ = Length of potentiometer wire
$V = \frac{{El}}{L} = \frac{{30E}}{{100}}$.
$ \Rightarrow $ $\frac{V}{E} = \frac{l}{L}$; where $V$ = $emf$ of battery, $ E$ = $emf$ of standard cell, $L$ = Length of potentiometer wire
$V = \frac{{El}}{L} = \frac{{30E}}{{100}}$.
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