(A) The balancing length $l_1$ for an open circuit cell is $52 \ cm$. When the cell is shunted by an external resistance $R = 5 \ \Omega$,the balancin

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(A) The balancing length $l_1$ for an open circuit cell is $52 \ cm$. When the cell is shunted by an external resistance $R = 5 \ \Omega$,the balancing length $l_2$ becomes $40 \ cm$. The internal resistance $r$ of the cell is given by the formula: $r = \left( \frac{l_1 - l_2}{l_2} \right) R$ Substituting the given values: $r = \left( \frac{52 - 40}{40} \right) \times 5$ $r = \left( \frac{12}{40} \right) \times 5$ $r = 0.3 \times 5 = 1.5 \ \Omega$.

Class 12 Physics Current Electricity Potentiometer

Medium

In a potentiometer experiment,it is found that no current passes through the galvanometer when the terminals of the cell are connected across $52 \ cm$ of the potentiometer wire. If the cell is shunted by a resistance of $5 \ \Omega$,a balance is found when the cell is connected across $40 \ cm$ of the wire. Find the internal resistance of the cell ........... $\Omega$.

JEE MAIN 2018 All JEE MAIN

A
$1.5$
B
$2$
C
$2.5$
D
$1$

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When a cell of $E.M.F.$ $E_1$ is connected to a potentiometer wire,the balancing length is $l_1$. Another cell of $E.M.F.$ $E_2$ $(E_1 > E_2)$ is connected such that the two cells oppose each other,then the balancing length is $l_2$. The ratio $E_1 : E_2$ is

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