The adjoining figure shows the connections of a potentiometer experiment to determine the internal resistance of a Leclanché cell. When the cell is on open circuit,the balancing length of the potentiometer wire is $3.4 \, m$,and on closing the key $K_2$,the balancing length becomes $1.7 \, m$. If the resistance $R$ through which current is drawn is $10 \, \Omega$,then the internal resistance of the cell is .............. $\Omega$.
- A$0.1$
- B$1$
- C$10$
- D$1.1$
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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 ........... $\Omega$.
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View Solution$A$ wire of length $10 \ m$ and resistance $30 \ \Omega$ is connected to a battery of $emf$ $2.5 \ V$ and internal resistance $5 \ \Omega$ through an external resistance $R$. If the potential gradient along the wire is $50 \ \mu V/mm$,then $R = $ ................. $\Omega$.
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