Two cells of emf $E_1$ and $E_2$ $(E_1 > E_2)$ are connected individually to a potentiometer and their corresponding balancing lengths are $625 \, cm$ and $500 \, cm$. Then the ratio $\frac{E_1}{E_2}$ is ...........

$A$ resistance of $R \; \Omega$ draws current from a potentiometer. The potentiometer has a total resistance $R_{0} \; \Omega$ (Figure). $A$ voltage $V$ is supplied to the potentiometer. Derive an expression for the voltage across $R$ when the sliding contact is in the middle of the potentiometer.

The arrangement as shown in the figure is called:

The potentiometer is best for measuring voltage,as

In a potentiometer experiment, the balancing length with a cell is $250 \, cm$. On shunting the cell with a resistance of $2 \, \Omega$, the balancing length becomes $125 \, cm$. The internal resistance of the cell is:

To measure the internal resistance of a battery,a potentiometer is used. For $R = 10 \ \Omega$,the balance point is observed at $\ell = 500 \ cm$ and for $R = 1 \ \Omega$,the balance point is observed at $\ell = 400 \ cm$. The internal resistance of the battery is approximately: (in $\Omega$)