The wire of a potentiometer has a resistance of $4 \Omega$ and a length of $1 \ m$. It is connected to a cell of emf $2 \ V$ and internal resistance $1 \Omega$. The current flowing through the potentiometer wire is: (in $A$)

$A$ uniform conducting wire $AB$ of length $5 \, m$ and resistance $5 \, \Omega$ is connected as shown in the circuit. If the balancing point is obtained at $3 \, m$ from $A$, then the value of $E$ is. (in $ \, V$)

The potentiometer is best for measuring voltage,as

Potentiometer measures the potential difference more accurately than a voltmeter because

The potential difference across the ends of a potentiometer wire is constant. Two cells are connected in such a way that they first assist each other and then oppose each other. They are balanced on the potentiometer wire at lengths of $120 \ cm$ and $60 \ cm$ respectively. The ratio of the electromotive forces $(EMF)$ of the cells is:

$A$ cell is connected to a potentiometer,and the balancing length is obtained at $125 \ cm$. When a resistor of $2 \ \Omega$ is connected in parallel with the cell,the balancing length is obtained at $100 \ cm$. What is the internal resistance of the cell in $\Omega$?