$A$ resistance of $4\,\Omega$ and a wire of length $5\,m$ and resistance $5\,\Omega$ are joined in series and connected to a cell of $e.m.f.$ $10\,V$ and internal resistance $1\,\Omega$. $A$ parallel combination of two identical cells is balanced across $300\,cm$ of the wire. The $e.m.f.$ $E$ of each cell is ........... $V$.

In the given circuit of a potentiometer,the potential difference $E$ across $AB$ ($10\, m$ length) is larger than $E_{1}$ and $E_{2}$ as well. For key $K_{1}$ (closed),the jockey is adjusted to touch the wire at point $J_{1}$ so that there is no deflection in the galvanometer. Now,the first battery $(E_{1})$ is replaced by the second battery $(E_{2})$ for working by making $K_{1}$ open and $K_{2}$ closed. The galvanometer then gives null deflection at $J_{2}$. The value of $\frac{E_{1}}{E_{2}}$ is $\frac{a}{b}$,where $a = \dots$ (Refer to the image for balancing lengths $l_{1}$ and $l_{2}$ from point $A$).

The arrangement as shown in the figure is called:

$A$ wire of length $100\, cm$ is connected to a cell of emf $2\, V$ and negligible internal resistance. The resistance of the wire is $3\, \Omega$. The additional resistance required to produce a potential difference of $1\, mV/cm$ is ............. $\Omega$.

$A$ $2\,V$ battery,a $15\,\Omega$ resistor,and a potentiometer of $100\,cm$ length are all connected in series. If the resistance of the potentiometer wire is $5\,\Omega$,then the potential gradient of the potentiometer wire is ............... $V/cm$.

In a potentiometer experiment, a cell of emf $1.25 \,V$ gives a balancing length of $30 \,cm$. If the cell is replaced by another cell, the balancing length is found to be $40 \,cm$. What is the emf of the second cell?