In the figure,the potentiometer wire of length $l = 100 \, cm$ and resistance $R = 9 \, \Omega$ is joined to a cell of emf $E_1 = 10 \, V$ and internal resistance $r_1 = 1 \, \Omega$. Another cell of emf $E_2 = 5 \, V$ and internal resistance $r_2 = 2 \, \Omega$ is connected as shown. The galvanometer $G$ will show no deflection when the length $AC$ is ............... $cm$.

To determine the internal resistance of a cell with a potentiometer,when the cell is shunted by a resistance of $5 \Omega$,the balancing length is $250 \ cm$. When the cell is shunted by $20 \Omega$,the balancing length of the potentiometer wire is $400 \ cm$. The internal resistance of the cell is: (in $\Omega$)

$A$ potentiometer wire has length $10\, m$ and resistance $20\,\Omega$. $A$ $2.5\, V$ battery of negligible internal resistance is connected across the wire with an $80\,\Omega$ series resistance. The potential gradient on the wire will be

$A$ cell of internal resistance $1.5\,\Omega$ and of $e.m.f.$ $1.5\,V$ balances at $500\,cm$ on a potentiometer wire. If a wire of $15\,\Omega$ is connected between the balance point and the cell,then the balance point will shift:

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 $A$).

Explain the comparison of the electromotive force (emf) of two cells using a potentiometer with a necessary diagram.