When current supplied by a cell to a circuit is $0.3 \,A$,its terminal potential difference is $0.9 \,V$. When the current supplied becomes $0.25 \,A$,its terminal potential difference becomes $1.0 \,V$. The internal resistance of the cell is ............ $\Omega$.

The value of the internal resistance of an ideal cell is:

$A$ battery consists of $n$ rows of cells connected in parallel,with each row containing $m$ cells in series. The maximum current is obtained through an external resistance of $3 \, \Omega$. If the total number of cells is $24$ and the internal resistance of each cell is $0.5 \, \Omega$,find the values of $m$ and $n$.

In the network shown,the potential difference between $A$ and $B$ is ................. $V$ $(R = r_1 = r_2 = r_3 = 1 \Omega, E_1 = 3 \, V, E_2 = 2 \, V, E_3 = 1 \, V)$.

Two cells $E_1$ and $E_2$ having equal $EMF$ $E$ and internal resistances $r_1$ and $r_2$ $(r_1 > r_2)$ respectively are connected in series. This combination is connected to an external resistance $R$. It is observed that the potential difference across the cell $E_1$ becomes zero. The value of $R$ will be

The $n$ rows,each containing $m$ cells in series,are joined in parallel. Maximum current is drawn from this combination across an external resistance of $3 \,\Omega$. If the total number of cells used is $24$ and the internal resistance of each cell is $0.5 \,\Omega$,then: