Two batteries with different $emf$ and different internal resistances are connected as shown in the figure. The voltage between the two terminals $A$ and $B$ is ........ $volt$.

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:

If the potential difference across the internal resistance $r_1$ is equal to the $emf$ $E$ of the battery,then

$A$ cell can supply currents of $1 \ A$ and $0.5 \ A$ via resistances of $2.5 \ \Omega$ and $10 \ \Omega$ respectively. The internal resistance of the cell is (in $\Omega$)

Four identical cells, each of $2 \, V$ e.m.f., are connected in parallel. They supply current to an external circuit consisting of two $15 \, \Omega$ resistors connected in parallel. The terminal voltage of the equivalent cell, as measured by an ideal voltmeter, is $1.6 \, V$. Calculate the internal resistance of each cell in $\Omega$.

$A$ current of $2\,A$ is flowing through a cell of $e.m.f.$ $5\,V$ and internal resistance $0.5\,\Omega$ from negative to positive electrode. If the potential of the negative electrode is $10\,V$,the potential of the positive electrode will be .............. $V$.