$A$ battery of $emf$ $10\,V$ is connected to resistances as shown in the figure. The potential difference between $A$ and $B$,$(V_A - V_B)$ is ................ $V$.

In the given circuit,the current in $8 \Omega$ resistance is $1.5 \text{ A}$. The total current $(I)$ flowing in the circuit is (in $\text{ A}$)

$A$ cell whose e.m.f. is $2\, V$ and internal resistance is $0.1\,\Omega$ is connected with a resistance of $3.9\,\Omega$. The voltage across the cell terminal will be ................ $V$.

$A$ constant voltage of $50 \ V$ is maintained between the points $A$ and $B$ of the circuit shown in the figure. The current through the branch $CD$ of the circuit is (in $A$)

$A$ $8 \Omega$ resistor is connected to a battery that has an internal resistance of $0.2 \Omega$. If the voltage across the battery (the terminal voltage) is $10 \ V$,then the emf of the battery is (in $V$)

In the circuit element given here,if the potential at point $B$ is zero $(V_B = 0)$,then the potentials of points $A$ and $D$ are: