The emfs of three cells connected in parallel are $E_1=5 \ V, E_2=8 \ V$ and $E_3=10 \ V$ and their internal resistances are $r_1=1 \ \Omega, r_2=2 \ \Omega$ and $r_3=3 \ \Omega$,respectively. By changing $E_3$ to $E_{3N}$,the equivalent emf is doubled. Then $E_{3N}$ in $V$ is:

The electromotive force of a primary cell is $2\,V$. When it is short-circuited,it gives a current of $4\,A$. Its internal resistance in $\Omega$ is:

When a battery is connected across a resistor of $16 \Omega$,the voltage across the resistor is $12 \ V$. When the same battery is connected across a resistor of $10 \Omega$,the voltage across it is $11 \ V$. The internal resistance of the battery in ohm is

Each of the six ideal batteries of emf $20 \, V$ is connected to an external resistance of $4 \, \Omega$ as shown in the figure. The current through the resistance is (in $ \, A$)

$A$ 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 .............. $V$.

Two cells are connected between points $A$ and $B$ as shown. Cell $1$ has an emf of $12 \, V$ and an internal resistance of $3 \, \Omega$. Cell $2$ has an emf of $6 \, V$ and an internal resistance of $6 \, \Omega$. An external resistor $R$ of $4 \, \Omega$ is connected across $A$ and $B$. The current flowing through $R$ will be $............. \, A$.