The current provided by a battery is maximum when:

Two cells of emf $1 \ V$ and $2 \ V$ and internal resistance $2 \ \Omega$ and $1 \ \Omega$,respectively,are connected in series with an external resistance of $6 \ \Omega$. The total current in the circuit is $I_1$. Now the same two cells in parallel configuration are connected to the same external resistance. In this case,the total current drawn is $I_2$. The value of $\left(\frac{I_1}{I_2}\right)$ is $\frac{x}{3}$. The value of $x$ 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

What is meant by the series connection of cells? Derive the equation for the equivalent electromotive force $(emf)$ of two cells with $emf$ $\varepsilon_1$ and $\varepsilon_2$ connected in series.

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:

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: