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 . . . . . . .

Infinite number of cells having $emf$ and internal resistance $(E, r)$,$(\frac{E}{n}, \frac{r}{n})$,$(\frac{E}{n^2}, \frac{r}{n^2})$,$(\frac{E}{n^3}, \frac{r}{n^3})$... are connected in series across an external resistance of $\frac{nr}{n+1}$. The current flowing through the external resistor is:

Twelve cells,each having emf $E$ volts,are connected in series and are kept in a closed box. Some of these cells are wrongly connected with positive and negative terminals reversed. This $12$-cell battery is connected in series with an ammeter,an external resistance $R$ ohms,and a two-cell battery (two cells of the same type used earlier,connected perfectly in series). The current in the circuit when the $12$-cell battery and $2$-cell battery aid each other is $3 \text{ A}$,and it is $2 \text{ A}$ when they oppose each other. Then,the number of cells in the $12$-cell battery that are connected wrongly is:

In the given circuit,if $V_{AB} = 4\ V$,then the value of resistance $X$ in ohms will be:

$A$ cell of internal resistance $r$ is connected to an external resistance $R$. The current will be maximum in $R$,if

$32$ cells,each of $emf$ $3V$,are connected in series and kept in a box. Externally,the combination shows an $emf$ of $84V$. The number of cells reversed in the combination is