$A$ group of $N$ cells whose $emf$ varies directly with the internal resistance as per the equation $E_N = 1.5\, r_N$ are connected in series as shown in the figure. The current $I$ in the circuit is ........... $A$.

$A$ cell of emf $1.2 \ V$ and internal resistance $2 \ \Omega$ is connected in parallel to another cell of emf $1.5 \ V$ and internal resistance $1 \ \Omega$. If the like poles of the cells are connected together,the emf of the combination of the two cells is (in $V$)

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

$12$ cells,each having the same emf $E$ and internal resistance $r$,are connected in series and kept in a closed box. Some of the cells are wrongly connected. This battery is connected in series with an ammeter and two cells,each identical to the previous cells. The current is $3 \, A$ when the external cells aid the battery and $2 \, A$ when the external cells oppose the battery. How many cells in the battery are wrongly connected?

Two cells $E_1$ and $E_2$ having equal $EMF$ $E$ and internal resistances $r_1$ and $r_2$ $(r_1 > r_2)$ respectively are connected in series. This combination is connected to an external resistance $R$. It is observed that the potential difference across the cell $E_1$ becomes zero. The value of $R$ will be

The circuit in the figure shows two cells connected in opposition to each other. Cell $E_1$ has an $emf$ of $6 \ V$ and an internal resistance of $2 \ \Omega$. Cell $E_2$ has an $emf$ of $4 \ V$ and an internal resistance of $8 \ \Omega$. Find the potential difference between the points $A$ and $B$.