The number of dry cells,each of $e.m.f.$ $1.5\,V$ and internal resistance $0.5\,\Omega$,that must be joined in series with a resistance of $20\,\Omega$ so as to send a current of $0.6\,A$ through the circuit is:

When two identical batteries of internal resistance $1 \Omega$ each are connected in series across a resistor $R$,the rate of heat produced in $R$ is $P_1$. When the same batteries are connected in parallel across $R$,the rate of heat produced is $P_2$. If $P_1 = 2.25 P_2$,then the value of $R$ is (in $Omega$)

When two identical cells are connected either in series or in parallel across a $2 \ \Omega$ resistor,they send the same current through it. The internal resistance of each cell is: (in $Omega$)

The internal resistance of a cell of emf $4 \text{ V}$ is $0.1 \ \Omega$. It is connected to an external resistance of $3.9 \ \Omega$. The terminal voltage across the cell will be . . . . . . . (in $\text{ V}$)

If the potential difference across the internal resistance $r_1$ is equal to the $emf$ $E$ of the battery,then

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 $............. \, A$.