If a resistance ${R_2}$ is connected in parallel with the resistance $R$ in the circuit shown,then the possible value of current through $R$ and the possible value of ${R_2}$ will be:

You are given several identical resistances each of value $R = 10\,\Omega$ and each capable of carrying a maximum current of $1\,A$. It is required to make a suitable combination of these resistances to produce a resistance of $5\,\Omega$ which can carry a current of $4\,A$. The minimum number of resistances of the type $R$ that will be required for this job is:

If the resistivity of all four wires connected in parallel is different as shown in the figure,and all other dimensions of the wires are the same,then the total current $i$ is (the resistance of a wire having resistivity $\rho$ is $R_0$):

The equivalent resistance of a group of resistances is $R$. If another resistance is connected in parallel to the group,its new equivalent becomes $R_1$,and if it is connected in series to the group,its new equivalent becomes $R_2$. We have:

There are $8$ equal resistances $R$. Two are connected in parallel,and such four groups are connected in series. The total resistance of the system will be:

$A$ wire of resistance $20 \Omega$ is divided into $10$ equal parts. $A$ combination of two parts are connected in parallel and so on. Now,the resulting pairs of parallel combinations are connected in series. The equivalent resistance of the final combination is . . . . . . $\Omega$.