Three resistances,each of $1\,\Omega$,are joined in parallel. Three such combinations are put in series,then the resultant resistance will be ............. $\Omega$.

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:

In the given circuit,the current flowing through the resistance $20 \ \Omega$ is $0.3 \ A$,while the ammeter reads $0.9 \ A$. The value of $R_1$ is . . . . . . $\Omega$.

From the combination of resistors with resistance values $R_1 = R_2 = R_3 = 5 \ \Omega$ and $R_4 = 10 \ \Omega$,which of the following combinations is the best circuit to get an equivalent resistance of $6 \ \Omega$?

$A$ wire of resistance $12 \,\Omega$ is bent into a circle. What is the equivalent resistance between two diametrically opposite points in $\Omega$?

$A$ current of $15\,mA$ flows in the circuit as shown in the figure. The value of the potential difference between the points $A$ and $B$ will be $...\,V$.