In the circuit shown in the figure,find the current through the branch $BD$ in $A$.

The equivalent resistance between points $A$ and $B$ is ............ $\Omega$.

In the following circuit,find the current through the path $ACB$ if each resistance $R = 4 \ \Omega$ and the battery voltage is $16 \ V$. (in $A$)

In the circuit shown in the figure,all the resistances are identical and each has the value $r \ \Omega$. The equivalent resistance of the combination between the points $A$ and $B$ will remain unchanged even when the following pairs of points marked in the figure are connected through a resistance $R$.

In the following network,the current through the galvanometer will:

For the circuit given below,the Kirchhoff's loop rule for the loop $BCDEB$ is given by the equation: