Consider the circuit shown. If all the cells have negligible internal resistance, what will be the current through the $2 \Omega$ resistor when steady state is reached (in $A$)?

In the cube shown,current $i$ enters at $H$ and leaves at $C$. If $i_{AB} = \frac{i}{6}$,$i_{DC} = \frac{2i}{3}$,$i_{HA} = \frac{i}{2}$,$i_{GF} = \frac{i}{6}$,and $i_{HE} = \frac{i}{6}$,choose the branch in which the current is zero.

Consider the following circuit shown below. All the resistors are identical. The ratio of $I / I^{\prime}$ is

The resistances in the Wheatstone bridge circuit shown in the diagram are all different,and the current through the galvanometer is zero. If all thermal effects are negligible,the current through the galvanometer will not be zero when:

In the circuit given here,the potentials at points $A$,$B$,and $C$ are $70 \,V$,$0 \,V$,and $10 \,V$ respectively. Then:

The equivalent resistance between the points $A$ and $B$ in the following circuit is $\frac{x}{5} \Omega$. The value of $x$ is . . . . . . .