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 $A$)?

What is a Wheatstone bridge? Explain its principle.

$A$ battery of $10 \;V$ and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of $12$ resistors each of resistance $1 \;\Omega$. Determine the equivalent resistance of the network and the current along each edge of the cube.

Find the current flowing through the resistance $R_1$ of the circuit shown in the figure,if the resistances are equal to $R_1=10 \Omega, R_2=20 \Omega$,and $R_3=30 \Omega$,and the potentials of points $1, 2$,and $3$ are equal to $\phi_1=10 \text{ V}, \phi_2=6 \text{ V}$,and $\phi_3=5 \text{ V}$. (in $\text{ A}$)

Why is the circuit configuration known as a Wheatstone bridge?

In the network shown in the figure,each of the resistance is equal to $2\,\Omega$. The resistance between the points $A$ and $B$ is .............. $\Omega$.