In the following circuit how can we say that the current flowing through both the $4\Omega$ would be same .
I know it would be same if in middle the $\frac{2}{3}\Omega$ is not there but here it is here . And for having same current flowing the potential difference should be equal between both the $4\Omega$ resistance
Symmetry.
At the left hand node the current can go through either the $2\Omega$ or the $4\Omega$ resistor and then the rest of the circuit, $\frac{2}{3}\Omega$, $2\Omega$ and $4\Omega$.
At the right hand node the situation is exactly similar in that the current comes from an identically arranged $\frac{2}{3}\Omega$, $2\Omega$ and $4\Omega$, and then passes through either a $2\Omega$ or $4\Omega$ resistor.
The same logic can be applied to the $2\Omega$ resistors.
