Figure below shows a network of resistors, cells, and a capacitor at steady state. What is the current through the resistance 4 $\Omega$ ?

At steady state, capacitor acts as open circuit $\Rightarrow$ no current through $3\mu F$ branch and the $2\Omega$ top-left resistor.

Simplified circuit has two active branches. Let left node be $V_A$, right node = 0 V (ground). The 6 V cell sets the middle wire at +6 V.

Current through middle branch (via 2 V cell and $1\Omega$):
$I_1 = \dfrac{(V_A + 2) - 6}{1} = V_A - 4$

Current through bottom branch (via 3 V cell and $4\Omega$):
$I_2 = \dfrac{(V_A - 3) - 0}{4} = \dfrac{V_A - 3}{4}$

KCL at node A: $(V_A - 4) + \dfrac{V_A - 3}{4} = 0$

$4V_A - 16 + V_A - 3 = 0 \Rightarrow 5V_A = 19 \Rightarrow V_A = 3.8$ V

$I_{4\Omega} = \dfrac{3.8 - 3}{4} = \dfrac{0.8}{4} = 0.2$ A