Question

Sol. [Key Concept: In steady state no current flows through the capacitor branch, so $V_{C}$ equals the drop across the parallel resistor, $i \times 2$.]

$${i = \frac{2}{6 + 2} = \frac{1}{4}\text{ }A }{V_{C} = V_{2\Omega} = i \times 2 = \frac{1}{2}\text{ }V = 0.5\text{ }V }$$