The stored charge in the capacitor in steady state of the following circuit is __________ $$\mu$$C.

Circuit simplification at steady state

In steady state, the capacitor branch is an open circuit. We simplify the resistance from right to left:

• Right loop series: $$4\ \Omega+4\ \Omega+2\ \Omega=10\ \Omega$$
• Parallel with middle $$10\ \Omega$$ resistor: $$\frac{10 \times 10}{10 + 10} = 5\ \Omega$$
• Middle loop series: $$5\ \Omega + 5\ \Omega + 2\ \Omega = 12\ \Omega$$
• Parallel with first $$12\ \Omega$$ resistor: $$\frac{12 \times 12}{12 + 12} = 6\ \Omega$$

Total current from the battery:

$$I_{total} = \frac{12\ V}{6\ \Omega} = 2\ A$$

• The $$2\ A$$ current splits equally at the first branch ($$12\ \Omega$$ each), so $$1\ A$$ enters the middle section.
• That $$1\ A$$ splits equally at the second branch ($$10\ \Omega$$ each), so $$0.5\ A$$ flows through the rightmost $$4\ \Omega$$ vertical resistor.

The voltage across the capacitor ($$V_{C}$$) is the same as the voltage across the $$4\ \Omega$$ vertical resistor:

$$V_{C} = 0.5\ A \times 4\ \Omega = 2\ V$$

Final charge ($$Q$$):

$$Q = C \times V_{C}$$

$$Q = 100\ \mu F \times 2\ V = 200\ \mu C$$