For the circuit shown in the figure, determine the charge in  $\left(\mu C\right)$ of the capacitor in steady state.

In steady state the capacitor is fully charged and is treated as open circuit, so no current flows through branch containing capacitor in steady state. So the circuit can be redrawn as:

Potential difference across the capacitor in steady state
$=V-6-V=-6V$ 
(-ve sign signifies that left hand plate is of negative polarity)
Charge  $=CV=1\times 6=6\mu C$