In an electrical circuit drawn below the amount of charge stored in the capacitor is ______ $$\mu C$$

At steady state (DC), the capacitor branch carries no current. So the 5 Ω branch is open.

step 1: reduce circuit

Only active loop:

battery (10 V) → 4 Ω → 6 Ω → back

Total resistance:

R=4+6=10 Ω

step 2: current in circuit

$$I=\frac{10}{10}=1A$$

step 3: voltage drop

Voltage drop across 4 Ω:

$$V_4=1\times4=4V$$

Voltage drop across 6 Ω:

$$V_6=6V$$

step 4: capacitor voltage

Capacitor is connected between left node and right node (same as ends of 6 Ω)

So voltage across capacitor = voltage across 6 Ω = 6 V

step 5: charge

$$Q=CV=10μF\times6=10\times10^{-6}\times6=60μC$$