Three capacitors are connected across a 45 V power supply as shown in fig. What is the charge on the 6 μF capacitor ?

It is obvious from figure that 6 μF and 12 μF capacitors are in parallel to each other and the combination is in series with 9 μF capacitor. So, we have 9 μF and 18 μF capacitors in series across 45 V power supply. Voltage drop across combination = 15 VNow   9×V1=V2×18So, charge on 6 μF capacitor =6 μF×15=90 μC