Charge on C2 = 40 μC Voltage on C2 = charge on C2 / capacitance = 40μC / 3μF = 16.67 V C1 and C2 are in parallel. Therefore, the same voltage exists at C1 and C2. In this case, the charge of a capacitor is directly proportional to its capacitance. Then press C1 = (40μC) x (6μF / 3μF) = 80μC Full load on C1 and C2 = 40μC + 80μC = 120μC. As C3 is in series with the combination of C1 and C2, press C3 = 120 μC Voltage on C3 = 120 μC / 5 μF = 24 V Vab = sum of the voltages on the combination C1 and C2 and C3 Vab = 16.67 + 24 = 40.67V