Capacitor of 6 microf is conected in parallel with the combination of 4 microf capacitor and a 3 microf capacitor that are in series.
Find the net capacitance of the entire combination. Find the potential difference across 3 microf capacitor when 20 V is maintained across the 6 microf capacitor.
Since the capacitors 3 microF and 4 microF are in series will be replaced by a capacitor that has the capacitance Cs.
1/Cs = 1/4 + 1/3
1/Cs = (3+4)/4*3
1/Cs = 7/12
Cs = 12/7
Now, the capacitor that has the capacitance Cs = 12/7 and the capacitor of 6 microF are in parallel.
The capacitors will be replaced by the capacitor that has the capacitance Cp.
Cp = 12/7 + 6
Cp = (12+7*6)/7
Cp = (12+42)/7
Cp = 54/7 microF
The net capacitance is Cp = 54/7 microF.
Since the capacitor of 6microF and the capacitor of 12/7 microF are in parallel, the potential difference of 20 V across 6 microF is the same across 12/7 microF, too.
Now, we'll determine the charge through 12/7 microF capacitor:
charge = voltage*capacitance
charge = 20V*(12/7) microF
charge = 240/7 microC
The charge on the capacitors of 4 and 3 microF will be the same, since they are in series.
Potential difference across 3 microF capacitor is:
240/7*3 = 240/21 = 80/7 = 11.4 V