Two capacitors 2 μF and 4 μF are connected in parallel. A third capacitor of 6 μF capacity is connected in series. The combination is then connected across a 12 V battery. The voltage across 2 μF capacity is

Resultant capacitance of condensers of capacity 2 μF and 4 μf when connected in parallel C′=2+4=6μFThis is connected in series with a capacitor of capacity 6 μF in series. The resultant capacity C is given by1C=16+16=13  or  C=3μFCharge on combinationq=3×10−6×(12)=36×10−6CLet the charge on 2 μF capacitor be q1, thenq12=q−q14  or  q1=q3∴ q1=12×10−6CNow potential across 2 μF condenser=q12×10−6=12×10−62×10−6=6V