Voltage across A and B,VAB=80V Let charge in 9µF,10µF and 5µF be Q1,Q2 and Q3. Net charge, Q=Q9µF+Q10µF+Q5µF As we know that, Parallel equivalent capacitance, Ceq=C1+C2+C2 ∴Ceq=9+10+5=24µF and series equivalent capacitance,
1
Ceq
=
1
Ca
+
1
Cb
+
1
Cc
⇒
1
Ceq
=
1
12
+
1
24
+
1
8
⇒
1
Ceq
=
2+1+3
24
⇒Ceq=
24
6
µF=4µF Since, C=
Q
V
∴ Q = CV =4×10−6×80=320µC Since, in series connection, V=V12µF+V10µF+V8µF ⇒80=
