AP EAPCET 19-Aug-2021 Shift 1 Paper

According to given circuit diagram, 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=

320

12

+V10µF+

320

8

⇒Vtouf=1333V
∴QQuf=V10µF×10
≈133µC