TS EAMCET 18-July-2022 Shift 1 Solved Paper

Given, C1=C2=C3=1µF=10−6F
V=9V
When switch is thrown to the right side, then left side of switch is open. Hence, capacitors C2 and C3 are uneffective.
Therefore, only capacitor C2 is in valid connection.
∴ Charge on capacitor C1,
Q1‌=C1V=1×10−6×9
‌=9×10−6C=9µC
Now, when switch is thrown to left, then capacitors C1,C2 and C3 are connected in series and charge of charged capacitor C1 is flowing towards uncharged capacitors C2 and C3.
Since, C1=C2=C3, hence VC1=VC2=VC3
Therefore, Q2=Q3=‌

Q1

3

=‌

9

3

=3µC