TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Given that, C1=5µF,C2=C3=10µF and E=20V
When the switch is connected to point A.
Then, situation of the circuit is shown below.Charge stored in C1,
Q=C1V1=C1E=5×20µC=100µC
When switch is connect to point B, then situation circuit is shown below.The charge Q1 will start to distribute through, C2 and C3 to make the potential difference V across all the three capacitors C1,C2 and C3.
Now, common potential difference,
V=‌

‌ Total charge ‌

‌ Total capacity ‌

=‌

C1V1+C2V2+C3V3

C1+C2+C3

where, V1=20V, but V2=V3=0V
∴‌‌V=‌

100

5+10+10

=‌

100

25

=4V
Now, charge on C3,q3=C3V
‌‌=10×4µC
‌‌=40µC