CGPET 2021 Solved Paper

Since, C1 and C2 are connected in series, therefore equivalent capacitance is

1

Cs

=

1

C1

+

1

C2

=

1

2

+

1

3

⇒Cs=

6

5

µF
Effective voltage, V=16−6=10V
New equivalent circuit is

Charges on each capacitor,
Q=Cs×V=

6

5

×10⇒Q=12µC
Thus, potential difference across 2µF capacitor is
VC1=

Q

C1

=

12

2

⇒VC1=6V