Question

Two capacitors of capacitance 8micro farad and 10 micro farad respectively are connected in series across a potential difference of 180 volt. The capacitors are are disconnected from the supply and are reconnected in parallel with each other. Calculate the new potential difference and charge on each capacitor.

Answer 1

Explanation:

Let C1=3 μF,C2=6 μF

C1 and C2 are connected in series with potential difference V=900 V acros the combination

Thus for the series combination Charge on both capacitor is same

∴C

1

V

1

=C

2

V

2

=Q ...1

∴

V

2

V

1

=

C

1

C

2

=2 μC ...2

also V

1

+V

2

=900 V ...3

From 2 and 3

V

2

=300 V and V

1

=600 V

Therefore charge on each capacitor Q=300×6=600×3=1800 μC

When these capacitors are disconnected and reconnected in parallel the charge is distributed and the potential difference is same

∴

C

1

Q

1

=

C

2

Q

2

=V ...4

∴

Q

2

Q

1

=

C

2

C

1

=

2

1

...5

and Q

1

+Q

2

=Q=1800 μC ...6

From 5 and 6

Q

1

=600 μC and Q

2

=1200 μC

Therefore the potential difference across the combination is V=

3

600