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Two capacitors of capacitances 3 mu F an...

Two capacitors of capacitances `3 mu F` and `6 mu F` are connected in series and connected to `120 V`. The potential difference across `3 mu F` is `V_(0)` and the charge here is `q_(0)`. We have
(A) `q_(0) = 40 mu C`
(B) `V_(0) = 60 V`
( C) `V_(0) = 80 V`
(D) `q_(0) = 240 mu C`.

A

A,C are correct

B

A,B are correct

C

B,D are correct

D

C,D are correct

Text Solution

Verified by Experts

The correct Answer is:
D
`Q=((C_(1)C_(2))/(C_(1)+C_(2)))V`.
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