A capacitor with capacitance C is charge...

A capacitor with capacitance C is charged to `Q_(0)` and then connected in the circuit shown. When key is closed, what is charge on this capacitor in steady state?

A

`(Q_(0))/(3)`

B

`(2Q_(0))/(3)`

C

`(Q_(0))/(2)`

D

`(3Q_(0))/(2)`

Verified by Experts

The correct Answer is:

B

`V_("common")=(Q_(0))/(C+(C)/(2))=(2Q_(0))/(3C)`
`q_("final")=CV_("common")=(2Q_(0))/(3)`

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