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In the circuit shown, left plate of the ...

In the circuit shown, left plate of the capacitor with capacitance C is given a charge `Q_(0)` and plates of capacitor are connected across the battery as shown. Amount of charge that will flow from battery to left plate of capacitor when key is closed is:

A

CV

B

`CV-(Q_(0))/(2)`

C

`CV+(Q_(0))/(2)`

D

`CV-Q_(0)`

Text Solution

Verified by Experts

The correct Answer is:
B
Key B closed

Charge `=((Q_(0))/(2) +CV)-Q_(0)`
`=CV -(Q_(0))/(2)`
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