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In the circuit shown in figure , initial...

In the circuit shown in figure , initially key `K_(1)` is closed and key `K_(2)` is open. Then `K_(1)` is opened and `K_(2)` is closed (order is important). [Take `Q_(1)^(')` and `Q_(2)^(')` as charges on `C_(1)` and `C_(2)` and `V_(1)` and `V_(2)` as voltage respectively].

Then

A

charge on `C_(1)` get redistributed such that `V_(1)=V_(2)`

B

charge on `C_(1)` gas redistributed such that `q_(1)'=q_(2)'`

C

charge on `C_(1)` gets redisributed such that `C_(1)V_(1)=C_(2)V_(2)=C_(1)V`

D

charge on `C_(1)` gets redistributed such that `q_(1)'+q_(2)'=2q`

Text Solution

Verified by Experts

The correct Answer is:
A
Form the figure where `K_(1)` is closed and `k_(2)` is open, `C_(1)` is charged to potential V acquiring a total charge `q = C_(1)V`
When `K_(1)` is open and `k_(2)` is closed , battery is cut off `C_(1)` and `C_(2)` are in parallel. The charge on `C_(1)` is shared between `C_(1)` and `C_(2)` such that `V_(1) = V_(2)`
As there is no loss of charge, `q_(1)+ q_(2) = q` .
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