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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)` gets redistributed such that `V_(1) = V_(2)`

B

charge on `C_(1)` gets redistributed such that `Q_(1)^(') = Q_(2)^(')`

C

charge on `C_(1)` gets redistributed such that `C_(1)V_(1) + C_(2)V_(2) = C_(1) E`

D

charge on `C_(1)` gets redistributed such that `Q_(1)^(') + Q_(2)^(') = Q`.

Text Solution

Verified by Experts

The correct Answer is:
A, D
Initially, when `K_(1)` is closed and `K_(2)` is opened, `C_(1)` gets charged to potential `E` possessing a charge `Q = C_(1) E`. When `K_(1)` and is opened and `K_(2)` is closed, battery gets disconnected from the circuit and `C_(1)` and `C_(2)` gets connected in parallel. Charge on `C_(1)` gets redistributed such that `V_(1) = V_(2)`. Also, as there in no loss of charge, `Q_(1)^(') + Q_(2)^(') = Q`.
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