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

Options

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)' = Q`
C. charge on `C_(1)` gets redistributed such that `C_(1) V_(1) + C_(2) V_(2) = C_(1) E`
D. Both A & B correct

Answer

D

Explanation

In Fig, when `K_(1)` is closed and `K_(2)` is open, `C_(1)` is charged to potential `E` acquiring a total chareg `Q = C_(1) E`
When `K_(1)` is opened and `K_(2)` is closed, battery is cut off. `C_(1)` and `C_(2)` are in parallel. The charges 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`
Choices (a) and (d) are correct.