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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
The cahrge stored by capacitor `C_(1)` gets redistributed between `C_(1) and C_(2)` till their potentials become same i.e., `V_(2)=V_(1)`. By law of conservation of charge, the charge stored stored in capacitor `C_(1)` when key `K_(1)` is closed and key `K_(2)` is open is equal to sum of charges on capacitors `C_(1) and C_(2)` when `K_(1)` is opened and `K_(2)` is clsoed i.e.,
`Q_(1)'+Q_(2)'=Q`.
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