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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, C
Initially, `K_(1)` is closed and `K_(2)` is open, thus the capacitor `C_(1)` gets fully charged. When `K_(1)` is opened and `K_(2)` is closed, the charged capacitor `C_(1)` is connected across an uncharged capacitor `C_(2)`. When a charged capacitor is connected across an uncharged capacitor, the charged capacitor supplies charge to uncharged capacitor till both the capacitors have common potential difference between their plates. Thus, charge on `C_(1)` gets redistributed such that `V_(1)=V_(2)`.
Charge acquired by `C_(1)` is `C_(1)E` when key `K_(1)` is closed. Now `K_(1)` is opened and `K_(2)` is closed hence this total charge will be redistributed between the two as per specified condition of `V_(1)=V_(2)`. As per charge conservation, total charge after redistribution must be same as before.
`C_(1)V_(1)+C_(2)V_(2)=C_(1)E`.
Hence, options (b) and (c ) are correct.
Q in option (d) is not defined, hence we cannot accept this option.
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