In the circuit shown in Fig. 2.4. initially key K₁ is closed and key K₂ is open. Then K₁ is opened and K₂ is closed (order is important).

[Take Q₁’ and Q₂’ as charges on C₁ and C₂ and V₁ and V₂ as voltage respectively.]

Then

In first case, K₁ is closed and K₂ is open. So the circuit is effectively

The capacitor C₁ gets charged due to the battery E. Let charge Q be there on C_1, then

In the second case, the key K₁ is open and K₂ is closed. The circuit now can be redrawn as

The capacitor C₂ will get charged due to the charge on C₁. If Q₁^’ and Q₂^’ are the final charges on C₁ and C₂ respectively, from conservation of charge, we can say that option (d) is correct.

Also, we know that Q₁^’=C₁V₁ and Q₂^’=C₂V₂, the sum of which is equal to Q which is equal to C₁E. Thus, option (c) is correct.

The potential across C₁ and across C₂ will also be same because they are connected in parallel to each other. Thus, option (a) is correct.

Option (b) can be correct if and only if C₁ and C₂ have the same value. Thus, in this case it is incorrect.