Question

In the circuit shown in figure 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

1. charge on C₁ gets redistributed such that V₁ = V₂
2. charge on C₁ gets redistributed such that Q₁′ = Q₂′
3. charge on C₁ gets redistributed such that C₁V1 + C₂V₂ = C₁E
4. charge on C1 gets redistributed such that Q₁′ + Q₂′ = Q

Options

• a and c

• a and d

• b and c

• c and d

Answer 1

a and d

Explanation:

Initially, key K₁ is closed and key K₂ is open, the capacitor C₁ is charged by the battery and capacitor C₂ is still uncharged. Now K₁ is opened and K₂ is closed, the capacitors C₁ and C₂ both are connected in parallel. The charge stored by capacitor C₁ gets redistributed between C₁ and C₂ till their potentials become the same, i.e., V₂ = V₁.

By law of conservation of charge, the charge stored in capacitor Cx is equal to the sum of charges on capacitors C₁ and C₂ when K₁ is opened and K₂ is closed, i.e., Q’₁ + Q’₂= Q