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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
- charge on C1 gets redistributed such that V1 = V2
- charge on C1 gets redistributed such that Q1′ = Q2′
- charge on C1 gets redistributed such that C1V1 + C2V2 = C1E
- charge on C1 gets redistributed such that Q1′ + Q2′ = Q
Options
a and c
a and d
b and c
c and d
Solution
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
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