Two capacitors, one that has a capacitance of 6 µF and one that has a capacitance of18 µF are first discharged and thenare connected in series. The series combination is then connectedacross the terminals of a 12-V battery. Next, they are carefully disconnected so that they are notdischarged and they are then reconnected to each other--positiveplate to positive plate and negative plate to negative plate.
(a) Find the potential difference across eachcapacitor after they are reconnected.
1Your answer is incorrect. V (6 µFcapacitor)
2Your answer is incorrect. V (18µF capacitor)

(b) Find the energy stored in the capacitor before they aredisconnected from the battery, and find the energy stored afterthey are reconnected
3 µJ (beforedisconnected)
4 µJ(reconnected)

(a) V1 = 9V, V2= 3V

(b)E = 324µJ, E1 = 243µJ, E2 = 81µJ.

Explanation:

Given C1 = 6 µF, C2 = 18 µF

V = 12V

Ceq = C1×C2/(C1 + C2) series connection

Ceq = 6×18/(6 + 18) = 108/24 = 4.5µF

Q = Ceq×V

Q = 4.5×12×10-⁶ = 54µC

The same amount of charge is on both capacitors as they are connected in series.

When the capacitors are reconnected,

V1 = voltage across capacitor 1

V2 = voltage across capacitor 2

V1 = Q/C1 = 54µC/ 6µF = 9V

V2 = Q/C2 = 54µC/ 18µF = 3V

(b)

Before Disconnection

E = 1/2×Ceq×V² = 1/2×4.5×10-⁶× 12² = 324µJ

After the reconnection,

E1 = energy stored in capacitor 1

E2 = energy stored in capacitor 2

E1 = 1/2C1V1² = 0.5× 6µF×9² = 243µJ

E2 = 1/2C2V2² = 0.5×18µF×3² = 81µJ

E = E1 +E2 = (243 + 81)µJ = 324µJ.

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