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)

Question

Physics

Eva Gallegos

11 September, 13:40

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)

+2

Answers (1)

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Doc · Answer 1

(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.