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 seri

Question

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)

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Thu Thảo 5 years 2021-08-12T21:52:16+00:00 1 Answers 11 views 0

Answers ( )

  1. bonexptip
    0
    2021-08-12T21:53:29+00:00
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    Answer:

    (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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