A 3.0-µF capacitor charged to 40 V and a 5.0-µF capacitor charged to 18 V are connected to each other, with the positive plate of each conne

Question

A 3.0-µF capacitor charged to 40 V and a 5.0-µF capacitor charged to 18 V are connected to each other, with the positive plate of each connected to the negative plate of the other. What is the final charge on the 3.0-µF capacitor? *

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Ngọc Khuê 3 weeks 2021-07-16T07:16:31+00:00 1 Answers 0 views 0

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    2021-07-16T07:18:22+00:00

    Answer:

    Explanation:

    Given that,

    Capacitor 1

    Capacitance C1 = 3.0-µF

    Voltage V1 = 40V

    Capacitor 2

    Capacitance C2 = 5.0-µF

    Voltage V2 = 18V

    The two capacitor are connected such that, the positive place is connected to the negative plate of the other, I.e series connection

    Charge on each capacitor before connection

    Q1 = C1V1 = 3.0-µF × 40

    Q1 = 120 -µC

    Q2 = C2V2 = 5.0-µF × 18

    Q2 =90 µC

    So, when they are connected together in series, the charge with combine together to have a single end equal charge since series connection have the same charge

    Therefore,

    1/Qeq = 1/Q1 + 1/Q2

    1/Qeq = (Q1+Q2)/Q1Q2

    Qeq = Q1Q2/(Q1+Q2)

    Qeq = 51.43 -µC

    So the charge on the will be

    Q = Qeq/2

    Q = 51.43/2

    Q = 25.71

    Q≈ 26 -µC

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