Given three capacitors, C1C1C_1 = 2.0 μFμF, C2C2C_2 = 1.5 μFμF, and C3C3C_3 = 3.0 μFμF, what arrangement of parallel and series connections

Question

Given three capacitors, C1C1C_1 = 2.0 μFμF, C2C2C_2 = 1.5 μFμF, and C3C3C_3 = 3.0 μFμF, what arrangement of parallel and series connections with a 12-VV battery will give the minimum voltage drop across the 2.0- μFμF capacitor?

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Latifah 3 years 2021-08-26T08:20:00+00:00 1 Answers 9 views 0

Answers ( )

  1. thugiang
    0
    2021-08-26T08:21:12+00:00
    Reply

    Answer:

    Explanation:

    If joined in parallel ,each will have potential of 12 volt

    so on 2 μF capacitor , voltage will be 12 V

    When joined in series , if C be equivalent capacitor

    1/C = 1/2 + 1/1.5 + 1/3

    = .5 + .667 + .333

    = 1.5

    C = .666 μF

    Charge on each capacitor

    = CV

    = .666 x 12

    = 8 μC

    Potential on 2 μF capacitor

    = charge / capacitance

    = 8 μC / 2 x μF

    = 4 V

    So in series , potential at 2μF capacitor is 4 V which is less than 12 V in parallel combination.

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