What is the potential difference between the plates of a 4.6-F capacitor that stores sufficient energy to operate a 75.0-W light bulb for on

Question

What is the potential difference between the plates of a 4.6-F capacitor that stores sufficient energy to operate a 75.0-W light bulb for one minute?
How do i get the answer to be 44.2 V?

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Calantha 3 years 2021-08-14T04:48:59+00:00 2 Answers 8 views 0

Answers ( )

  1. Nem
    0
    2021-08-14T04:50:12+00:00
    Reply

    Answer:

    44.2v

    Explanation:

    The energy stored in capacitor, which is equal to energy required to operate a 75W bulb for one minute (60s) is

    Energy = Pt =(75W)(60s) = 4500J

    Solving equation energy = 1/2CV^2 for V we get

    V = √2(Energy)/C

    V = √2(4500J)/4.6F = 44.2V

  2. minhkhang
    0
    2021-08-14T04:50:52+00:00
    Reply

    Answer:

    V = 44.2 V

    Explanation:

    We have time = 1 minute = 60 sec, C = 4.6 F, P = 75.0 W

    Solution:

    Electric Potential Energy U Formula in capacitor for the given data.

    U = 1/2 CV²

    V = \sqrt{(2U/C)}

    So to find energy U = Pt = 75.0 W × 60 sec

    U = 4500 J

    now V =\sqrt{2 x 4500j /4.6 F}

    V = 44.232 = 44.2 V

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