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Yesterday, 18:53

Three capacitors C1 = 10.1 µF,

C2 = 23.0 µF,

and

C3 = 29.4 µF

are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.

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Answers (1)
  1. Yesterday, 21:16
    0
    energy is 0.14 J

    Explanation:

    given data

    capacitor C1 = 10.1 µF

    capacitor C2 = 23.0 µF

    capacitor C3 = 29.4 µF

    charged = 125 V

    to find out

    maximum energy stored

    solution

    we will apply here formula that is

    energy = 1/2 Q²/c ... 1

    here c = 1 / (1/c1 + 1/c2 + 1/c3)

    c = 1 / (1/10.1 + 1/23 + 1/29.4)

    c = 5.66 µF

    and Q = c1v1

    Q = 10.1 * 125

    Q = 1262.5 µC

    so from equation 1

    energy = 1/2 * 1262.5² / 5.66 = 140804.439

    so energy is 0.14 J
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