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1 February, 07:15

A capacitor has a peak current of 310 μa when the peak voltage at 300 khz is 3.0 v.

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  1. 1 February, 07:19
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    I think the question for this problem, is to find the capacitance. The working equation would be:

    C = Q/V,

    where

    Q = It

    Since f = 1/t, t = (1/300 kHz) (1 kHz/1,000 Hz) = 3.33*10⁻⁶ s

    So,

    Q = (310*10⁻⁶ A) (3.33*10⁻⁶ s)

    Q = 1.033*10⁻⁹ C

    Thus,

    C = (1.033*10⁻⁹ C) / (3 V)

    C = 3.44*10⁻¹⁰ F
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