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9 July, 16:35

6-3 The current through a 0.1-μF capacitor is a rectangular pulse with an amplitude of 2 mA and a duration of 5 ms. Find the capacitor voltage at the end of the pulse when the capacitor voltage at the beginning of the pulse is - 1 V.

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  1. 9 July, 16:50
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    999 V

    Explanation:

    Charge deposited on the capacitor

    = current x time

    = 2 x 10⁻³ A x 5 x 10⁻³

    = 10⁻⁵ C

    increase in volt = charge deposited / capacitance

    10⁻⁵ /.1 x 10⁻⁶

    = 10² V

    = 100 V

    voltage in the beginning = - 1 V

    voltage at the end

    = - 1 + 100

    99 V.
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