A capacitor of capacitance 4 µF is discharging through a 2.0-MΩ resistor. At what time will the energy stored in the capacitor be one-third

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A capacitor of capacitance 4 µF is discharging through a 2.0-MΩ resistor. At what time will the energy stored in the capacitor be one-third of its initial value?

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  1. Answer:

    Explanation:

    Given that,

    The capacitance of the capacitor is

    C = 4 µF

    Discharging through a resistor of resistance

    R = 2 MΩ

    Time the energy stored in the capacitor be one-third of its initial energy

    i.e. u(E)_ final = ⅓u(E)_initial

    U / Uo = ⅓

    Energy stored in a capacitor (discharging) can be determined using

    U = Uo•RC•exp(-t/RC)

    U / Uo = -RC exp(-t/RC)

    U / Uo = ⅓

    RC = 2 × 10^6 × 4 × 10^-6 = 8s

    ⅓ = 8 exp( -t / 8)

    Divide both side by -8

    1/24 = exp(-t/8)

    Take In of both sides

    In(1/24) = In•exp(-t/8)

    -3.1781 = -t / 8

    t = -3.1781 × -8

    t = 25.42 seconds

    It will take 25.42 seconds for he capacitor to be ⅓ of it’s initial energy

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