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    Working with time constant RC

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    A capacitor is used in the electronic flash unit of a camera. A small battery with a constant voltage of 6 V is used to charge the capacitor with a constant current of 10 microamperes. How long does it take to charge the capacitor when C = 10 microfarads? My Answer:
    V=6V
    i=10mA
    C=10microF
    Wc=1/2CV^2*(t)
    Wc=1/2(10^-5)6^2*(t)
    Wc=18*10^-5*(t)
    Wc=6V*.01A=.06J
    Wc=.06J=18*10^-5*(t)
    t=.06/(18*10^-5)=333.33
    Final answer: 333.33 s

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    https://brainmass.com/engineering/electronic-engineering/working-with-time-constant-rc-37632

    Solution Preview

    We know that charge, Q = current * time = I * t

    Also, for a capacitor, C = ...

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    The answer with good explanation on why the student answered it wrongly.

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