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Physics

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A fully charged 90-µF capacitor is discharged through a 60-Ω resistor. How long will it take for the capacitor to lose 80% of its initial energy?

I know energy is given by E(t)=q(t)^2, but beyond that I really don't know how to solve this. Any suggestions are appreciated.

Thank you!

  • Physics -

    Q(t)=Q₀•exp(-t/RC)
    E~Q² =>
    E=E₀•exp(-2•t/RC)
    100-80=20%
    0.2•E₀=E₀•exp(-2•t/R•C)
    ln0.2 =- 2•t/R•C
    t= - ln0.2• R•C/2 =- (-1.6) •60•90•10⁻⁶/2 =0.0043 s.

  • Physics -

    E = (1/2) c v^2
    c = 90 * 10^-6
    q = c v
    so dq/dt = c dv/dt = -i
    v = -i r = r c dv/dt
    so
    dv/dt = (-1/rc) v

    v = Vi e^kt
    dv/dt = k Vi e^kt = (-1/rc)Vi e^kt
    k = -1/rc
    v = Vi e^-t/rc
    E = constant * v^2
    so v^2 at t = .8 v^2 at t = 0
    so
    v at t = sqrt (.8) v at t = 0
    v at t = .64 Vi
    .64 = e^-(1/rc)t
    ln .64 = -.446 = -(1/rc) t

    t = .446 * 90 * 10^-6 * 60

  • misread -

    sqrt .2 not .8
    so
    .04 = e^-(1/rc) t
    -3.22 = -(1/rc) t
    t = 3.22 * 90 * 10^-6 * 60

  • Physics -

    thank you!

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