Physics

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A charged capacitor,C=10e-6 F, with an initial stored energy of 3.5J, is discharged across a resistor, R. The charge on the capacitor drops to 40% of its maximum value in time, t=8.2s, from the instant the switch is closed (t=0).
Find the current through the resistor at a time t=10s.
How much thermal energy is dissipated across the resistor after two time constants?
Determine the electrostatic energy on the capacitor after two time constants.

  • Physics -

    C = Q/V
    V = (1/C) integral i dt
    dV/dt = (1/C) i
    but
    i = -V/R
    so
    dV/dt = (1/C) -V/R
    or
    dV/V = (-1/RC) dt (I bet you knew that)
    so
    ln V = (-1/RC) t
    V = Vi e^(-t/RC) (which you also know)
    and Q = CV = CVi e^(-t/RC)
    at t = 8.2
    Q(8.2) = Cv = .4 CVi = CVi e^(-8.2/RC)
    so
    .4 = e^(8.2/{R*10^-5} )
    ln .4 = -8.2*10^5/R
    solve for R

    but how much energy is stored at V ?
    E = (1/2) C V^2
    that should get you started

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