A charged capacitor is discharged through a resistor. The current I(t) through t

Question

A charged capacitor is discharged through a resistor. The current I(t) through this resistor, determined by measuring the voltage VR(t) = I(t)R with an oscilloscope, is shown in the graph. The total energy dissipated in the resistor is 1.90 10-4 J.
http://www.webassign.net/web/Student/Assignment-Responses/submit?dep=4494917
a) Find the capacitance C, the resistance R, and the initial charge Q0 on the capacitor. [Hint: You will need to solve three equations simultaneously for the three unknowns. You can find both the initial current and the time constant from the graph.]
Okay so for this part a I got the right answers
C=156.5e-6
R=16.402
Q0=2.4e-4 but I need help with part b
(b) At what time is the stored energy in the capacitor 4.80 10-5 J? 0.001767

Explanation / Answer

energy stored = 0.5q^2/C = 4.8 * 10^-5 where q = Q0(1-e^(-t/RC)) q^2 = 9.6 * 156.5 * 10^-11 q=sqrt(9.6 * 156.5 * 10^-11) = 1.2 * 10^-4 -t/RC = ln(1-q/Q0) = ln(1/2) t = 16.402 * 156.5 * 10^-6 * ln(2) = 1.8 ms

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