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 1.60 10-4 J. (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.] C = F R = ? Q0 = C (b) At what time is the stored energy in the capacitor 5.10 10-5 J? s

Explanation / Answer

Io = 95*10^-3 [It is not clear from figure approx. it is]

I(t)=Io*e^-t/tou

40*10^-3 = 95*10^-3e^-8*10^-3/tou

tou = 9.249*10^-3 sec

(a)

Resistance = 2*Uo/(Io^2*tou) = 2*1.60 10-4/(95*10^-3^2*9.249*10^-3) = 3.834 ohm

Capacitance =tou/R = 2.413*10^-3 F

Charge =sqrt(2*Uo*C) = 8.787*10^-4 C

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