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 2.60 10-4 J.

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(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 5.70 10-5 J?
___ s

Explanation / Answer

The current is defined by i0*(e^-(t/t)) where t is the time constant = RC (time constant is the time for the current to dissipate to 0.368 of it max value) Reading from the graph the value of t when I = 36.8mA is approximately 12ms So 0.012 = R*C so C = 0.012/205 = 5.85x10^-5F = 58.5 µF The total energy stored is the energy dissipated U = 1/2*C*V^2 = 1/2*5.85x10^-5*(20.5)^2 = 0.0123J = 12.3mJ b) The energy is 1/2 when the voltage is sqrt(.5)*V0 = 0.707V0 So when e^-(t/t) = 0.707 so -t/0.012 = ln(0.707) => t = -0.012*ln(.707) = 0.00416s = 4.16ms

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