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

Question

A charged capacitor is discharged through a resistor. The current I(t) through this resistor, determined by measuring the voltage

C =      F R =      ? Q0 =      C 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 2.40 10-4 J. 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.] At what time is the stored energy in the capacitor 5.70 times 10-5 J?

Explanation / Answer

The current is defined by i0*(e^-(t/?)) where ? 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 12m/s

So 0.012 = R*C so C = 0.012/205 = 5.85x10^-5F = 58.5

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