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 1.70 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 4.70 10-5 J? __s

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 1.70 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 4.70 10 - 5 J? __s

Explanation / Answer

I=Ioe^-t/RC dW=I^2Rdt W=Io^2R^2C/2 40mA=Ioe^-2/RC (From Graph) 20mA=Ioe^-4/RC (From Graph) e^-2/RC=.5 =>Io=80*10^-3=8*10^-2 2ms/RC=.69 =>RC=2.89*10^-3 1.7*10^-4=64*10^-4*RC*R/2 R=18.38 ohm RC=2.89*10^-3=>C=1.57*10^-4 Qo=I0RC=2.3 *10^-4 C C=1.57*10^-4 F R=18.38 ohm Qo=2.3*10^-4 C b)W=1.7*10^-4e^-2t/RC=4.7*10^-5 2t/RC=1.28 t=1.86 *10^-3 sec

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