Capacitor, Charge, and Sinewave, please

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SUMMARY

The discussion centers on a circuit involving a capacitor charged to 70.0 V, an inductor with an inductance of 4.03×10-2 H, and resistance RL. Participants analyze the energy in the circuit after 16 periods and the time required for 86% of the initial energy to dissipate. A critical error identified is the incorrect calculation of the period, as the sine wave does not pass through zero at t=0, indicating that the time between t=0 and t=0.7 ms does not represent a complete period. The correct frequency is determined to be 12.5 Hz.

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Homework Statement



A capacitor C is charged to an initial potential of 70.0 V, with an initial charge of Q0. It is in a circuit with a switch and an inductor with inductance L = 4.03×10-2 H and resistance RL.

At t=0, the switch is closed, and the curve below shows the potential V across the capacitor as a function of time t.

https://s4.lite.msu.edu/cgi-bin/plot.png?file=jonesb63_msu_1427749730_13825679_plot.data

Calculate the energy in the circuit after a time of 16 periods. Note that the curve passes through a grid intersection point.

Calculate the time required for 86% of the initial energy to be dissipated.

The Attempt at a Solution



media%2F8f4%2F8f40d080-e8e7-46ef-8ed5-b79a27de2fea%2FphpPOHOPO.png


media%2F3cf%2F3cfa6b6a-758d-4580-a16b-852e1d994a33%2FphpsCHPxR.png


media%2F9de%2F9dea4368-d50e-47bb-9e2d-dc043892efa2%2FphpUpSWtr.png

Not sure what is wrong with this.
 
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I haven't looked through everything, but your calculation of the period is clearly incorrect. The sine wave passes through zero at t=0.7ms, but it is not zero at t=0. So the time between t=0 and t=0.7 is not a whole number of periods.
 
phyzguy said:
I haven't looked through everything, but your calculation of the period is clearly incorrect. The sine wave passes through zero at t=0.7ms, but it is not zero at t=0. So the time between t=0 and t=0.7 is not a whole number of periods.
oh 12.5 per second yeah my bad, thanks
 

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