Reducing Energy in an RC Capacitor: Solving for Discharge Time

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SUMMARY

The discussion focuses on calculating the time after discharge when the energy stored in an RC capacitor is reduced to half its initial value. Given an RC time constant of 7 ms, the relevant equations include \(E = 0.5CV^2\) and \(Q_f = Q_i \cdot e^{-t/RC}\). The final charge required to achieve half the energy is determined to be \(Q_f = \frac{Q_i}{\sqrt{2}}\). The solution concludes that the time for the energy to halve is approximately 2.43 ms.

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Homework Statement


At what time (in ms) after the discharge begins is the energy stored in the capacitor reduced to half its initial value?

RC=7 ms

Homework Equations


RC=7ms
E=.5CV^2
E=.5Q^2/C
E=.5QV
Q_f = Q_i * e^(-t/RC)

The Attempt at a Solution


I don't even know where to go from here. I know that since the problem involves RC, the equation to solve the problem must have some form of Q_f = Q_i * e^(-t/RC) but I'm stuck.
 
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hover said:

Homework Statement


At what time (in ms) after the discharge begins is the energy stored in the capacitor reduced to half its initial value?

RC=7 ms

Homework Equations


RC=7ms
E=.5CV^2
E=.5Q^2/C
E=.5QV
Q_f = Q_i * e^(-t/RC)

The Attempt at a Solution


I don't even know where to go from here. I know that since the problem involves RC, the equation to solve the problem must have some form of Q_f = Q_i * e^(-t/RC) but I'm stuck.

Since energy is proportional to the square of charge, by what fraction of the maximum charge should the final charge be to produce half the energy stored with maximum charge?
 
Last edited:
xcvxcvvc said:
Since energy is proportional to the square of charge, by what fraction of the maximum charge should the final charge be to produce half the energy stored with maximum charge?

Ah! I see now! When one works the math, the final charge after half the energy is gone is q/2^.5. the initial charge is q. From there it is just plugging into get an the answer 2.43 ms.

Thanks! :D
 

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