Time for Capacitor A to Fall 10% After Discharge

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    Capacitor Voltage
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Discussion Overview

The discussion revolves around calculating the time it takes for a 22 nF capacitor, initially charged to 10V, to discharge to 90% of its voltage when connected to a 100kΩ resistor. The focus is on the application of the time constant in the context of capacitor discharge.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents the equation v = v₀e^(–t/τ) for capacitor discharge and questions its appropriateness.
  • Another participant confirms that the equation is correct and emphasizes the importance of its application.
  • A participant calculates the time constant τ as 2.2 ms and derives the time t for the voltage to fall by 10% as approximately 232 µs.
  • A subsequent reply agrees with the calculated time, suggesting that the solution is correct.

Areas of Agreement / Disagreement

Participants generally agree on the correctness of the equation and the calculations presented, with no significant disagreement noted.

Contextual Notes

None noted.

Who May Find This Useful

Individuals interested in capacitor discharge calculations, electrical engineering students, and those studying circuit theory may find this discussion relevant.

luigihs
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A 22 nF capacitor is initially charged to 10V. It is then discharged by connecting a 100k resistor across it. Approximately how long does it take for the voltage on the capacitor to fall by 10%?



v = v₀e^(–t/τ)
RC = τ = time constant



Im not sure if this the right equation what do you think guys?
 
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The equation looks fine. How you make use of it is what's important :smile:
 
τ = 22nF x 100k = 2200µs = 2.2 ms

then
v/v₀ = 0.9 = e^(–t/2.2m)
–t/2.2m = ln 0.9 = –0.1054
t = 0.0002318 = 232 µs <-- answer!

is that alright?
 
Last edited:
Yes, that's fine.
 

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