Time taken to charge capacitor to 2/3 V in

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SUMMARY

The discussion focuses on calculating the time taken to charge a capacitor to approximately 2/3 of the input voltage (Vin) using the formula Vc = Vin * (1 - e^(-t/RC)). Given a capacitance (C) of 0.047 microfarads and a resistance (R) of 2.2K ohms, the time constant (RC) is calculated as 1.034E-4 seconds. The user successfully derives the time (t) using the equation t = -RC * ln(1 - Vc/Vin) to find the specific time required to reach 2/3 Vin.

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  • Understanding of capacitor charging principles
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  • Knowledge of natural logarithms and their application in equations
  • Basic proficiency in algebra for solving equations
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mogley76
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Homework Statement


C= 0.047 micro farads
R= 2.2K
using the formula given, calcultae the time taken to charge the capacitor to 2/3 V in.


Homework Equations



Vc = Vin * (1-e^-t/rc)

The Attempt at a Solution



rc= 2.2E3 * 0.047 E-6 = 1.034E-4 seconds ( 1 time constant , 63% of vin )

i can't figure out figure out after that, i know 1 time constant is 63% and 2/3 is slightly higher than that around 67% of vin but how do i work out the time taken to reach that?
 
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You should be able to solve the equation for t.
 
t = -rc* ln(1-vc/vn)

got it thanks
 

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