What Percent of Final Charge Does a Capacitor Reach After 4.5 Time Constants?

AI Thread Summary
After 4.5 time constants in an RC circuit, a capacitor reaches approximately 98.2% of its final charge. The formula used to calculate the charge is Q = Qo [1 - e^(-t/RC)], where Qo is the initial charge. By substituting t with 4.5RC, the expression simplifies to e^(-4.5). This calculation confirms that the capacitor is nearly fully charged at this point. Understanding this concept is crucial for analyzing charging behavior in RC circuits.
itryphysics
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Homework Statement


Consider an RC circuit in which a capacitor is being charged by a battery connected in a circuit. Initially the capacitor is uncharged. Then the switch is closed.
After a time equal to 4.5 time constants, what percent of the final charge is present on the capacitor?


Homework Equations


Q = Qo [1 - e^ (-t/RC)]



The Attempt at a Solution



I picked a value of 100 for Qo
so
Q= 100 [1 - e^ (-t/4.5)]
my approach was going to be to find Q and then see wht percent that is of the initial charge Qo
but I don't know wht to put in for the time "t"
please help me out
 
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I think you would take t = 4.5RC
 
so for the expression e^(-t/RC) I do e^(-4.5) ?
 
4.5 time constants means a time of 4.5*R*C.
R*C comes out in units of time.
 
itryphysics said:
so for the expression e^(-t/RC) I do e^(-4.5) ?

Yes that should do it.
 
Awsome! Thank you so much!
 

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