Question

Assume a circuit like that in the picture is used to charge a 470-uF capacitor to 50 volts...?
which is then discharged through a 1.8-k ohhm resistor. How do I find the capacitor voltage after the switch is opened for 3 time constants...The time constant is 0.846 so i think i have to multiply that by three then do something else but i'm not sure...

Answer 1

http://user-content.enotes.com/b664caf15141c6753b3bc6dae03e108baac123b0.jpg

Answer 2

Use the formula \[V _{c}=50^{-t/\tau}\]50 Is the fully charged voltage, t=time and tau is the time constant since it is given that t = to 3 tau the equation can be rewritten:
\[V _{c}=50/e ^{3}\] e being the base for natural logs.

Answer 3

i don't understand

Answer 4

Ok, Kirchhoff's loop rule:
\[\frac{ dq }{ dt }R+\frac{ q }{ C }=0 \Rightarrow q=q_0e^{-\frac{ t }{ RC }}=q_0e^{-\frac{ t }{ \tau }}\]\[q=VC\]\[q_0=V_0C\]thus\[V=V_0e^{-\frac{ t }{\tau }}\]
t has to be 3 times tau(time constant) so
\[V=V_0e^{-3}\]so\[V=50e^{-3}=24.94 \]

Answer 5

Do you agree with that?

Answer 6

I know it's the right answer but i'm still not understanding how you got it. I don't think i understand the variables :-/

Answer 7

Lets review your problem. You state that the time constant (tau) is .846. Do you know how that was obtained? RC time constant tau or the symbol\[\tau \]
RC=1.8 K 470 UFarad or\[\tau = 1.8 \times 10^{3} \times 470 \times 10^{-6}=846 \times 10^{-3}=.846\] That is tau or Now lets look at the equation for voltage discharge, First the equation is in both post, in Fifcial post it is \[V=V _{o}e ^{-3}\] In my post it was shown as\[V _{c}=50^{-e t/\tau}\]Now maybe you are confused over the variables:
The \[V _{o}\]is the voltage across the capacitor it will be 50 volts initially and then bleed off as time goes by at rate determined by the time constant.
The 50 is the value of the charged capacitor voltage at the time the switch is closed.

e is a mathematical constant used as the base of natural log it is called Euler's number and is approximately 2.71828

t is for time.\[ \tau = \]as stated earlier is for the RC time constant

Now you were given t as 3 time constants that means \[t =3 \tau\]so the exponent of e is -3tau/tau or simply -3. I may have rambled here, but next time when you are "confused" please be more specific. That way I can address the point of confusion more directly.

Answer 8

Now as far as the answer goes @Fifciol may have misplaced a decimal, I believe the capacitor has discharged further down than that. A rule of thumb 5 time constants and the capacitor will almost fully discharged. (Theoretcally it never discharges completely but we need to be more practical).

Answer 9

Ok i think i understand now...So the voltage should be 2.5 V right?