
#1
Oct2605, 02:20 PM

P: 1,629

Hello everyone, I almost got all the parts to this problem:
In an RC series circuit, = 11.0 V, R = 1.10 M, and C = 1.80 µF. (a) Calculate the time constant. 1.98 s (b) Find the maximum charge that will appear on the capacitor during charging. 19.8 µC (c) How long does it take for the charge to build up to 9.0 µC? s I tried the following: q(t) = Q[1e^(t/RC)] q(t) = 9.0e6[1e^(1.98)] = 7.75e6s, which is wrong; Since i know t = RC, and i found RC = 1.98, would it be q(t) = 9.0e6[1e^(1)] ? I just tried this and its also wrong. 



#2
Oct2605, 03:52 PM

PF Gold
P: 867

In your equation q(t) should equal 9.0mc since that is the charge that has built up after time t. and Q which is the total charge should be 19.8mc, then just solve for time.




#3
Oct2605, 06:05 PM

P: 1,629

thanks! i'm stuck on how i'm suppose to solve for t...
i have (19.8E6)(e^(t/1.98)); do i take the natural log of both sides or somthing? but u can't take the natural log of a negative number! 



#4
Oct2705, 01:30 AM

PF Gold
P: 867

ow long does it take for the charge to build up to 9.0 µC? RC circuits
you know that:
9.0E6 = 19.8E6[1e^(t/1.98)] = 19.8E6  19.8E6*e^(t/1.98) so: 10.8E6 = 19.8E6*e^(t/1.98) /1 10.8E6 = 19.8E6*e^(t/1.98) now you can take the natural log of both sides 



#5
Oct2705, 07:52 AM

P: 1,629

Hey thanks for the responce!
I took the ln of both sides and got: 11.44 = (t/1.98)*ln(19.8E6); I end up with a negative number of 10.8, did I do this wrong? Sorry its been awhile since I did stuff with ln 



#6
Oct2705, 08:15 AM

P: 255

you didn't take the natural logarithm of each side in the correct manner.
First divide through by 19.8*10^6 to get rid of that term on the RHS of the expression. So you would then have: 0.54 = e^(t/1.98) Take the natural log of that and you have: ln(0.54) = t/1.98 This will give you a positive number 



#7
Oct2705, 09:17 AM

P: 1,629

Thanks!! But i submitted that as my answer: t = .311 and it was also incorrect!! hm...




#8
Oct2705, 09:28 AM

P: 1,629

ohh my bad hah thanks big man, it worked great!!!



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