Finding R in a RC circuit given battery, capacitance, and time.

AI Thread Summary
To find the resistance R in an RC circuit with a 10.0 µF capacitor charged by a 10-V battery, the voltage across the capacitor reaches 4.00 V in 3.00 seconds. The relevant equations include the charge equation q = εC(1-e(-t/RC)) and the voltage equation Vt = V0(1-e(-t/RC)). By substituting known values into the voltage equation, it simplifies to 0.6 = e(-3/RC), leading to the calculation of RC as approximately 5.9. Given the capacitance, R is calculated to be 590,000 ohms. The discussion emphasizes the importance of correctly using logarithmic and exponential functions in calculations.
twisted079
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Homework Statement



A 10.0 uF capacitor is charged by a 10-V battery through a resistance R. The capacitor reaches a potential difference of 4.00-V in a time interval of 3.00 s after charging begins. Find R.

Homework Equations



C = Q/ΔV
q = εC(1-e(-t/RC)

The Attempt at a Solution



Im not really sure where to begin. Q is not given. But I know I can relate capacitance and potential difference using C = Q/ΔV. I cannot find q using q = εC(1-e(-t/RC) because R is not given. I am missing a variable and I do not know how to solve since q nor Q were given. I follow how my teacher got the equation of q = εC(1-e(-t/RC) and in all cases 2 variables are unknown. I am assuming I should set up a system of 2 equations to account for the variables, but I do not know how or where to begin. Could someone please help?
 
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The equation for charge at time t is the same as the equation for V at time t q = εC(1-e(-t/RC)
Vt = V0 (1-e(-t/RC))
You know V0 = 10, Vt = 4 and t =3. So you should be able to calculate (RC)
You know C ...you should be OK
 
Here is what I did:

4 = 10(1-e(-3/RC))
6/10 = e(-3/RC))
Here is where I think I lose it

6/10 = -ln(3) -ln(RC)

My algebra is rusty :(
 
You are OK, just tricky maths.
4 = 10(1-e(-3/RC))
4/10 = 1-e(-3/RC)
0.4 = 1-e(-3/RC)
0.4 -1 = -e(-3/RC)
-0.6 = -e(-3/RC)
0.6 = e(-3/RC)
ln0.6 = -3/RC
-0.51 = -3/RC
RC = 3/0.51 = 5.9
So5.9 = RC... C = 10 x 10^-6F
Gives R =5.9/10 x 10^-6
R = 590,000 ohms
Check that you know how to use your calculator ln and e^x
 
Thank you so much! I knew I was using ln and e wrong (I was hoping it wouldn't come back to haunt me, but it found a way). Just so much to remember in so little time.
 
in my posts e(-3/RC) means e^(-3/RC).
I have not worked out how to post these numbers yet!
 
Right above where you enter these messages is a few options involving bold, italics, etc... if you look over more to the right you will see X2.

Or you can type [.SUP.] "what you want supscripted" [./SUP] without the "." in the brackets.
 
Brilliant... thank you
 
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