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

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Homework Help Overview

The problem involves a 10.0 µF capacitor being charged by a 10-V battery through a resistor R, with the capacitor reaching a potential difference of 4.00 V after 3.00 seconds. The objective is to find the resistance R.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster expresses uncertainty about how to start, noting the lack of a given charge Q and questioning how to relate capacitance and potential difference. Some participants suggest using the equations for charge and voltage over time, while others attempt to manipulate these equations to isolate variables.

Discussion Status

Participants are actively engaging with the mathematical relationships involved, with some providing algebraic manipulations and others confirming the correctness of these steps. There is a mix of attempts to clarify the use of logarithmic functions and exponential decay in the context of the problem.

Contextual Notes

There is a noted concern about the original poster's algebra skills and the challenge of remembering mathematical concepts under pressure. The discussion also includes references to formatting issues in posting equations.

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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