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

In summary, a capacitor is charged by a battery through a resistance. The capacitor reaches a potential difference of 4.00-V in a time interval of 3.00 s after charging begins. Find R.
  • #1
twisted079
25
1

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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  • #2
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
 
  • #3
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 :(
 
  • #4
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
 
  • #5
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.
 
  • #6
in my posts e(-3/RC) means e^(-3/RC).
I have not worked out how to post these numbers yet!
 
  • #7
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.
 
  • #8
Brilliant... thank you
 

1. How do I find the resistance in a RC circuit?

The resistance in a RC circuit can be found using the formula R = t/ (C ln (V0/Vt)), where t is the time, C is the capacitance, V0 is the initial voltage, and Vt is the voltage at time t.

2. Can I use any battery for this calculation?

Yes, as long as the battery provides a constant voltage throughout the circuit. However, it is recommended to use a battery with a known voltage for more accurate results.

3. What if I don't know the capacitance of the circuit?

If you don't know the capacitance, you can measure it using a capacitance meter or calculate it using the formula C = Q/V, where Q is the charge on the capacitor and V is the voltage across the capacitor.

4. Is there an easier way to find the resistance in a RC circuit?

Yes, you can use a multimeter to directly measure the resistance in the circuit. Simply set the multimeter to the resistance mode and connect it across the circuit to get the resistance value.

5. Can I use this formula for any type of RC circuit?

Yes, this formula can be used for any type of RC circuit, as long as the voltage is constant and there are no other components affecting the circuit.

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