# An RC Circuit

1. Jul 17, 2009

### peaceandlove

1. The problem statement, all variables and given/known data
A 3 µF capacitor is charged to 76 V and is then connected across a 698 Ω resistor. (a) What is the initial charge on the capacitor? Answer in units of µC. (b) What is the initial current just after the capacitor is connected to the resistor? Answer in units of A. (c) What is the time constant of this circuit? Answer in units of ms. (d) How much charge is on the capacitor after 5.122 ms? Answer in units of µC.

2. Relevant equations
1) Q= CV
2) I= V/R
3) (TC) = RC
4) V = (V_0)(1-(e^(-t/RC)))

3. The attempt at a solution
a) Q=CV; Q=3 uF * 76 V; Q=228 uC
b) I = V/R; I=76 V/ 698 Ohm; I=0.109 A
c) TimeConstant TC=RC; TC=698 * 3 uF; TC=2.094 mSec

However, I can't figure out (d). I used equation 4 to figure out the voltage at 5.122 ms (=0.005122 s) and then plugged in the numbers to equation 1. I got 227.44298 µC, but that value is incorrect.

2. Jul 17, 2009

### Staff: Mentor

I think your equation in #4 may not be the right form. The "1-e" form is for when you have a voltage rising to some value. If it's starting at 76V and falling to zero, what form should you use...?

3. Jul 17, 2009

### Staff: Mentor

Another sanity check is that when you are out a couple time constants (5.122ms versus the 2.094ms time constant), the voltage on the cap should be pretty low...

4. Jul 17, 2009

### peaceandlove

Oh... so if the capacitor is discharging I would use the equation: V = (V_0)*(e^(-t/RC)). So then V = 6.58435. Plugging that into equation 1, I would then get 1.9753*10^-5 C (=19.753 microC).

5. Jul 17, 2009

### Staff: Mentor

Right and almost right. The units of the answer are uC, so 19.75uC.