Solve Physics Prob: Q, I, TC & Charge on C after 5.122ms

[peaceandlove]

Homework Statement

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.

Homework Equations

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

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.

 

[berkeman]

Homework Statement

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.

Homework Equations

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

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.

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...?

 

[berkeman]

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...

 

[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).

 

[berkeman]

Oh... so if the capacitor is discharging I would use the equation: V = (V_0)*(e^(-t/RC)). Is that the equation you're talking about? If that is, then V = 6.58435. Plugging that into equation 1, I would then get 1.9753*10^-5 C (=19.753). Right?

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