# RC Circuits charge homework

1. Mar 23, 2008

### hokie1020

Strictly speaking, the equation q=Q_0e^{-t/RC} implies that an infinite amount of time is required to discharge a capacitor completely. Yet for practical purposes, a capacitor may be considered to be fully discharged after a finite length of time. To be specific, consider a capacitor with capacitance C connected to a resistor R to be fully discharged if its charge q differs from zero by no more than the charge of one electron.

I could do everything except explaining why Part C's answer is correct
Part A
Calculate the time required to reach this state if C = 0.910 microF, R = 690 kilo ohms, and Q_0 = 6.80 microC.
t=19.7s
Part B
How many time constants is this?
31.4 time constants
Part C
For a given Q_0, is the time required to reach this state always the same number of time constants, independent of the values of C and R?
yes
Part D
Why or why not? (in response to Part C)

Part D is what i cant explain. Can anyone help?

2. Mar 23, 2008

### dynamicsolo

If you rearrange the charge equation (which is the solution to the differential equation produced by Kirchkoff's circuit rules) for the situation where the capacitor discharges to q(T) = e (one fundamental charge), we have

e/Q_0 = e^(-T/RC) .

I think what they're saying is this: for any RC circuit with varying choices of R and C, and starting with a fully-charged capacitor, the ratio of this defined discharge time to the circuit time constant is simply related to the ratio of e/Q_0 .
[EDIT: Re-reading the question once again, I agree with their answer; the equation above shows why. Try it with numbers to satisfy yourself -- I had to...]

Last edited: Mar 23, 2008