Physics Problem RC circuits?

Question

An RC circuit has one resistor R = 6.7kohms and one capacitor C = 6.0micro farads (F). The capacitor is at voltage V0 at t=0, when the switch is closed. How long does it take the capacitor to discharge to 1.0 percent of its initial voltage?

Help is appreciated.

Answer 1

I don't know why the other answers say the capacitor will charge when the switch closes. You have no voltage source to cause that. All RC charging/discharging is exponential. For this case you want a decreasing function. So that is:

V = V0 e^(-t/RC)

Set V = 0.01V0 and solve for t.

Answer 2

I think your question is wrongly worded. I think you mean how long will it take to become 99% charged. If the capacitor is discharging scroll down

The time constant (time to become 63% charged) for an RC circuit is T=RC so in this case T=6700 x 6E-6 =0.0402sec. (don't forget to include the prefixes in your calculation)

At t=n T (n time constants), V= V[n-1]+(.63*(100- V[n-1]))
At t=0 the capacitor voltage will be 0%
At t=1T the capacitor voltage will be 63%
At t=2T the capacitor voltage will be 86.5%
At t=3T the capacitor voltage will be 94.9%
At t=4T the capacitor voltage will be 98.1%
At t=5T the capacitor voltage will be 99.3%

So after 5 time constants the charge is 99% of total in this case T= 0.0402sec so V= 99% of max at 5*(0.0402) = 0.201sec

If the capacitor is discharging the time is the same as

The capacitor discharges by 63% of the remainder each time constant so

At t= nT (n time constants), V=0.37^n x Vo

At t=0, V= Vo
At t=1T, V=0.37Vo
At t=2T, V=0.136Vo
At t=3T, V=0.050Vo
At t=4T, V=0.018Vo
At t=5T, V=0.007Vo

Answer 3

The voltage will discharges and decreases at an exponential rate.
0.1*Vo=Vo*exp(-t/RC).
which can be simplified to 0.1=exp(-t/RC)

Just solve for this equation.

Answer 4

The EE is correct. Use the discharge formula