# Finding the time in which a potential drop is equal to 3V

1. Dec 4, 2012

### grace85233

1. The problem statement, all variables and given/known data
Capacitor, C1, is initially charged so that it has a potential difference of 12V. At time t=0, switch S1 is closed allowing the capacitor to discharge through resistor R1. At t=5, the potential across the capacitor has fallen to 6V. At what time will the potential across the capacitor reach 3V?

2. Relevant equations
V=Voe^(-t/RC)

3. The attempt at a solution
6=12e^(-5/RC)
The RC is confusing to me. I'm not sure how what I'm supposed to do with it.

2. Dec 4, 2012

### Mute

If 6=12e^(-5/RC), what does RC have to be to make that equation true?

3. Dec 4, 2012

### grace85233

It would equal 5ln(2), but I don't know how to use that for the answer. If I plug it in, there are lots of natural logs.

4. Dec 4, 2012

### Mute

Not quite - you made a division error. Double-check the time again. Also, you should keep track of the units.

Once you've found RC, it may involve a natural log, but sometimes things just involve natural logs. At any rate, it's just a number, so you can now solve for the time at which V = 3 volts, and then you can plug in the number for RC to get a numerical answer for the time at which V = 3 V. Does that make sense?

5. Dec 4, 2012

### grace85233

I went from
6=12e^-5(RC)
ln(1/2)=-5/RC
RC=7.21

Is that correct?

6. Dec 4, 2012

### Mute

Yes, that is correct. Note that that is 5/ln(2), whereas before you wrote 5*ln(2).

So RC = 7.21... what? What are the units? (Did your problem tell you the units? You didn't list any in your problem statement).

Anywho, now that you have RC = 7.21, can you solve for the time at which V = 3 volts?

Last edited: Dec 4, 2012
7. Dec 4, 2012

### grace85233

RC is in s.
So then it would just be
3=12e^(-t/7.21)
t=10s

8. Dec 4, 2012

### Mute

Yep, looks good.

9. Dec 4, 2012

Thank you!