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

[grace85233]

Homework Statement

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?

Homework Equations

V=Voe^(-t/RC)

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.

 

[Mute]

Homework Statement

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?

Homework Equations

V=Voe^(-t/RC)

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.

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

 

[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.

 

[Mute]

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.

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?

 

[grace85233]

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

Is that correct?

 

[Mute]

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

Is that correct?

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?

 

[grace85233]

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

 

[Mute]

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

Yep, looks good.

 

[grace85233]

Thank you!