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

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Homework Help Overview

The discussion revolves around a capacitor discharging through a resistor, with a focus on determining the time at which the potential across the capacitor reaches 3V after initially being charged to 12V. The problem involves concepts from electrical circuits and exponential decay.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between voltage, time, and the RC time constant, with some expressing confusion about how to manipulate the equation involving natural logarithms. There are attempts to derive the value of RC and its implications for finding the time when the voltage reaches 3V.

Discussion Status

The discussion is progressing with participants verifying calculations and clarifying the units of RC. Some have provided guidance on how to approach the problem, while others are still working through the implications of their findings.

Contextual Notes

There is a mention of potential confusion regarding the units of RC and the need to ensure they are consistent with the problem statement. Participants are also checking for accuracy in their calculations and assumptions about the exponential decay model.

grace85233
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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.
 
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grace85233 said:

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?
 
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.
 
grace85233 said:
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?
 
I went from
6=12e^-5(RC)
ln(1/2)=-5/RC
RC=7.21

Is that correct?
 
grace85233 said:
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?
 
Last edited:
RC is in s.
So then it would just be
3=12e^(-t/7.21)
t=10s
 
grace85233 said:
RC is in s.
So then it would just be
3=12e^(-t/7.21)
t=10s

Yep, looks good.
 
Thank you!
 

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