Finding the Resistance in an RC Circuit

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Homework Help Overview

The problem involves an RC circuit where a capacitor is charged by a battery. The original poster seeks to determine the resistance value based on the capacitor's voltage after a specific time period.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster calculates the charge on the capacitor at 4 volts but expresses uncertainty about the timing of this voltage. They consider using multiple equations to solve for unknowns but encounter difficulties.
  • Some participants clarify that the time at which the capacitor reaches 4 volts is indeed 3 seconds, as stated in the problem.

Discussion Status

Participants are actively engaging with the problem, with some providing clarifications on the timing aspect. There is a recognition of the original poster's confusion regarding the approach, but no consensus or resolution has been reached yet.

Contextual Notes

The original poster mentions having incomplete information, which affects their ability to solve the problem effectively.

hamhamt
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Homework Statement


A 10-µF capacitor is charged by a 10-V battery through a resistance R. The capacitor reaches a potential difference of 4 V in a period of 3 s after the charging began. Find the value of R.

Homework Equations



c=q/v
q=CE(1-e^-t/RC)
i=(E/R)e^(-t/RC)

The Attempt at a Solution



I found the charge on the capacitor when its at 4 volts and got 4.0 * 10^-5 C. I then tried to plug this into the equation for a capacitor charging, but I don't know the time at which the capactior is at 4 volts.

At this point, I attempted finding another equation, so I could attempt a system of 2 equations with 2 unknowns. I tried to use the current in a capactor charging equation, but I ended up getting infinitely many solutions.

I am not entirely sure how to approach this problem.
 
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hamhamt said:
I don't know the time at which the capactior is at 4 volts.

I thought...

hamhamt said:
The capacitor reaches a potential difference of 4 V in a period of 3 s after the charging began.
 
Use c = q/v to find the charge. Then plug that into the second equation at t = 3 s, because the capacitor begins charging at t = 0. So since it reaches 4 V after 3 s of charging, t = 3 s when the capacitor is charged to 4 V.
 
Jesus. I had half of the question cut off at the most important part. Thank you both lol
 

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