Finding the Resistance in an RC Circuit

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SUMMARY

The discussion focuses on calculating the resistance (R) in an RC circuit where a 10-µF capacitor is charged by a 10-V battery. The capacitor reaches a potential difference of 4 V after 3 seconds. The relevant equations used include q = CE(1 - e^(-t/RC)) and i = (E/R)e^(-t/RC). The key conclusion is that the time at which the capacitor reaches 4 V is indeed 3 seconds, allowing for the calculation of resistance using the charge derived from the voltage and capacitance.

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  • Understanding of RC circuits and their components
  • Familiarity with capacitor charging equations
  • Knowledge of exponential decay in electrical circuits
  • Basic algebra for solving equations
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  • Learn about the time constant (τ = RC) in RC circuits
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Students studying electrical engineering, educators teaching circuit theory, and hobbyists working with capacitors and resistors in electronic projects.

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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