Solve RC Circuit Problem: Calculate Charge on Capacitor

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Homework Help Overview

The discussion revolves around calculating the charge on a capacitor in an RC circuit, utilizing Kirchhoff's rule and understanding the relationships between voltage, charge, and resistance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the application of Kirchhoff's rule and the implications of current flow in a fully charged capacitor. Questions arise regarding the correct interpretation of time constants and unit conversions in the context of capacitor discharge.

Discussion Status

The discussion is active, with participants sharing their attempts to calculate the charge and time for discharge. Some guidance has been provided regarding the relationships between voltage, charge, and resistance, but there is still uncertainty about unit conversions and the correct application of formulas.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is a noted confusion regarding the units of capacitance and time in the calculations.

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Homework Statement


What is the charge on the capacitor?


Homework Equations


Kirchoff's rule
change in V =q/c


The Attempt at a Solution


I want to use Kirchoff rule but I don't know if this is the right way to go.
 

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When the capacitor is completely charges, there is no current in that branch. The voltage across the capacitor is the potential difference across 40 ohm resistor.
 
rl.bhat said:
When the capacitor is completely charges, there is no current in that branch. The voltage across the capacitor is the potential difference across 40 ohm resistor.

Ya, I was able to figure out that the charge on the resistor would be 80 microCoulombs . The voltage difference between the plates of the capacitor was 40 volts and from there it was plugging in. Now I face another problem where I can't find how long it will take the capacitor to discharge to 25% of 80 microCoulombs. I derived that it should be -RC*ln(1/4)=t. I plug in R= 50, c = 2 and get the wrong answer :/
 
hover said:
Ya, I was able to figure out that the charge on the resistor would be 80 microCoulombs . The voltage difference between the plates of the capacitor was 40 volts and from there it was plugging in. Now I face another problem where I can't find how long it will take the capacitor to discharge to 25% of 80 microCoulombs. I derived that it should be -RC*ln(1/4)=t. I plug in R= 50, c = 2 and get the wrong answer :/
t = 2 μF.
 
rl.bhat said:
t = 2 μF.

How does t = 2 MICROFARADS? Don't you mean seconds or milliseconds? I don't quite understand your logic though. Doesn't the capacitor discharge through 2 resistors that become 50 ohms?
 
hover said:
How does t = 2 MICROFARADS? Don't you mean seconds or milliseconds? I don't quite understand your logic though. Doesn't the capacitor discharge through 2 resistors that become 50 ohms?
Sorry.It is typo. I mean C = 2 μF. Your R is correct. You didn't mention your answer.
 
rl.bhat said:
Sorry.It is typo. I mean C = 2 μF. Your R is correct. You didn't mention your answer.

Well then it would be -RC*ln(1/4) = -(50)(2)*ln(1/4)= 138.63 ms but that is somehow wrong.. :/
 
Oh stupid me! I freakin' forgot to convert the units... it should be 138.63 microseconds or .1386 milliseconds.

Thanks for the help! :P
 

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