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RC circuit problem

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  • #1
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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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  • #2
rl.bhat
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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.
 
  • #3
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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.
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 :/
 
  • #4
rl.bhat
Homework Helper
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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.
 
  • #5
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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?
 
  • #6
rl.bhat
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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.
 
  • #7
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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.. :/
 
  • #8
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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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