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- #2

rl.bhat

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- #3

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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 :/

- #4

rl.bhat

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t = 2 μF.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 :/

- #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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Sorry.It is typo. I mean C = 2 μF. Your R is correct. You didn't mention your answer.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?

- #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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Thanks for the help! :P

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