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Finding rate of change of capacitor discharge?

travispickle

May 31, 2015
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A 22kilaohm resistor is connected in series with a 100nF capacitor under a 5v supply voltage.

Vc = V (1 - e^(-t/T))

The capacitor is fully charged to the supply voltage, use the discharging equation to show the value of the capacitor voltage as a percentage of voltage V after it has discharged for t = T (one time constant?)

What I have done so far:

Vc / V = 1 - e^(-t/T)

Vc = 1 - e^-1 = 1-(1/e)

100 x (1-(1/e)) = 63.2%

Im pretty sure this isnt meeting the questions criteria, as im not including any figures for capacitance, voltage, and resistance, but its the only way I can think to do this? (this is a smaller part of a much larger question)

Do I need to use T=CR somewhere? multiply the capacitance and resistance, and use this figure in place of T to give a different percentage?

Any help would be extremely helpful! thankyou!!
 

LvW

Apr 12, 2014
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Why do you think that you can use the same equation as for the charging process?
 

Arouse1973

Adam
Dec 18, 2013
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Welcome to EP Travis. Well I am sorry to say..........your absolutely right :) Because the time your interested in equals Tau (R*C) then this is just 1-e^(-1) = 0.632 this is because RC/RC = 1. See (-t) is the time your interested in, as in after a certain amount of time the voltage will be.......... and Tau is you RC. So you see you have taken into account R and C because it's Tau and the input voltage (5 V) is what you multiply by to get the fraction of voltage on the capacitor after (-t).

"Do I need to use T=CR somewhere" not when (-t) = Tau because you always get 1. You do need to work out Tau when (-t) and Tau are not equal, which as you know is just R*C.

Thanks
Adam
 

LvW

Apr 12, 2014
604
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Why do you think that you can use the same equation as for the charging process?
Ask yourself what the output - according to the following equation - will be for a discharging time t approaching infinite. Is the result OK?

Vc / V = 1 - e^(-t/T)
 

Ratch

Mar 10, 2013
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A 22kilaohm resistor is connected in series with a 100nF capacitor under a 5v supply voltage.

Vc = V (1 - e^(-t/T))

The capacitor is fully charged to the supply voltage, use the discharging equation to show the value of the capacitor voltage as a percentage of voltage V after it has discharged for t = T (one time constant?)

What I have done so far:

Vc / V = 1 - e^(-t/T)

Vc = 1 - e^-1 = 1-(1/e)

100 x (1-(1/e)) = 63.2%

Im pretty sure this isnt meeting the questions criteria, as im not including any figures for capacitance, voltage, and resistance, but its the only way I can think to do this? (this is a smaller part of a much larger question)

Do I need to use T=CR somewhere? multiply the capacitance and resistance, and use this figure in place of T to give a different percentage?

Any help would be extremely helpful! thankyou!!

To further your understanding, note that there are an infinite number of R and C combinations that give the same time constant RC. As long as the time constant is the same, the capacitor will energize and de-energize to the same percentage at the same time. Can you explain what the difference would be in the energize/de-energize event if the resistance were high and the capacitance were low and vice versa, assuming the RC time constant stayed the same?

Like LvW said, you should check the de-energizing equation you are using.

Ratch
 

travispickle

May 31, 2015
2
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May 31, 2015
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I might need all this in simpler terms, or I should have added this is a maths question with an electronics twist (this is more a long calculus math question) what was I right with?

Are you saying 63.2% as a charge time after one time constant is universal? A change in capacitor voltage and resistor value wouldnt effect what I'm asking as the time constant is 1?

Thanks!
 

LvW

Apr 12, 2014
604
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I might need all this in simpler terms, or I should have added this is a maths question with an electronics twist (this is more a long calculus math question) what was I right with?

Are you saying 63.2% as a charge time after one time constant is universal?

Yes - each R-C lowpass charges UP to 63.2% of the final value within a time t=T=RC.
However, you were asking for a discharging process!
It is even possible (without calculating) to make a good estimate of the result.
 

Arouse1973

Adam
Dec 18, 2013
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I might need all this in simpler terms, or I should have added this is a maths question with an electronics twist (this is more a long calculus math question) what was I right with?

Are you saying 63.2% as a charge time after one time constant is universal? A change in capacitor voltage and resistor value wouldnt effect what I'm asking as the time constant is 1?

Thanks!

You have to remember that because it is discharging you are loosing 63.2% of the charge on the capacitor and your left with 36.8%.
Thanks
Adam
 
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