# First Order RC Circuits

Discussion in 'Homework Help' started by LightAce, Sep 19, 2010.

1. ### LightAce Thread Starter New Member

Sep 19, 2010
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Question:If a 100μF capacitor is initially charged to 100 V and it is desired that 90% of that voltage remain stored after the source of that charge has been removed for one minute, what is the maximum value for the leakage resistance that can be tolerated?

So I'm assuming your trying to find what resistor would allow the capacitor to have 90 V by the end of one minute. I'm really not sure what to do here or what formulas to use, so if anyone can explain in detail on how to get started with this, it would be much appreciated.

2. ### Georacer Moderator

Nov 25, 2009
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1,281
Here we have, as you already said, a first order RC circuit. It involves a capacitor and a resistance connected in a loop. The time constant of this circuit is T=RC, where R the resistance and C the capacitance.

The voltage of the capacitor as a function of time is described by the equation $V_c(t)=V_0 (1-e^{-\frac{t}{RC}})$

In order to solve your problem, substitute time for 60 seconds, the capacitance, $V_c$ with the desired final voltage and $V_0$ with the initial voltage, and solve for R.

Any questions?

Refer to "RC circuits" in Wikipedia for more reading.

3. ### LightAce Thread Starter New Member

Sep 19, 2010
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Thank you so much, that was more than clear enough for me to understand.

Apr 14, 2005
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5. ### Georacer Moderator

Nov 25, 2009
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Yes, it seems I did. The correct one is $V_c(t)=V_0 e^{-\frac{t}{RC}}$

I never learned them by heart. Thanks for the correction!

6. ### Ghar Active Member

Mar 8, 2010
655
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Yeah I never get them right on the first try either... you need to plug in t = 0 and see what it gives you, nice fast and simple check.

7. ### Ron H AAC Fanatic!

Apr 14, 2005
7,012
681
Same here. That's how I discovered the error.
Here is the equation that always works:

$V(t)=V_f+(V_i-V_f)*e^{\frac{-t}{R*C}}$
Where

$V(t)$is the instantaneous voltage across the capacitor,

$V_i$ is the initial voltage across the capacitor,

and $V_f$ is the final (target) voltage.

I memorized it in college, about 50 years ago, when capacitors were Leyden jars and resistors were all wirewound.

8. ### LightAce Thread Starter New Member

Sep 19, 2010
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Glad this assignment isn't due till Thursday. Thanks for noticing the error but is it possible that the resistance could be as high as 5692K ohms, because that is what I had gotten for the resistance

Last edited: Sep 21, 2010