What is the Capacitance of a Charging RC Circuit with Known Time and Resistance?

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SUMMARY

The capacitance of a charging RC circuit can be calculated using the formula Vc(t) = Vs(1-e^(-t/τ)), where τ = RC. In this discussion, a resistor of 20kΩ and a time interval of 15 seconds were provided, with the capacitor voltage rising from 10% to 90% of its final value. The initial calculation yielded a capacitance of approximately 0.000326 F, but the analysis revealed the need for additional equations to account for the absolute timing of the voltage measurements. The correct approach involves setting up two equations based on the voltage at the start and end of the interval.

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Kashuno
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Homework Statement


"A capacitor is being charged from a battery and through a resistor of 20kΩ. It is observed that the voltage on the capacitor rises from 10% of its final value to 90%, in 15 seconds. Calculate the capacitor's capacitance.

t = 15s
R = 20kΩ
Vi = 10% of max
Vf = 90% of max

Homework Equations


Vc(t) = Vs(1-e^(-t/τ))
τ = RC

The Attempt at a Solution


Vc(t) = 90% = .9
Vs = 100% = 1.0
.9 = 1.0(1-e^(-t/τ))
.9 = 1-e^(-t/τ)
e^(-t/τ) = .1

From there it has been a while since I took calculus, but I continued with

ln(e^-t/τ) = ln(.1)
-t/τ = ln(.1)
-15/20000C = ln(.1)
-.00075 = ln(.1)C
C = .000326 F

I got an answer, but I feel like I am missing something to do with the voltage on the capacitor at t = 0 being 10%. I have no way to check if this answer is correct, so I guess I'm really just asking for someone to check my work (and if it's wrong, give me a push in the right direction)
 
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Kashuno said:

Homework Statement


"A capacitor is being charged from a battery and through a resistor of 20kΩ. It is observed that the voltage on the capacitor rises from 10% of its final value to 90%, in 15 seconds. Calculate the capacitor's capacitance.

t = 15s
R = 20kΩ
Vi = 10% of max
Vf = 90% of max

Homework Equations


Vc(t) = Vs(1-e^(-t/τ))
τ = RC

The Attempt at a Solution


Vc(t) = 90% = .9
Vs = 100% = 1.0
.9 = 1.0(1-e^(-t/τ))
.9 = 1-e^(-t/τ)
e^(-t/τ) = .1

From there it has been a while since I took calculus, but I continued with

ln(e^-t/τ) = ln(.1)
-t/τ = ln(.1)
-15/2000C = ln(.1)
-.0075 = ln(.1)C
C = .00326 F

I got an answer, but I feel like I am missing something to do with the voltage on the capacitor at t = 0 being 10%. I have no way to check if this answer is correct, so I guess I'm really just asking for someone to check my work (and if it's wrong, give me a push in the right direction)

Hi Kashuno, Welcome to Physics Forums.

Yes, something's missing. You're given the time between two points on the charging curve, but you're not given an absolute time for either point. In other words, you can't fix either the start or the end of the interval in question without doing a bit of fancy footwork.

Why don't you write expressions for both of the given amplitudes, letting the first occur at time t and the second at time t + 15s. How many variables and equations will that leave you with?
 
Wow! I can't believe I didn't see that! I was having a hard time finding an equation that incorporated percentages that I didn't even think of multiple equations. :rolleyes:

Set it up and get two equations with two unknowns, solve and substitute, done. Thank you so much!
 

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