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

AI Thread Summary
The discussion revolves around calculating the capacitance of a charging RC circuit where a capacitor's voltage rises from 10% to 90% over 15 seconds with a resistor of 20kΩ. The voltage across the capacitor is modeled using the equation Vc(t) = Vs(1-e^(-t/τ)), where τ is the time constant (RC). The initial calculations led to a capacitance of approximately 0.00326 F, but the participant expressed uncertainty about the effect of the initial voltage at t=0. A response highlighted the need for additional equations to account for the two voltage points over time, suggesting a more complex approach to solve for both time and capacitance. The conversation emphasizes the importance of correctly framing the problem to derive accurate results.
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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