Finding the time for charging Voltage

[Mohamed Abdul]

Homework Statement

1. The characteristic decay constant for a capacitor C discharging through a resistor R is the product of R times C. If the initial voltage at time t = 0 is Vo = 13Volts and what is the voltage at time t = 2RC?

2. A capacitor C = 9microFarads is charged from 0Volts to 7Volts through a resistor R = 39kOhms. How much time will it take to charge to 59% of full voltage? Express your answer in milliseconds.

Homework Equations

Charging => V = Vo(1-e^-t/Tau)
Discharge => V = Vo(e^-t/Tau)

Tau = RC

The Attempt at a Solution

For the first question I plugged in 2RC for time into the discharge equation, and canceled the RC with Tau to get 13*(e^-2) = 1.76

For the second question I'm thinking of plugging the same variables, but am unsure of what my initial and final voltages are. I thought they could be 0 and 7, but if that was the case V would equal 0 because Vo would equal 0, which I don't think is the case.

 

[TSny]

Homework Equations

Charging => V = Vo(1-e^-t/Tau)
Discharge => V = Vo(e^-t/Tau)

Tau = RC

The Attempt at a Solution

For the first question I plugged in 2RC for time into the discharge equation, and canceled the RC with Tau to get 13*(e^-2) = 1.76

OK. Don't forget the units.

For the second question I'm thinking of plugging the same variables, but am unsure of what my initial and final voltages are. I thought they could be 0 and 7, but if that was the case V would equal 0 because Vo would equal 0, which I don't think is the case.

For the second question use the discharging charging equation listed in your Relevant Equations section. But note that $V_0$ in that equation is not the initial potential. Consult your textbook or notes.

 

[Mohamed Abdul]

OK. Don't forget the units.

For the second question use the discharging equation listed in your Relevant Equations section. But note that $V_0$ in that equation is not the initial potential. Consult your textbook or notes.

I'm confused, what else would V0 be besides the initial voltage? My notes say the same as well.

 

[TSny]

I'm confused, what else would V0 be besides the initial voltage? My notes say the same as well.

Sorry, I meant to say use the formula for charging the capacitor for the second problem. In this equation $V_0$ is not the initial voltage. This cannot be zero since the equation would not make sense. As you noted, the initial voltage is zero. If your notes are saying $V_0$ is the initial voltage in the charging equation, then your notes are mistaken. Do a quick web search for "charging a capacitor".

Or, you can think about what the equation reduces to in the limit of t going to infinity.

 

[Mohamed Abdul]

Sorry, I meant to say use the formula for charging the capacitor for the second problem. In this equation $V_0$ is not the initial voltage. This cannot be zero since the equation would not make sense. As you noted, the initial voltage is zero. If your notes are saying $V_0$ is the initial voltage in the charging equation, then your notes are mistaken. Do a quick web search for "charging a capacitor".

Or, you can think about what the equation reduces to in the limit of t going to infinity.

In that case would V0 be my max voltage, or the 59% of the full voltage?

 

[TSny]

In that case would V0 be my max voltage, or the 59% of the full voltage?

Your charging equation is written as $V = V_0 \left( 1- e^{-t / \tau }\right)$.

What is the limit of the right side for $t$ → $\infty$? This should tell you the meaning of $V_0$.

 

[Mohamed Abdul]

Your charging equation is written as $V = V_0 \left( 1- e^{-t / \tau }\right)$.

What is the limit of the right side for $t$ → $\infty$? This should tell you the meaning of $V_0$.

So that would mean when t approaches infinity it would be near zero, so V would equal Vo, right?

 

[TSny]

So that would mean when t approaches infinity it would be near zero, so V would equal Vo, right?

Yes, $e^{-t/\tau}$ approaches zero as t → ∞. So, yes, V approaches V₀ as t → ∞.

 

[Mohamed Abdul]

Yes, $e^{-t/\tau}$ approaches zero as t → ∞. So, yes, V approaches V₀ as t → ∞.

Doing that I set
.59(7) = 7(1-e^(-t/.351))
.59 = 1- e^-t/.351
.41 = e^-t/.351
ln(.41) = -t/.351
t=312.95 milliseconds

Is that the right line of thought for that, in terms of where I put my max voltage and where I put the 59% voltage?

 

[TSny]

Doing that I set
.59(7) = 7(1-e^(-t/.351))
.59 = 1- e^-t/.351
.41 = e^-t/.351
ln(.41) = -t/.351
t=312.95 milliseconds

Is that the right line of thought for that, in terms of where I put my max voltage and where I put the 59% voltage?

Yes, that looks good.