Capacitor problem

KFS

New member
Joined
Jun 15, 2018
Messages
34
The problem says: The equation R(dQ/dt)+Q/C=E describes the charge Q on a capacitor, where R, C, and E are constants. (a) Find Q as a function of time if Q=0 at t=0. (b) How long does it take for Q to attain 99% of its limiting charge?
I solved (a) and I got Q=-exp(-t/RC+ln(EC))+EC. But I don't know how to solve (b), I don't know what to equal Q to solve for t. I tried making Q=-exp(-t/RC+ln(EC))+EC=EC but I can't take logarithms to find t.
 
I tried making Q=-exp(-t/RC+ln(EC))+EC=EC but I can't take logarithms to find t.

It would be easier to provide help if you can flesh out your working a little (you can post a photo of your work on paper if you don't know LaTeX). I assume that you're saying...

[math]\lim_{t\to \infty}Q(t) [/math]
[math]= \lim_{t\to \infty}\left(-e^{-\left(\frac{t}{RC}+ln(EC)\right)}+EC\right) [/math]
[math]= EC[/math]
...I agree with the above work, but I don't understand where you are stuck, can you please explain? Remember to use the "99%" in the question.
 
You did not say how you got q as a function of t. I guess you did

[MATH]r * \dfrac{dq}{dt} + \dfrac{q}{c} = E \implies[/MATH]
[MATH]\dfrac{dq}{dt} = \dfrac{cE - q}{cr} \implies[/MATH]
[MATH]\dfrac{dq}{cE - q} = \dfrac{1}{cr} * dt \implies[/MATH]
[MATH]- ln(cE - q) = \dfrac{t}{cr} + \lambda.[/MATH]
[MATH]\text {Given } q = 0 \text { if } t = 0 \implies \lambda = - ln(cE).[/MATH]
[MATH]\therefore - ln(cE - q) = \dfrac{t}{cr} - ln(cE).[/MATH]
Now from there you CAN solve for q by going

[MATH]ln(cE - q) = - \dfrac{t}{cr} + ln(cE) \implies cE - q = exp \left ( - \dfrac{t}{cr} + ln(cE) \right ) \implies[/MATH]
[MATH]q = - exp \left ( - \dfrac{t}{cr} + ln(cE) \right ) + cE.[/MATH]
I personally would not do it that way, but it works. And I certainly would not try to find a formula for q and then find a formula for t out of that quite cumbersome formula for q. Instead, how about

[MATH]- ln(cE - q) = \dfrac{t}{cr} - ln(cE) \implies t = cr * ln \left ( \dfrac{cE}{cE - q} \right ).[/MATH]
 
As Cubist told you, the "limiting charge" is Q= EC. 99% of that is 0.99EC. You want to solve \(\displaystyle Q=-exp(-t/RC+ln(EC))+EC= 0.99EC\). That is, of course, the same as \(\displaystyle exp(-t/RC+ln(EC))= 0.01EC\) and you certainly can take the logarithm of that!
 
Top