1. ## physics related calc

Well I am once again at a stand still. Its basically a simple circuit problem, but not so simple to me. Heres where im at:

e^(t/RC) q = CV e^(t/RC) + K

All letters in capital are constant. So what i need to do from here is show that if q(0) = 0, then K = -CV.

It is probably easy but im new to the natural base honestly so if someone could explain the process a little bit, id really appreciate it.
This is the 1st of 2 parts
Thanks guys
Chris

2. Originally Posted by chrsr345
Well I am once again at a stand still. Its basically a simple circuit problem, but not so simple to me. Heres where im at:

e^(t/RC) q = CV e^(t/RC) + K

All letters in capital are constant. So what i need to do from here is show that if q(0) = 0, then K = -CV.

It is probably easy but im new to the natural base honestly so if someone could explain the process a little bit, id really appreciate it.
This is the 1st of 2 parts
Thanks guys
Chris
begin by dividing through by $e^{t/RC}$:

$\Rightarrow q(t) = CV + Ke^{-t/RC}$

now plug in t = 0 and q(0) = 0 and solve for K

3. Boy i feel silly, thanks Jhevon.
Welp from there i had to derive:

CV(1 - e^(-t/RC))

What id did was plug in -K and executed the product rule. From there i used the quotient rule on the riased power of (-t/RC) to get t/RC.

My final derivative was dq/dt = K(e^(-t/RC) t/RC)

Did i execute this correctly?

4. Originally Posted by chrsr345
Boy i feel silly, thanks Jhevon.
Welp from there i had to derive:

CV(-1 + e^(1-t/RC)
check this to make sure you typed the equation correctly. you are missing parentheses and i don't know what else.

What id did was plug in -K and executed the product rule. From there i used the quotient rule on the riased power of (-t/RC) to get t/RC.

My final derivative was dq/dt = K(e^(1-t/RC) t/RC)

Did i execute this correctly?
why would you do that. we went through all the trouble of finding out that K = -CV. shouldn't we plug that in for K and see what we get?

5. Ahh sorry, i corrected my last post. both equations were written wrong.

But i plugged K back in so i would have 1 constant to work with for the product rule rather than 2 constants. But maybe should plug the -CV for K at the end for further simplification?

dq/dt = K(e^(-t/RC) t/RC) --------> dq/dt = -V(e^(-t/RC) t/R)

what do you think?

6. Originally Posted by chrsr345
Ahh sorry, i corrected my last post. both equations were written wrong.

But i plugged K back in so i would have 1 constant to work with for the product rule rather than 2 constants. But maybe should plug the -CV for K at the end for further simplification?

dq/dt = K(e^(-t/RC) t/RC) --------> dq/dt = -V(e^(-t/RC) t/R)

what do you think?
i think it's wrong.

$q(t) = CV - CVe^{-\frac t{RC}}$

$\Rightarrow \frac {dq}{dt} = \frac {CV}{RC}e^{- \frac t{RC}} = \frac {V}{R}e^{- \frac t{RC}}$

7. Hmm, yes your right...i think i actually took a tougher way which apparently is the wrong way. I see what you did.
So for t > 0, dq/dt is getting larger and more negative?

8. Originally Posted by chrsr345
Hmm, yes your right...i think i actually took a tougher way which apparently is the wrong way. I see what you did.
So for t > 0, dq/dt is getting larger and more negative?
i don't know. it depends on what signs R and C have. (and to some extent the sign of V matters as well, not as t goes to infinity though)

9. double post, woops

10. Hmm, or maybe its possible to figure out from the q(t)

$q(t) = CV - CVe^{-\frac t{RC}}$

Where C is capacitance, V is voltage and R is resistance. All these values should be positive i believe. The dq/dt is actually the formala for i(t) which is the current at time (t). Does this make any more sense to you? Im certainly no electrical engineer..

11. Originally Posted by chrsr345
Hmm, or maybe its possible to figure out from the q(t)

$q(t) = CV - CVe^{-\frac t{RC}}$

Where C is capacitance, V is voltage and R is resistance. All these values should be positive i believe. The dq/dt is actually the formala for i(t) which is the current at time (t). Does this make any more sense to you? Im certainly no electrical engineer..
if all the variables are positive, then dq/dt gets small as t gets large, since e raised to a negative number goes to zero as the negative number gets more negative

12. Thank you again Dan. I didnt have enough time last night to thoroughly try out one of your answers. I tried it now, and actually i dont see the process..
What did you do to get
$\Rightarrow \frac {dq}{dt} = \frac {CV}{RC}e^{- \frac t{RC}} = \frac {V}{R}e^{- \frac t{RC}}$

Im deriving the 1st CV as 1, then the rest as a prodect rule but not getting the above answer...

13. Originally Posted by chrsr345
Thank you again Dan.
Ummm... No problem. But Jhevon's helping you on this one.

-Dan

14. Originally Posted by topsquark
Ummm... No problem. But Jhevon's helping you on this one.

-Dan
...well, i guess i should be flattered if someone is mistaking me for you... it must mean i'm doing something right, after all

15. Originally Posted by chrsr345
Thank you again Dan. I didnt have enough time last night to thoroughly try out one of your answers. I tried it now, and actually i dont see the process..
What did you do to get
$\Rightarrow \frac {dq}{dt} = \frac {CV}{RC}e^{- \frac t{RC}} = \frac {V}{R}e^{- \frac t{RC}}$

Im deriving the 1st CV as 1, then the rest as a prodect rule but not getting the above answer...
t is the variable, so CV is a constant, it's derivative is zero

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