Question

A RC circuit containing a battery, a capacitor and a resistor:
Given that R*dQ/dt + Q/C = V
Solve the differential equation for Q(t).
A little help here?

Answer 1

\[R * dQ/dt +Q/ C = V \]
You may or may not recognize this as a first order, linear differential equation,
http://www.sosmath.com/diffeq/first/lineareq/lineareq.html

Answer 2

Gonna try to solve it, one moment.

Answer 3

Okay!

Answer 4

Yeah One Moment Up d00d, Your time is Up ! :P :D

Answer 5

I've just started studying ODEs, don't be hard on me :(

Answer 6

Hey I was pulling your leg d00d, try it. Soon you become expert at it! Good luck!

Answer 7

All right, here's what I did.
I separated the variables and got\[\frac{ R*dQ }{ dt } = V - \frac{ Q }{ C } \leftarrow \rightarrow \frac{ dQ }{ V - C } = \frac{ dt }{ R }\]
Integrating both sides I got\[\ln |V-\frac{ Q }{ C }| = \frac{ t }{ -c*R } + c\]
I guess it's pretty easy noe :P

Answer 8

How did u get that?

Answer 9

Just Ignore that "-c"

Answer 10

\[ \frac{ R*dQ }{ dt } +\frac{ Q }{C }=V\]
\[ \frac{ R*dQ }{ dt }= V-\frac{ Q }{C }\]
\[ \frac{ R*dQ }{ dt }= \frac{ VC-Q }{C }\]
\[ \frac{ dQ }{ VC-Q }= \frac{ dt }{CR }\]

Answer 11

Oh! Noticed what went wrong... damn, pretty dumbish error lol

Answer 12

Hmm! it happens sometimes, never mind it!

Answer 13

Thanks for the help! I've reached the answer.

Answer 14

If you are really interested in learning ODE's I suggest this video
http://www.youtube.com/watch?v=CogfMjKUGc0
[ the video also contains some programming in between, never mind it , just concentrate on theory he is saying]

Answer 15

That will help, thanks.

Answer 16

My pleasure
But be patient while hearing that lecture, coz that lecture might be too long for you!