Question

Hi people, what is the solution of q capacitor diff equation : dq/dt+1/RC * q = 0

Answer 1

Solve this:\[
y' + c y = 0
\]

Answer 2

Let \(c= 1/RC\)

Answer 3

This is an easy diff eq.

Answer 4

\[
q'=-cq
\]So \[
\frac{q'}{q}=-c
\]And we get: \[
\ln(q) = -ct+ K
\]Put it all together: \[
q(t) = Ke^{(1/RC) t}
\]

Answer 5

Notice \[
q(0) = Ke^{0} \implies K = q(0) = q_0
\]So \[
q(t) = q_0 e^{(1/RC)t}
\]

Answer 6

@wio and what is q0 ?

Answer 7

It is \(q\) when \(t=0\).

Answer 8

i can't understand why it's CEe^-t/yau
dvc/dt+1/RC*vc=0
the same equation but the solutions is : Ee^-t/tau

Answer 9

I might have forgotten a negative sign.

Answer 10

yes, you dropped a negative sign, wio, and I'm not sure exactly what you mean @AntarAzri , the solution to the equation you posted is\[Q(t)=Q_0e^{-t/\tau}=C\mathcal E e^{-t/\tau}=CV(t)\]as you can see you can get an equation for the voltage across the capacitor as a function of time by cancelling \(C\) from both sides, which gives your second equation\[V(t)=\mathcal E e^{-t/\tau}\]note that you could have canceled the \(C\) before solving the equation, and instead solved\[\frac{dV}{dt}+{V\over RC}=0;~~~V(0)=\mathcal E\]