First order responce; RC circuit

[sandy.bridge]

Homework Statement

Assume there is a voltage source in series with a resistor and a capacitor. Thus,
$$V_S=i(t)R+v_C(t)=CR\frac{dv_C}{dt}+v_C\rightarrow{}dt/(RC)=dv_C/(V_S-v_C)$$
From this point I understand that one has to apply a negative sign to both sides before integrating, but why is it essential to do this mathematically?
Why can one simply not do
$$e^{t/(RC)}e^D=V_S-v_C(t)$$ where D is some constant determined by the state of the circuit before?

 

[gneill]

The integrals to be performed are definite integrals:
$$\int_0^{V_c} \frac{1}{V_s - V_c}dV_c = \int_0^t \frac{1}{R C} dt$$
As such the constants of integration cancel out, and you're left with:
$$ln(V_s) - ln(V_s - V_c) = \frac{t}{R C}$$
$$ln(\frac{V_s - V_c}{V_s}) = -\frac{t}{R C}$$
and so on.

 

[sandy.bridge]

Nevermind, thanks for the help!

 

[sandy.bridge]

@ gneil
My textbook does not analyze it as a definite integral, but it certainly makes sense to do it in that way. Just a few things, when analyzing differential circuits in this way, if the capacitor had an initial charge one would have the initial charge as the lower limit, with the charge on the voltage source being the upper?

 

[gneill]

@ gneil
My textbook does not analyze it as a definite integral, but it certainly makes sense to do it in that way. Just a few things, when analyzing differential circuits in this way, if the capacitor had an initial charge one would have the initial charge as the lower limit, with the charge on the voltage source being the upper?

Sure. That would work.