# RC Circuit Step Response Derivation

This is the step response of an RC circuit: $$V(t)=V_S+(V_0-V_S)e^{-\frac{t}{RC}}$$ Where $V(t)$ is the voltage across the capacitor with respect to time, $V_S$ is the height of the step in supply voltage, and $V_0$ is the initial voltage across the capacitor at $t=0$. But there's something in its derivation that I don't really agree with, so I'll write its derivation all the way there. $$V_S-V_R-V(t)=0$$ $$V_S-IR-V(t)=0$$ $$V_S-RC\frac{dV}{dt}-V(t)=0$$ $$-RC\frac{dV}{dt}=V(t)-V_S$$ $$\frac{1}{V(t)-V_S}\frac{dV}{dt}=-\frac{1}{RC}$$ And then we take the integral of both sides from $0$ to $t$ with respect to $t$ (???): $$\int_0^t{\frac{1}{V(t)-V_S}\frac{dV}{dt}dt}=\int_0^t{-\frac{1}{RC}dt}$$ $$\ln(V(t)-V_S)-\ln(V(0)-V_S)=-\frac{t}{RC}$$ $$\ln\left(\frac{V(t)-V_S}{V_0-V_S}\right)=-\frac{t}{RC}$$ $$\frac{V(t)-V_S}{V_0-V_S}=e^{-\frac{t}{RC}}$$ $$V(t)=V_S+(V_0-V_S)e^{-\frac{t}{RC}}$$ It works, but what confused me of course was the integration part. I thought it wasn't a valid move to take the integral of something with respect to a variable with its bounds depending on that same variable. How does this derivation work with that?

• Try doing the integration with $T$ for the time elapsed instead of $t$. That should clear your doubt (the variable for total time in the upper limit is just a dummy variable used as a bound and has nothing to do with the differential inside). Mar 10, 2018 at 15:23
• A math problem?
– jim
Mar 10, 2018 at 18:21

1. Use a different, but similar variable as a bound (as mentioned by Yuzuriha). $$\int_0^T f(t)dt$$
2. Use an apostrophe to distinguish the bound and integration variable $$\int_0^t f(t')dt'$$
3. Completely ignore the problem. This one is the most convenient and if you have seen it before it will be understandable, but you still introduce a possible point of confusion. $$\int_0^t f(t)dt$$ The writer of the derivation used (3), which can be justified in some cases, but you are right to be confused.