Homogeneous Solution

Consider an RC circuit consisted of a resistor and a capacitor in series connected to an AC voltage source $v_s(t)=\cos(\omega t)$, find the voltage $v_C(t)$ across $C$. The governing DE describing the circuit can be obtained based on KVL:

$$\tau\frac{d}{dt}v_C(t)+v_C(t)=v_s(t) =\left\{\begin{array}{c}0\\ 1\\ \cos(\omega t)\end{array}\right.$$ (73)

When the external input is zero $v_s(t)=0$, the DE is homogeneous (zero on the right-hand side):

$$\tau\frac{d}{dt}v_C(t)+v_C(t)=0,\;\;\;\;\;v_C(t)\vert _{t=0}=v_C(0)=V_0$$ (74)

The non-trivial solution of this homogeneous equation is due to some non-zero initial value, the voltage $v_C(0)=V_0\ne 0$ across the capacitor before $t=0$. The homogeneous solution needs to be a function whose derivative takes the same form as the function itself, an exponential function:

$$v_C(t)=A e^{st},\;\;\;\; \;\;\;\;\;\;\; \frac{d}{dt}v_C(t)=s A e^{st}$$ (75)

Substituting them into the DE, we get

$$\tau sA e^{st} + Ae^{st}=(\tau s+1) Ae^{st}=0$$ (76)

As we are not interested in the trivial solution $Ae^{st}=0$, we must have

$$\tau s+1=0,\;\;\;\;\; \;\;\;\;\;\;s=-\frac{1}{\tau}$$ (77)

Now we get $v_C(t)=Ae^{-t/\tau}$ for $t\ge 0$. The constant coefficient $A$ can be obtained by the initial condition $v_C(0)=V_0$:

$$v_C(0)=A e^{-t/\tau}\bigg\vert _{t=0}=A e^{0}=V_0, \;\;\;\;\; \;\;\;\;\;A=V_0$$ (78)

Now the homogeneous solution is found to be

$$v_C(t)=V_0 e^{-t/\tau}$$ (79)

which decays to zero as $t\rightarrow\infty$:

$$\lim_{t\rightarrow\infty} V_0 e^{-t/\tau}=0$$ (80)

The same result can also be obtained by the Laplace transform method.

The current through $R$ and $C$ is

$$i(t)=C\frac{d}{dt}\;v_C(t)=C\frac{d}{dt}\left(V_0 e^{-t/\tau}\right) =-V_0 \frac{C}{\tau} e^{-t/\tau}=-\frac{V_0}{R} e^{-t/\tau}$$ (81)

The voltage across $R$ is

$$v_R(t)=i(t)R=-V_0 e^{-t/\tau}=-v_C(t)$$ (82)

This result can be verified by KVL: $v_C(t)+v_R(t)=v_s(t)=0$.

The time constant $\tau$ can be identified on the time plot of the general first order response $v(t)=V_0 e^{-t/\tau}=e^{-t\tau}$ with $V_0=1$. Specifically consider the derivative of $v(t)$ evaluated at $t=0$:

$$\frac{d}{dt}v(t)\big\vert _{t=0}=-\frac{1}{\tau}e^{-t/\tau}\bigg\vert _{t=0}=-\frac{1}{\tau}$$ (83)

This is the slope of the tangent line of $v(t)$ at $t=0$, and $\tau$ is its intersection with the horizontal axis. We conclude that the voltage across a capacitor cannot change instantaneously due to the time constant $\tau>0$, unless $R=0$ and therefore $\tau=RC=0$. (The same can be said for the current through an inductor.)