To find the natural response of an RC circuit, the following is required:
-
The initial capacitor voltage \(v(0)\).
Since the capacitor voltage cannot change instantaneously,
\(\boxed{v\left(0^-\right) = v\left(0^+\right) = v\left(0\right) = V_0}\)
where \(v\left(0^-\right)\) is the voltage across the capacitor just before switching and \(v\left(0^+\right)\) is its voltage immediately after switching.
-
The time constant \(\tau\).
The time constant \(\tau\) for an RC circuit is
\(\boxed{\tau = RC}\)
where \(R\) is the thevenin resistance at the capacitor terminals and \(C\) is the equivalent capacitance.
A source-free RC circuit occurs when its DC source is suddenly disconnected. The energy already stored in the capacitor is released to the resistors.
Assume \(v\left(t\right)\) is the voltage across the capacitor. Since the capacitor is initially charged, the initial voltage at time \(t = 0\) is
\(\boxed{v\left(0\right) = V_0}\)
with the corresponding value of the energy stored as
\(\boxed{w\left(0\right) = \frac{1}{2} CV_{0}^2}\)
Applying KCL at the top node
\(\displaystyle i_C + i_R = 0\)
\(\displaystyle C\frac{dv}{dt} + \frac{v}{R} = 0\)
\(\displaystyle \frac{dv}{dt} + \frac{v}{RC} = 0\)
This is a first-order differential equation. To solve it, arrange the terms as
\(\displaystyle \frac{dv}{v} = -\frac{1}{RC} dt\)
Integrating both sides
\(\displaystyle \ln{v} = -\frac{t}{RC} + \ln{A}\)
where \(\ln{A}\) is the integration constant.
\(\displaystyle \ln{v} - \ln{A} = -\frac{t}{RC}\)
\(\displaystyle \ln{\frac{v}{A}} = -\frac{t}{RC}\)
\(\displaystyle \frac{v}{A} = e^{\displaystyle -\frac{t}{RC}}\)
\(\displaystyle v\left(t\right) = Ae^{\displaystyle -\frac{t}{RC}}\)
But from the initial conditions, \(v\left(0\right) = V_0\)
\(\displaystyle v\left(t\right) = Ae^{\displaystyle -\frac{t}{RC}}\)
\(\displaystyle V_0 = Ae^{\displaystyle -\frac{0}{RC}}\)
\(\displaystyle V_0 = Ae^0 = A\)
\(\boxed{v\left(t\right) = V_0\,e^{\displaystyle -\frac{t}{RC}} = V_0\,e^{\displaystyle -\frac{t}{\tau}}}\)
where time constant \(\tau = RC\)
This shows that the natural response of the RC circuit is an exponential decay of the initial voltage \(V_0\).
The energy that was initially stored in the capacitor is eventually dissipated in the resistor.
The power dissipated in the resistor is
\(\displaystyle i_R \left(t\right) = \frac{v}{R} = \frac{V_0}{R} e^{\displaystyle -t/\tau}\)
\(\boxed{p\left(t\right) = v i_R = \frac{V_{0}^{2}}{R} e^{\displaystyle -2t/\tau}}\)
The energy absorbed by the resistor up to time \(t\) is
\(\displaystyle w_{R}\left(t\right) = \int_{0}^{t} p\left(\lambda\right)\,d\lambda = \int_{0}^{t} \frac{V_{0}^{2}}{R} e^{\displaystyle -2\lambda/\tau}\,d\lambda\)
\(\displaystyle w_{R}\left(t\right) = -\left.\frac{\tau V_{0}^{2}}{2R} e^{\displaystyle -2\lambda/\tau}\right|_{0}^{t},\quad \tau = RC\)
\(\boxed{w_{R}\left(t\right) = \frac{1}{2} CV_{0}^{2} \left(1 - e^{\displaystyle -2t/\tau}\right)}\)
where time constant \(\tau = RC\)