Natural Response of an RC circuit

The natural response describes how a circuit behaves on its own after all external sources are removed.

For an RC circuit, this means we look at how the voltage across the capacitor changes when the circuit is left to discharge or charge only through its internal components.

Fig-1: Natural Response of an RC circuit (switching event at ${\displaystyle t_o}$)

1. Current-Voltage Relationship

As we have alredy discussed in our article on Capacitors and their behaviour, the most important equation for circuit analysis defines the current (I) flowing through the capacitor. The current is proportional to the rate of change of the voltage across it, not the voltage itself.

${\displaystyle I(t)=C\cdot <!--\; \cdot \; -->\frac{dV(t)}{t}}$

2. Steady-State Requirement

Before we begin analyzing the natural response of an RC circuit, we assume that the switching event took place at time ${\displaystyle t_0}$. The circuit is assumed to have been in DC operation for time ${\displaystyle t<0}$, with all sources and switch positions fixed, for a duration sufficient that all natural (transient) terms have decayed to zero.

Thus the capacitor voltage at time 𝑡 = 0⁻ is equal to its DC steady-state value (${\displaystyle V_s}$) and can be used as the initial condition for analyzing the natural response for ${\displaystyle t>0}$.

Formally, the voltage across the capatitor is equal to the source DC voltage:

${\displaystyle \underset{t\to <!--\; \to \; -->0^{-<!--\; -\; -->}}{lim}{v}_C(t)=V_{DC}}$

or, we can also say that the current through the capacitor is zero:

${\displaystyle \underset{t\to <!--\; \to \; -->0^{-<!--\; -\; -->}}{lim}{i}_C(t)=0}$

3. Building the Differential Equation

In order to derive the natural response, we apply KCL at the capacitor node. Because the capacitor, C, and the resistor, R, are in series, the sum of the current entering the node and the current leaving the node is equal to zero (0).

We know that both the capacitor and the resistor are in series, therefore, same current is flowing through them. Let's call it "i". Formally,

${\displaystyle i_R=i_C=i}$

Then applying KCL at the node gives us:

${\displaystyle i_R+i_C=0}$

At the capacitor node:

a) Current through the resistor, ${\displaystyle i_R}$:

${\displaystyle \frac{(v_C-<!--\; -\; -->0)}{R}}$

b) Current through the capacitor, ${\displaystyle i_C}$:

${\displaystyle C\frac{d{v}_c}{dt}}$

Subtituting the values in the KCL equation, we get the first-order ODE:

${\displaystyle \frac{v_C}{R}+C\frac{d{v}_C}{dt}=0}$

4. Solving the first-order ODE

We begin the solution be rewriting the ODE above in standard first-order ODE form.

Step 1: For this, we divide by C

${\displaystyle \frac{d{v}_C}{dt}+\frac{1}{RC}{v}_C=0}$

Step 2:

Let, ${\displaystyle \mathrm{\alpha <!--\; \alpha \; -->}=\frac{1}{RC}}$

${\displaystyle \frac{d{v}_C}{dt}+\mathrm{\alpha <!--\; \alpha \; -->}{v}_C=0}$

Step 3: Separating the variables, we get:

${\displaystyle \frac{1}{v_C}d{v}_C=-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}dt}$

Step 4: Integrating both sides:

${\displaystyle \int <!--\; \int \; -->\frac{1}{v_C}d{v}_C=\int <!--\; \int \; -->-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}dt}$

or,

${\displaystyle \Rightarrow <!--\; \Rightarrow \; -->\ln |v_C(t)|=\mathrm{\alpha <!--\; \alpha \; -->}t+K}$

Step 5: Exponentiate to solve for v_C:

${\displaystyle v_C(t)=e^K\cdot <!--\; \cdot \; -->e^{-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}t}}$

Step 6: Apply the initial condition for the voltage across the capacitor: ${\displaystyle v_C(0)=V_0}$

${\displaystyle V_0=v_C(0)=e^K\cdot <!--\; \cdot \; -->e^{-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}0}}$

Therefore,

${\displaystyle V_0=v_C(0)=e^K}$

Step 7: Final solution (natural response)

${\displaystyle v_C(t)=V_0{e}^{-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}t}}$

or,

${\displaystyle v_C(t)=V_0{e}^{-<!--\; -\; -->\frac{t}{RC}}}$