Section 3.5 Capacitor Charging and Discharging Behaviors
Let’s start examining circuits containing resistors, capacitors, and inductors. We’ll start with the circuit in Figure 3.5.1. We want to evaluate the voltage across the capacitor (and resistor) as a function of time if the switch starts closed and opens suddenly at time \(t=0\text{.}\) Assume that the internal battery resistance is much smaller than the circuit resistor so that \(r_e \ll R\) and \(\left(\Delta V\right)_C\approx V_0\) at \(t=0\text{.}\)
We can use KVL to examine the circuit. Since the switch is opened at \(t=0\text{,}\) we only have a single KVL equation
\begin{equation*}
V_C - IR = 0
\end{equation*}
where \(V_C\) is the voltage difference between the capacitor plates. For a capacitor, \(Q=CV_c\text{,}\) so
\begin{equation*}
\frac{Q}{C} - IR = 0\text{.}
\end{equation*}
If we take the time derivative of both sides and recognize that \(I=dQ/dt\text{,}\) we get
\begin{equation*}
\frac{I}{C} - R\frac{dI}{dt} = 0
\end{equation*}
or, after rearranging:
\begin{equation*}
\frac{dI}{dt}=-\frac{1}{RC} I(t)\text{.}
\end{equation*}
By inspection, the solution to this differential equation is
\begin{equation*}
I(t)=I_0 e^{-t/RC}
\end{equation*}
where \(I_0=I(t=0)\text{.}\) Now, this current can be plugged back into our original equation \(V_C - IR=0\) to find
\begin{equation*}
V_C(t)=I_0 R e^{-t/RC} = V_0 e^{-t/\tau}
\end{equation*}
where \(V_C(t=0)=V_0 = I_0 R\) at the instant that the switch opens and we’ve defined something called the RC time constant \(\tau=RC\) which is a characteristic response time associated with this RC circuit. We plot this discharging behavior in Figure 3.5.2.
Similarly, we can examine the behavior of a charging capacitor by examing the behavior of the circuit in Figure 3.5.3. Here, we do not worry about the battery’s internal resistance \(r_e\) as \(r_e \ll R\) typically, meaning that the charging behavior will be determined by \(R+r_e\approx R\text{.}\)
Applying the loop law, we find that
\begin{equation*}
V_0-IR-V_C = 0
\end{equation*}
where \(V_C\) is the voltage difference across the capacitor. Remembering that
\begin{equation*}
I=C\frac{\text{d}V}{\text{d}t}
\end{equation*}
we find that
\begin{equation*}
V_0 - RC\frac{\text{d}V}{\text{d}t} - V_C = 0
\end{equation*}
or
\begin{equation*}
V_0 - V_C = RC\frac{\text{d}V}{\text{d}t}\text{.}
\end{equation*}
Since \(V_0\) is a constant, we can write this as
\begin{equation*}
V_0 - V_C = RC\frac{\text{d}}{\text{d}t}\left(V_C - V_0\right)\text{.}
\end{equation*}
If we define \(V'=V_C - V_0\text{,}\) then this becomes
\begin{equation*}
\frac{\text{d}V'}{\text{d}t}=-\frac{1}{RC}V'\text{.}
\end{equation*}
This has an exponential solution of the form
\begin{equation*}
V'=V_C - V_0 = A e^{-t/RC}
\end{equation*}
which can be rearranged into
\begin{equation*}
V_C = V_0\left(1-e^{-t/RC}\right)\text{.}
\end{equation*}
This shows that the capacitor charges from 0% to 63% in \(\Delta t=RC\text{.}\)
When charging a capacitor, energy is being stored in the electric field
\begin{equation*}
U_C = \frac{1}{2} C V^2 = \frac{Q^2}{2C} = \frac{1}{2} QV\text{.}
\end{equation*}
Likewise, inductors can store energy. As the current through an inductor increases, energy is stored i nthe magnetic field
\begin{equation*}
U_L = \frac{1}{2}LI^2\text{.}
\end{equation*}
