A simple circuit for charging and discharging a capacitor consists of a capacitor $C$, a resistor, and a switch. When the switch is set to the charging position, the charges from the battery flow through the resistor $R_a$ into the capacitor.
According to Kirchhoff’s voltage law, the sum of the voltages across the resistor and the capacitor must at any time equal the battery voltage $U_0$:
\[
U_R(t) + U_C(t) = U_0
\]
Substituting Ohm’s law and the definition of capacitance gives:
\[
I(t)R_a + \frac{Q(t)}{C} = U_0
\]
With $I(t) = \dot{Q}(t)$, this becomes
\[
\dot{Q}(t)R_a + \frac{Q(t)}{C} = U_0
\]
The solution is an exponential function:
\[
Q(t) = CU_0\left(1 - e^{-\frac{t}{R_aC}}\right)
\]
Instead of the exponential constant, it is common to use the time constant
\[
\boxed{\tau = R_aC}
\]
which describes the time at which the exponential has reached $1 - 1/e \approx 63.2\%$ of its final value.
Dividing by $C$ gives the voltage across the capacitor:
\[
\boxed{U(t) = U_0\left(1 - e^{-\frac{t}{R_aC}}\right)}
\]
Thus, the voltage rises quickly at first and then more slowly, asymptotically approaching $U_0$ but never reaching it exactly.
The charging current follows from differentiating $Q(t)$ with respect to time:
\[
\boxed{I(t) = I_0 e^{-\frac{t}{R_aC}}}, \quad \text{with } I_0 = \frac{U_0}{R_a}
\]
The current decreases exponentially as the electric field inside the capacitor builds up, binding more and more charges to the plates.
When the switch is flipped to the discharging position, the capacitor releases its stored charge through a resistor $R_e$. In this case, Kirchhoff’s law requires the sum of the voltages across the capacitor and resistor to be zero:
\[
\dot{Q}(t)R_e + \frac{Q(t)}{C} = 0
\]
The solution is
\[
Q(t) = CU_0 e^{-\frac{t}{R_eC}}
\]
and thus the discharging voltage becomes
\[
\boxed{U(t) = U_0 e^{-\frac{t}{R_eC}}}
\]
The current has the same functional form:
\[
\boxed{I(t) = I_0 e^{-\frac{t}{R_eC}}}
\]
During discharge, voltage and current decay proportionally, both following the same exponential law.
Example:
A capacitor of $C = 1\,\text{mF}$ is discharged through a resistor of $27\,\text{k}\Omega$. After what time has the voltage dropped to 10% of its initial value?
Solution:
\[
U(t) = U_0 e^{-\frac{t}{R_eC}} \quad \Rightarrow \quad \frac{U(t)}{U_0} = 0.1 = e^{-\frac{t}{R_eC}}
\]
Taking the natural logarithm:
\[
t = -R_eC \ln(0.1) = 27{,}000 \cdot 0.001 \cdot 2.3026 \approx 62.2 \,\text{s}
\]
After about 62 seconds, the voltage has decayed to 10% of its original value.