Calculate the reciprocal of each capacitance, sum the reciprocals, and then take the reciprocal of the sum: $\frac{1}{C_{\text{eq, s}}} = \sum_{i} \frac{1}{C_i}$.

Sum the individual capacitances: $\C_{\text{eq, p}} = \sum_{i} C_i$.

Initially, the uncharged capacitor allows easy charge flow, acting like a wire. As it charges, the charge on the plates increases, the current decreases, and the stored electric potential energy increases, approaching steady-state asymptotically.

Charge and stored energy decrease, and current decreases over time. After a time much greater than $\tau$, the circuit reaches a steady state.

Simplify the circuit by finding equivalent capacitances for series and parallel combinations.

The single capacitance that has the same effect as a combination of capacitors in a circuit.

The time required for a charging capacitor to reach approximately 63% of its maximum charge or for a discharging capacitor to drop to about 37% of its initial charge. It is given by $\tau = RC$

The ability of a component or circuit to collect and store energy in the form of an electrical charge.

A measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms ($\Omega$).

The voltage generated by a battery or other power source.

Series: Same charge on each capacitor; equivalent capacitance is smaller than the smallest individual capacitance. Parallel: Same voltage across each capacitor; equivalent capacitance is the sum of individual capacitances.

t=0: Current is at its maximum (limited only by the resistance). t=$\infty$: Current is zero (capacitor is fully charged and blocks current flow).

t=0: Voltage across the capacitor is zero (initially uncharged). t=$\infty$: Voltage across the capacitor is equal to the voltage of the source (fully charged).

Capacitor: Stores energy in an electric field; opposes changes in voltage. Resistor: Dissipates energy as heat; opposes current flow.

Charging: Capacitor accumulates charge, voltage increases, current decreases. Discharging: Capacitor loses charge, voltage decreases, current decreases.