Main variables:

Main Equations:

The capacitance \(C\) in terms of the charge and the voltage.

\begin{equation*}
Q = C V.
\end{equation*}

For a parallel-plate capacitor, the capacitance is:

\begin{equation*}
C = \epsilon_0 \frac{A}{d}.
\end{equation*}

If the capacitors are connected in series, the equivalent capacitance is:

\begin{equation*}
\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots
\end{equation*}

If the capacitors are connected in parallel, the equivalent capacitance is:

\begin{equation*}
C_{eq} = C_1 + C_2 + C_3 + \dots
\end{equation*}

The energy stored in a capacitor:

\begin{equation*}
U = \frac{1}{2} C V^2.
\end{equation*}

A dielectric capacitor, the capacitance is:

\begin{equation*}
C = K C_0.
\end{equation*}

where \(C_0\) is the capacitance of the empty capacitor.

If resistors are connected in series, the equivalent resistance is:

\begin{equation*}
R_{eq} = R_1 + R_2 + R_3 + \dots
\end{equation*}

If resistors are connected in parallel, the equivalent resistance is:

\begin{equation*}
\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots
\end{equation*}

Ohm’s law states:

\begin{equation*}
V = IR.
\end{equation*}

Kirchhoff’s voltage law for closed path the sum of voltages \(V_i\) is:

\begin{equation*}
\sum_{\text{closed path}} V_i=0.
\end{equation*}

The power dissipated by resistor \(R\) due to a voltage drop \(V\) is:

\begin{equation*}
P = VI = I^2 R = \frac{V^2}{R}.
\end{equation*}

Capacitor in charge:

\begin{equation*}
Q = C V ( 1 – e^{-t/RC}).
\end{equation*}

The current for a capacitor in charge:

\begin{equation*}
I = \frac{dq}{dt} = \frac{V}{R} e^{-t/RC}.
\end{equation*}

Capacitor in discharge:

\begin{equation*}
Q = Q_0 e^{-t/RC}.
\end{equation*}

The current for a capacitor in discharge:

\begin{equation*}
I = \frac{dq}{dt} = -\frac{Q_0}{RC} e^{-t/RC}.
\end{equation*}