Mathematical Equation for Capacitor Charging Voltage

Considering the circuit below to find out the charging current. When the switch is closed, the capacitor will start to charge.

Applying Kirchhoff’s Voltage Law, we get

$\mathit{E}-{\mathit{v}}_{\mathit{R}}-{\mathit{v}}_{\mathit{C}}=0\dots \dots \dots \dots \dots \dots \left(\mathit{a}\right)$
$\mathit{o}\mathit{r},\mathit{E}-{\mathit{v}}_{\mathit{C}}={\mathit{v}}_{\mathit{R}}={\mathit{i}}_{\mathit{C}}\mathit{R}$$=\mathit{R}\frac{\mathit{d}\mathit{q}}{\mathit{d}\mathit{t}}=\mathit{R}\frac{\mathit{d}}{\mathit{d}\mathit{t}}\left(\mathit{C}{\mathit{v}}_{\mathit{C}}\right)$$=\mathit{R}\mathit{C}\frac{\mathit{d}}{\mathit{d}\mathit{t}}{\mathit{v}}_{\mathit{C}}$
$\mathit{o}\mathit{r},\frac{\mathit{d}\mathit{t}}{\mathit{R}\mathit{C}}=\frac{\mathit{d}{\mathit{v}}_{\mathit{C}}}{\mathit{E}-{\mathit{v}}_{\mathit{C}}}$
$\mathit{o}\mathit{r},\frac{\mathit{t}}{\mathit{R}}\mathit{C}+{\mathit{k}}_1=-\mathit{l}\mathit{n}\mathit{E}-{\mathit{v}}_{\mathit{C}}$ $\left[\mathit{I}\mathit{n}\mathit{t}\mathit{e}\mathit{g}\mathit{r}\mathit{a}\mathit{t}\mathit{i}\mathit{n}\mathit{g}\mathit{b}\mathit{o}\mathit{t}\mathit{h}\mathit{t}\mathit{h}\mathit{e}\mathit{s}\mathit{i}\mathit{d}\mathit{e}\mathit{s}\right]$
$\mathit{o}\mathit{r},\frac{-\mathit{t}}{\mathit{R}\mathit{C}}+\mathit{k}=\mathit{l}\mathit{n}\mathit{E}-{\mathit{v}}_{\mathit{C}}\dots \dots \dots .\mathit{b}$

Where, k is constant of integration. It can be calculated by putting initial time, t = 0 when capacitor voltage, ${\mathit{v}}_{\mathit{C}}=0\mathit{V}$ then, $\mathit{k}=\mathit{l}\mathit{n}\mathit{E}$

Substituting $\mathit{k}=\mathit{l}\mathit{n}\mathit{E}$ on (b) we get,

$\frac{-\mathit{t}}{\mathit{R}\mathit{C}}+\mathit{l}\mathit{n}\mathit{E}=\mathit{l}\mathit{n}\left(\mathit{E}-{\mathit{v}}_{\mathit{C}}\right)$
$\mathit{o}\mathit{r},\frac{-\mathit{t}}{\mathit{R}\mathit{C}}=\mathit{l}\mathit{n}\left(\mathit{E}-{\mathit{v}}_{\mathit{C}}\right)-\mathit{l}\mathit{n}\mathit{E}$
$\mathit{o}\mathit{r},\frac{-\mathit{t}}{\mathit{R}\mathit{C}}=\mathit{l}\mathit{n}\left(\frac{\mathit{E}-{\mathit{v}}_{\mathit{C}}}{\mathit{E}}\right)$
$\mathit{o}\mathit{r},{\mathit{e}}^{-\frac{\mathit{t}}{\mathit{R}\mathit{C}}}=\frac{\mathit{E}-{\mathit{v}}_{\mathit{C}}}{\mathit{E}}$
$\mathit{o}\mathit{r},\mathit{E}{\mathit{e}}^{-\frac{\mathit{t}}{\mathit{R}\mathit{C}}}=\mathit{E}-{\mathit{v}}_{\mathit{C}}$
$\mathit{o}\mathit{r},{\mathit{v}}_{\mathit{C}}=\mathit{E}-\mathit{E}{\mathit{e}}^{-\frac{\mathit{t}}{\mathit{R}}\mathit{C}}$
$\mathit{o}\mathit{r},{\mathit{v}}_{\mathit{C}}=\mathit{E}\left(1-{\mathit{e}}^{-\frac{\mathit{t}}{\mathit{R}\mathit{C}}}\right)$
$\therefore {\mathit{v}}_{\mathit{C}}=\mathit{E}\left(1-{\mathit{e}}^{-\frac{\mathit{t}}{\mathit{\tau }}}\right)$$\left[\mathit{W}\mathit{h}\mathit{e}\mathit{r}\mathit{e},\mathit{\tau }=\mathit{R}\mathit{C}=\mathit{T}\mathit{i}\mathit{m}\mathit{e}\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{t}\right]\dots \dots \dots ..\mathit{c}$

After a several period of time constant, the capacitor will be as charged as the applied potential E. It is approximately 5 times of the time constant, τ to be fully charged. After 5τ{or >5τ} the capacitor voltage will be 20V as per the circuit shown above.