Find the total capacitance of the combination of capacitors shown below

Capacitors in Series: When capacitors with capacitances C₁, C₂, … are joined in series combination, the equivalent capacitance C_eq is given as-

Capacitors in Parallel: When capacitors with capacitances C₁, C₂, … are joined in parallel combination, the equivalent capacitance C_eq is given as-

Combination of capacitors are given in the above figures and we are supposed to figure out what the equivalent capacitance of the capacitors in Figure is.

The $5.0\mathrm{\mu F}$and $3.5\mathrm{\mu F}$capacitors shown in the blue rectangle in Figure 1 are in series and their equivalent capacitance is found from Equation (1):

The $0.75\mathrm{\mu F}$and $15\mathrm{\mu F}$capacitors in Figure 1, that is shown in red rectangle are in parallel, and their equivalent capacitance may be calculated using Equation (2):

$\begin{array}{rcl}{\mathbf{C}}_{\mathbf{eq}}& \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{75}\mathbf{}\mathbf{\mu F}\mathbf{+}\mathbf{15}\mathbf{}\mathbf{\mu F}\\ & \mathbf{=}& \mathbf{15}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{\mu F}\end{array}$

The $1.5\mathrm{\mu F}$and $15.8\mathrm{\mu F}$capacitors in Figure 2, that is shown in black rectangle are connected in series, and their equivalent capacitance may be calculated using Equation (l):

$\begin{array}{rcl}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{eq}}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\mu F}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{15}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{\mu F}}\\ {\mathbf{C}}_{\mathbf{eq}}& \mathbf{=}& \frac{\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\mu F}\mathbf{)}\mathbf{(}\mathbf{15}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{\mu F}\mathbf{)}}{\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\mu F}\mathbf{+}\mathbf{15}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{\mu F}}\\ & \mathbf{=}& \mathbf{1}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{\mu F}\end{array}$

The $2.1\mathrm{\mu F}$, localid="1655170157245" $8.0\mathrm{\mu F}$, and $1.4\mathrm{\mu F}$capacitors in Figure 3 are in parallel, and their equivalent capacitance may be calculated using Equation (2):

$\begin{array}{rcl}{\mathbf{C}}_{\mathbf{eq}}& \mathbf{=}& \mathbf{2}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{\mu F}\mathbf{+}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu F}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{4}\mathbf{}\mathbf{\mu F}\\ & \mathbf{=}& \mathbf{11}\mathbf{}\mathbf{\mu F}\end{array}$

Therefore, the total capacitance of the combination of capacitor is ${\mathrm{C}}_{\mathrm{eq}}=11\mathrm{\mu F}$.