Find the total capacitance of the combination of capacitors shown in Figure 19.34 .

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-

The \({\rm{0}}{\rm{.30\mu F}}\) and$10\mathrm{\mu F}$capacitors shown in the red rectangle in Figure 1 are in series and their equivalent capacitance is found from Equation (1):

\(\begin{array}{c}\frac{{\rm{1}}}{{{{\rm{C}}_{{\rm{eq}}}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{0}}{\rm{.30\mu F}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{10\mu F}}}}\\{{\rm{C}}_{{\rm{eq}}}}{\rm{ = }}\frac{{{\rm{(0}}{\rm{.30\mu F)(10\mu F)}}}}{{{\rm{0}}{\rm{.30\mu F + 10\mu F}}}}{\rm{ = 0}}{\rm{.29\mu F}}\end{array}\)

The $0.29\mathrm{\mu F}$and $2.5\mathrm{\mu F}$capacitors shown in the black rectangle in Figure 2 are in parallel and their equivalent capacitance is found from Equation (2) :

$\begin{array}{rcl}{\mathbf{C}}_{\mathbf{eq}}& \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{29}\mathbf{}\mathbf{\mu F}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\mu F}\\ & \mathbf{=}& \mathbf{2}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{\mu F}\end{array}$

Therefore, capacitance obtained is role="math" localid="1655166662635" ${\mathrm{C}}_{\mathrm{eq}}=2.8\mathrm{\mu F}$.