Question

Find the total capacitance of the combination of capacitors in Figure 19.33.

Short answer

The total capacitance for the combination of capacitors is ${\mathrm{C}}_{\mathrm{eq}}=0.29\mathrm{\mu F}$.

Step-by-step solution

Concept Introduction

Capacitors in Parallel: When capacitors with capacitances C₁, C₂, ,,… are connected in parallel, the equivalent capacitance C_eq equals the sum of the individual capacitances –

${\mathbf{C}}_{\mathbf{eq}}\mathbf{=}{\mathbf{C}}_{\mathbf{1}}\mathbf{+}{\mathbf{C}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Capacitors in Series: When capacitors with capacitances C₁, C₂, ,,… are connected in series, the reciprocal of the equivalent capacitance C_eqequals the sum of the reciprocals of the individual capacitances –

$\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{eq}}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{1}}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{2}}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Breaking down of the image

The $10\mathrm{\mu F}$and $2.5\mathrm{\mu F}$capacitors shown in the red rectangle in Figure l are in parallel and their equivalent capacitance is found from Equation (1):

$\begin{array}{rcl}{\mathbf{C}}_{\mathbf{eq}}& \mathbf{=}& \mathbf{10}\mathbf{}\mathbf{\mu F}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\mu F}\\ & \mathbf{=}& \mathbf{12}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\mu F}\end{array}$

Calculation of the capacitance

The $12.5\mathrm{\mu F}$and $0.30\mathrm{\mu F}$capacitors shown in the black rectangle in Figure 2 are in series and their equivalent capacitance is found from Equation (2):

$\begin{array}{rcl}\frac{\mathbf{1}}{{\mathbf{C}}_{\mathbf{eq}}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{12}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\mu F}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{0}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{\mu F}}\\ {\mathbf{C}}_{\mathbf{eq}}& \mathbf{=}& \frac{\mathbf{(}\mathbf{12}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\mu F}\mathbf{)}\mathbf{\times }\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{\mu F}\mathbf{)}}{\mathbf{12}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{\mu F}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{\mu F}}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{29}\mathbf{}\mathbf{\mu F}\end{array}$

Therefore, the value for capacitance is ${\mathrm{C}}_{\mathrm{eq}}=0.29\mathrm{\mu F}$.