Q61PE - College Physics (Urone)

Question

Find the total capacitance of the combination of capacitors shown below

                                              

Step-by-Step Solution

Answer

The required solution is ${\mathrm{C}}_{\mathrm{eq}}=11 \mathrm{\mu F}$.

Step 1: Concepts and Principles

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.

Step 2: Drawing the figure

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):

Step 3: Finding Parallel equivalent capacitance

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):

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

Step 5: Finding Series equivalent capacitance

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 $2.1 \mathrm{\mu F}$, $8.0 \mathrm{\mu F}$, 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 ${\mathrm{C}}_{\mathrm{eq}}=11 \mathrm{\mu F}$.