Question

What is the equivalent capacitance of the three capacitors in Figure P23.36?

36. Il What is the equivalent capacitance of the three capacitors in Figure P23.36? 20 μF FIGURE P23.36 FIGURE P23.37

Answer 1

Concepts and reason

This question is based on the concept of equivalent capacitance of series and parallel combination of capacitors.

First calculate the equivalent capacitance of first arm and then find the equivalent capacitance of both parallel capacitors of both arms.

Fundamentals

Capacitor is the device which stores the energy in from of electric field. Capacitance is the property of capacitor by which it stores the charge. Its unit is Farad.

In any electrical circuit two or more than two capacitors connected in series connection can be represented by a capacitor having equivalent capacitance. The magnitude of equivalent capacitance can be represented as follows;

$\frac{1}{{{C_{{\rm{eq}}}}}} = \frac{1}{{{C_1}}} + {\frac{1}{C}_2} + ...\frac{1}{{{C_{\rm{n}}}}}$

Here ${C_{{\rm{eq}}}}$ is equivalent capacitance, ${C_1},{C_2},{C_3}{\rm{ and }}{C_{\rm{n}}}$ are capacitance of capacitors in series connection.

In any electrical circuit two or more than two capacitors connected in parallel can be represented by a capacitor having equivalent capacitance. The magnitude of equivalent capacitance can be represented as follows;

${C_{{\rm{eq}}}} = {C_1} + {C_2} + {C_3}{\rm{ }}... + {C_{\rm{n}}}$

Here ${C_{{\rm{eq}}}}$ is equivalent capacitance, ${C_1},{C_2},{C_3}{\rm{ and }}{C_{\rm{n}}}$ are capacitance of capacitors in parallel connection.

(a)

The given circuit is as follows;

Calculate the equivalent capacitance of the first arm as follows;

Substitute $20{\rm{ }}\mu {\rm{F}}$ for ${C_1}$ and $30{\rm{ }}\mu {\rm{F}}$ for ${C_2}$ in the equation $\frac{1}{{{C_{12}}}} = \frac{1}{{{C_1}}} + {\frac{1}{C}_2}$ to calculate the equivalent capacitance of first arm.

$\begin{array}{c}\\\frac{1}{{{C_{12}}}} = \frac{1}{{20{\rm{ }}\mu {\rm{F}}}} + \frac{1}{{30{\rm{ }}\mu {\rm{F}}}}\\\\\frac{1}{{{C_{12}}}} = \frac{{\left( {30{\rm{ }}\mu {\rm{F}}} \right) + \left( {20{\rm{ }}\mu {\rm{F}}} \right)}}{{\left( {20{\rm{ }}\mu {\rm{F}}} \right)\left( {30{\rm{ }}\mu {\rm{F}}} \right)}}\\\\{C_{12}} = 12{\rm{ }}\mu {\rm{F}}\\\end{array}$

Hence the equivalent capacitance of first arm is $12{\rm{ }}\mu {\rm{F}}$ . The circuit can be redrawn as follows;

Calculate he equivalent capacitance of the circuit as follows:

Substitute $12{\rm{ }}\mu {\rm{F}}$ for and ${C_{12}}$ $25{\rm{ }}\mu {\rm{F}}$ for ${C_3}$ in the equation ${C_{eq}} = {C_{12}} + {C_3}$ to calculate equivalent capacitance.

$\begin{array}{c}\\{C_{eq}} = 12{\rm{ }}\mu {\rm{F}} + 25{\rm{ }}\mu {\rm{F}}\\\\ = 37{\rm{ }}\mu {\rm{F}}\\\end{array}$

Hence the equivalent capacitance of the circuit is ${\rm{37 }}\mu {\rm{F}}$ .

(b)

The given circuit is as follows;

Calculate the equivalent capacitance of the first arm as follows;

Substitute $20{\rm{ }}\mu {\rm{F}}$ for ${C_1}$ and $60{\rm{ }}\mu {\rm{F}}$ for ${C_2}$ in the equation ${C_{12}} = {C_1} + {C_2}$ to calculate the equivalent capacitance of first arm.

$\begin{array}{c}\\{C_{12}} = 20{\rm{ }}\mu {\rm{F}} + 60{\rm{ }}\mu {\rm{F}}\\\\ = 80{\rm{ }}\mu {\rm{F}}\\\end{array}$

Hence the equivalent capacitance of first loop is $80{\rm{ }}\mu {\rm{F}}$ . The circuit can be redrawn as follows;

Calculate he equivalent capacitance of the circuit as follows:

Substitute $80{\rm{ }}\mu {\rm{F}}$ for ${C_{12}}$ and $10{\rm{ }}\mu {\rm{F}}$ for ${C_3}$ in the equation $\frac{1}{{{C_{{\rm{eq}}}}}} = \frac{1}{{{C_{12}}}} + \frac{1}{{{C_{\rm{3}}}}}$ to calculate equivalent capacitance.

$\begin{array}{c}\\\frac{1}{{{C_{{\rm{eq}}}}}} = \frac{1}{{80{\rm{ }}\mu {\rm{F}}}} + \frac{1}{{10{\rm{ }}\mu {\rm{F}}}}\\\\ = \frac{9}{{80{\rm{ }}\mu {\rm{F}}}}\\\\{C_{{\rm{eq}}}} = 8.89{\rm{ }}\mu {\rm{F}}\\\end{array}$

Hence the equivalent capacitance of the circuit is ${\rm{8}}{\rm{.89 }}\mu {\rm{F}}$ .

Ans: Part a

The equivalent capacitance of the three capacitors is ${\rm{37 }}\mu {\rm{F}}$ .

Part b

The equivalent capacitance of the three capacitors is ${\rm{8}}{\rm{.89 }}\mu {\rm{F}}$ .