Chapter 22, Problem 49GP

(a)

To determine

To Explain: Whether the capacitors are connected in series or in parallel.

Answer to Problem 49GP

Capacitors are connected in parallel.

Explanation of Solution

Introduction:

An electronic device used to store energy is called a capacitor. A simple capacitor device can be made by using two conducting plates separated by some distance and connected to a battery.

According to the given figure, the capacitors are connected in parallel, because this can change the capacitance simply by turning the capacitors to different angles.

Conclusion:

Capacitors are connected in parallel.

(b)

To determine

To Find: The range of the capacitance values.

Answer to Problem 49GP

The range is given as: $8.9\text{ pF}\le \mathrm{C}\le 80\text{ pF}$

Explanation of Solution

Given:

Separation of the plates, $\mathrm{d}=1.1\text{ mm}=1.1\times {10}^{-3}\text{m}$

Minimum area, ${\mathrm{A}}_{\min }=1.0{\text{ cm}}^2=1\times {10}^{-4}{\text{m}}^2$

Maximum area, ${\mathrm{A}}_{\max }=9.0{\text{ cm}}^2=9\times {10}^{-4}{\text{m}}^2$

Formula used:

Capacitance can be given as:

  $\mathrm{C}=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}$

Here,

  $\mathrm{C}$ is the capacitance

  ${\mathrm{\epsilon }}_0=8.854\times {10}^{-12}{\text{C}}^2\text{/N}\cdot {\text{m}}^2$ (Permittivity of free space)

  $\mathrm{A}$ is the area of cross section.

  $\mathrm{d}$ is separation of plates.

Calculation:

Range can be calculated as follows:

Firstly, for minimum area the capacitance will be:

  $\begin{array}{l}{\mathrm{C}}_{\min }=\frac{11\left(8.854\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\left(1\times {10}^{-4}{\text{m}}^2\right)}{1.1\times {10}^{-3}\text{m}}\\ {\mathrm{C}}_{\min }=8.9\times {10}^{-12}\text{F}\\ {\mathrm{C}}_{\min }=8.9\text{ pF}\end{array}$

For maximum area, the capacitance can be given as:

  $\begin{array}{l}{\mathrm{C}}_{\max }=\frac{11\left(8.854\times {10}^{-12}{\text{C}}^2/\text{N}{\text{.m}}^2\right)\left(9\times {10}^{-4}{\text{m}}^2\right)}{1.1\times {10}^{-3}\text{m}}\\ {\mathrm{C}}_{\max }=80\times {10}^{-12}\text{F}\\ {\mathrm{C}}_{\max }=80\text{ pF}\end{array}$

So, the range is given as: $8.9\text{ pF}\le \mathrm{C}\le 80\text{ pF}$

Conclusion:

The range is given as: $8.9\text{ pF}\le \mathrm{C}\le 80\text{ pF}$

(c)

To determine

To find: Value of inductor needed.

Answer to Problem 49GP

The value of inductor needed can be: $1.05\text{ mH}\le \mathrm{L}\le 1.22\text{ mH}$

Explanation of Solution

Given:

Frequency range of AM stations: 550 kHz to 1600 kHz.

Formula used:

Frequency can be given as:

  $\mathrm{f}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{1}{\mathrm{L}\mathrm{C}}}$

Here,

  $\mathrm{f}$ is the frequency.

  $\mathrm{L}$ is the inductance.

  $\mathrm{C}$ is the capacitance.

Calculation:

For lowest frequency, the value of inductor is given by:

  $\begin{array}{l}{\mathrm{f}}_1=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{1}{{\mathrm{L}}_1{\mathrm{C}}_{\max }}}\\ {\mathrm{L}}_1=\frac{1}{4{\mathrm{\pi }}^2{\mathrm{f}}_1^2{\mathrm{C}}_{\max }}\\ {\mathrm{L}}_1=\frac{1}{4{\left(3.14\right)}^2{\left(550\times {10}^3\text{Hz}\right)}^2\left(80\times {10}^{-12}\text{F}\right)}\\ {\mathrm{L}}_1=1.05\times {10}^{-3}\text{H}\end{array}$

So, the resonant frequency is given by:

  $\begin{array}{l}{\mathrm{f}}_{\max }=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{1}{{\mathrm{L}}_1{\mathrm{C}}_{\min }}}\\ {\mathrm{f}}_{\max }=\frac{1}{2\left(3.14\right)}\sqrt{\frac{1}{\left(1.05\times {10}^{-3}\text{H}\right)\left(8.9\times {10}^{-12}\text{F}\right)}}\\ {\mathrm{f}}_{\max }=1.64\times {10}^6\text{Hz}\end{array}$

For highest frequency, the value of inductor can be given as:

  $\begin{array}{l}{\mathrm{L}}_2=\frac{1}{4{\mathrm{\pi }}^2{\mathrm{f}}_{\max }^2{\mathrm{C}}_{\max }}\\ {\mathrm{L}}_2=\frac{1}{4{\left(3.14\right)}^2{\left(1.64\times {10}^6\text{Hz}\right)}^2\left(80\times {10}^{-12}\text{F}\right)}\\ {\mathrm{L}}_2=0.12\times {10}^{-3}\text{H}\end{array}$

Conclusion:

Hence, the value of inductor needed can be: $0.12\text{ mH}\le \mathrm{L}\le 1.05\text{ mH}$