Chapter 20, Problem 93E

(a)

To determine

The capacitance of the circuit.

Answer to Problem 93E

The capacitance of the circuit is $1.29\times {10}^{-9}\text{F}$.

Explanation of Solution

Write the expression for the resonant frequency of the circuit.

    $f_0=\frac{1}{2\pi \sqrt{LC}}$

Here, $f_0$ is the resonant frequency, $L$ is the inductance and $C$ is the capacitance.

Rearrange the above equation.

    $C=\frac{1}{4{\pi }^2{f}_0^{2}L}$        (1)

Conclusion:

Substitute $1.4\times {10}^6\text{Hz}$ for $f_0$ and ${10}^{-5}\text{H}$ for $L$ in equation (1).

    $\begin{array}{c}C=\frac{1}{4{\pi }^2{\left(1.4\times {10}^6\text{Hz}\right)}^2\left({10}^{-5}\text{H}\right)}\\ \approx 1.29\times {10}^{-9}\text{F}\end{array}$

Thus, the capacitance of the circuit is $1.29\times {10}^{-9}\text{F}$.

(b)

To determine

The reactance at the frequency above $1$ percent of $f_0$.

Answer to Problem 93E

The reactance of the circuit is $1.75\text{ohm}$.

Explanation of Solution

Write the expression for the reactance of the circuit.

    $X=X_L-X_C$        (2)

Here, $X$ is the reactance, $X_L$ inductive impedance and $X_C$ is capacitive impedance.

Write the expression for inductive impedance.

    $X_L=2\pi fL$        (3)

Write the expression for capacitive impedance.

    $X_C=\frac{1}{2\pi fC}$        (4)

Conclusion:

The value of the new frequency is above $1$ percent of $f_0$.

Write the expression for $f_0$.

    $f=f_0+\frac{f_0}{100}$

Substitute $\left(1.4\times {10}^6\text{Hz}\right)$ for $f_0$ in the above equation.

    $\begin{array}{c}f=\left(1.4\times {10}^6\text{Hz}\right)+\frac{\left(1.4\times {10}^6\text{Hz}\right)}{100}\\ =1.414\times {10}^6\text{Hz}\end{array}$

Substitute $1.414\times {10}^6\text{Hz}$ for $f$ and ${10}^{-5}\text{H}$ for $L$ in equation (3).

    $\begin{array}{c}{X}_L=2\pi \left(1.414\times {10}^6\text{Hz}\right)\left({10}^{-5}\text{Hz}\right)\\ \approx 88.8442\text{ohm}\end{array}$

Substitute $1.29\times {10}^{-9}\text{F}$ for $C$ and $1.414\times {10}^6\text{Hz}$ for $f$ in equation (4).

    $\begin{array}{c}{X}_C=\frac{1}{2\pi \left(1.414\times {10}^6\text{Hz}\right)\left(1.29\times {10}^{-9}\text{F}\right)}\\ \approx 87.2531\text{ohm}\end{array}$

Substitute $88.8442\text{ohm}$ for $X_L$ and $87.2531\text{ohm}$ for $X_C$ in equation (2).

    $\begin{array}{c}X=88.8442\text{ohm}-87.2531\text{ohm}\\ =1.75\text{ohm}\end{array}$

Thus, the reactance of the circuit is $1.75\text{ohm}$.

(c)

To determine

The value of the resistance.

Answer to Problem 93E

The value of the resistance is $0.405\text{ohm}$.

Explanation of Solution

Write the expression for the impedance for the frequency $f_0$.

    $Z_{f_0}=\sqrt{R^2+X_{f_0}^2}$        (5)

Here, $Z_{f_0}$ is the impedance, $R$ is the resistance and $X_{f_0}$ is the reactance.

Write the expression for the impedance for the frequency $f$.

    $Z_f=\sqrt{R^2+X_f^2}$        (6)

Here, $Z_f$ is the impedance, $R$ is the resistance and $X_f$ is the reactance.

Divide equation (5) by equation (6)

    $\frac{Z_f}{Z_{f_0}}=\frac{\sqrt{R^2+X_f^2}}{\sqrt{R^2+X_{f_0}^2}}$        (7)

Write the expression for the reactance of the circuit.

    $X_f=X_L-X_C$

Write the expression for inductive impedance.

    $X_L=2\pi fL$

Write the expression for capacitive impedance.

    $X_C=\frac{1}{2\pi fC}$

Conclusion:

Substitute $1.4\times {10}^6\text{Hz}$ for $f$ and ${10}^{-5}\text{H}$ for $L$ in equation (3).

    $\begin{array}{c}{X}_L=2\pi \left(1.4\times {10}^6\text{Hz}\right)\left({10}^{-5}\text{Hz}\right)\\ \approx 87.964\text{ohm}\end{array}$

Substitute $1.29\times {10}^{-9}\text{F}$ for $C$ and $1.4\times {10}^6\text{Hz}$ for $f$ in equation (4).

    $\begin{array}{c}{X}_C=\frac{1}{2\pi \left(1.4\times {10}^6\text{Hz}\right)\left(1.29\times {10}^{-9}\text{F}\right)}\\ \approx 88.125\text{ohm}\end{array}$

Substitute $88.8442\text{ohm}$ for $X_L$ and $87.2531\text{ohm}$ for $X_C$ in equation (2).

    $\begin{array}{c}{X}_f=88.125\text{ohm}-87.964\text{ohm}\\ =0.161\text{ohm}\end{array}$

Substitute $\left(\frac{1}{4}\right)$ for $\left(\frac{Z_f}{Z_{f_0}}\right)$ in equation (7).

    $\left(\frac{1}{4}\right)=\frac{\sqrt{R^2+X_f^2}}{\sqrt{R^2+X_{f_0}^2}}$

Substitute $0.161\text{ohm}$ for $X_f$ and $1.75\text{ohm}$ for $X_{f_0}$ in the above equation.

    $\frac{1}{4}=\frac{\sqrt{R^2+{\left(0.161\text{ohm}\right)}^2}}{\sqrt{R^2+{\left(1.75\text{ohm}\right)}^2}}$

Solved the above equation.

    $R=0.405\text{ohm}$

Thus, the value of the resistance is $0.405\text{ohm}$.