Chapter 20, Problem 82E

(a)

To determine

The value of $\omega$ for greatest current.

Answer to Problem 82E

The value of $\omega$ for the greatest current is $10{}^3\text{rad/s}$.

Explanation of Solution

Write the expression of $\omega$ for greatest current for a $RLC$ circuit.

    $\omega =\frac{1}{\sqrt{LC}}$        (1)

Here, $\omega$ is the resonant frequency, $L$ is the inductance and $R$ is the resistance.

Conclusion:

Substitute $0.01\text{H}$ for $L$ and ${10}^{-4}\text{F}$ for $C$ in equation (1).

    $\begin{array}{c}\omega =\frac{1}{\sqrt{\left(0.01\text{H}\right)\left({10}^{-4}\text{F}\right)}}\\ =10{}^3\text{rad/s}\end{array}$

Thus, the value of $\omega$ for the greatest current is $10{}^3\text{rad/s}$.

(b)

To determine

The magnitude of the frequency.

Answer to Problem 82E

The corresponding frequency is $1570.8\text{Hz}$.

Explanation of Solution

Write the expression for the frequency for a series $RLC$ circuit.

    $f=\frac{\omega }{2\pi }$        (2)

Here, $f$ is corresponding frequency.

Conclusion:

Substitute $10{}^3\text{rad/s}$ for $\omega$ in equation (2).

    $\begin{array}{c}f=\frac{{10}^3\text{rad/s}}{2\pi }\\ =1570.8\text{Hz}\end{array}$

Thus, the corresponding frequency is $1570.8\text{Hz}$.

(c)

To determine

The power that is dissipated at the frequency.

Answer to Problem 82E

The power dissipated is $20\text{W}$.

Explanation of Solution

Write the expression for the impedance of a series $RLC$ circuit.

    $Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$        (3)

Here, $Z$ is the impedance of the circuit, $R$ is the resistance, $X_L$ inductive impedance and $X_C$ is capacitive impedance.

Write the expression for inductive impedance.

    $X_L=\omega L$        (4)

Write the expression for capacitive impedance.

    $X_C=\frac{1}{\omega C}$        (5)

Write the expression for rms value of current.

    $i_e=\frac{v_e}{Z}$        (6)

Here, $i_e$ is the rms current and $v_e$ is the supply voltage.

Write the expression for the power dissipation in the circuit.

    $P=i_e{v}_e\left(\frac{R}{Z}\right)$        (7)

Here, $P$ is the power.

Conclusion:

Substitute $0.01\text{H}$ for $L$ and ${10}^3\text{rad/s}$ for $\omega$ in equation (4).

    $\begin{array}{c}{X}_L=\left(0.01\text{H}\right)\left({10}^3\text{rad/s}\right)\\ =10\text{ohm}\end{array}$

Substitute ${10}^{-4}\text{F}$ for $C$ and ${10}^3\text{rad/s}$ for $\omega$ in equation (5).

    $\begin{array}{c}{X}_C=\frac{1}{\left({10}^{-4}\text{F}\right)\left({10}^3\text{rad/s}\right)}\\ =10\text{ohm}\end{array}$

Substitute $5\text{ohm}$ for $R$, $10\text{ohm}$ for $X_L$ and $10\text{ohm}$ for $X_C$ in equation (3)

    $\begin{array}{c}Z=\sqrt{{\left(5\text{ohm}\right)}^2+{\left(10\text{ohm}-10\text{ohm}\right)}^2}\\ =5\text{ohm}\end{array}$

Substitute $5\text{ohm}$ for $Z$ and $10\text{V}$ for $v_e$ in equation (6).

    $\begin{array}{c}{i}_e=\frac{10\text{V}}{5\text{ohm}}\\ =2\text{A}\end{array}$

Substitute $2\text{A}$ for $i_e$, $10\text{V}$ for $v_e$, $5\text{ohm}$ for $R$ and $5\text{ohm}$ for $Z$.

    $\begin{array}{c}P=\left(2\text{A}\right)\left(10\text{V}\right)\left(\frac{5\text{ohm}}{5\text{ohm}}\right)\\ =20\text{W}\end{array}$

Thus, the power dissipated is $20\text{W}$.