Chapter 21, Problem 86P

(a)

To determine

The resonant frequency of the circuit.

Answer to Problem 86P

The resonant frequency of the circuit is $\underset{\_}{445\text{Hz}}$.

Explanation of Solution

Write the expression for the resonant frequency of the circuit.

$f_0=\frac{1}{2\pi \sqrt{LC}}$

Here, $L$ is the inductance of the circuit and $C$ is the capacitance of the circuit.

Conclusion:

Substitute $0.146\text{H}$ for $L$ and $877\times {10}^{-9}\text{F}$ for $C$ in the above equation.

$\begin{array}{c}{f}_0=\frac{1}{2\pi \sqrt{\left(0.146\text{H}\right)\left(877\times {10}^{-9}\text{F}\right)}}\\ =445\text{Hz}\end{array}$

Therefore, the resonant frequency of the circuit is $\underset{\_}{445\text{Hz}}$.

(b)

To determine

Whether the current or voltage will lead.

Answer to Problem 86P

The current leads the voltage.

Explanation of Solution

The frequency applied is less than the resonant frequency. At the resonance, the capacitive reactance is equal to the inductive reactance. The applied frequency is small, so the capacitive reactance will be greater due to the relation, $X_{\text{C}}\propto f^{-1}$.

The inductive reactance is small using the relation $X_{\text{L}}\propto f$. The condition when voltage lags the current is when $X_{\text{C}}>X_{\text{L}}$. Therefore, the current leads the voltage.

(c)

To determine

The phase angle of this circuit.

Answer to Problem 86P

The phase angle of this circuit is $\underset{\_}{-67°}$.

Explanation of Solution

Write the expression of phase angle of the circuit.

$\varphi ={\text{tan}}^{-1}\left(\frac{\omega L-\frac{1}{\omega C}}{R}\right)$

Substitute $0.5{\omega }_0$ for $\omega$ in the above equation.

$\varphi ={\text{tan}}^{-1}\left(\frac{0.5{\omega }_{0}L-\frac{1}{0.5{\omega }_{0}C}}{R}\right)$

Substitute $\frac{1}{\sqrt{LC}}$ for ${\omega }_0$ in the above equation.

$\varphi ={\text{tan}}^{-1}\left(\frac{\frac{0.5L}{\sqrt{LC}}-\frac{\sqrt{LC}}{0.5C}}{R}\right)$

Conclusion:

Substitute $255\Omega$ for $R$, $0.146\text{H}$ for $L$ and $877\times {10}^{-9}\text{F}$ for $C$ in the above equation.

$\begin{array}{c}\varphi ={\text{tan}}^{-1}\left(\frac{\frac{0.5\left(0.146\text{H}\right)}{\sqrt{\left(0.146\text{H}\right)\left(877\times {10}^{-9}\text{F}\right)}}-\frac{\sqrt{\left(0.146\text{H}\right)\left(877\times {10}^{-9}\text{F}\right)}}{0.5\left(877\times {10}^{-9}\text{F}\right)}}{255\Omega }\right)\\ =-67°\end{array}$

Therefore, the phase angle of this circuit is $\underset{\_}{-67°}$.

(d)

To determine

The rms current in this circuit.

Answer to Problem 86P

The rms current in this circuit is $\underset{\_}{0.51\text{A}}$.

Explanation of Solution

Write the expression for rms current.

$I_{\text{rms}}=\frac{{\epsilon }_{\text{m}}\text{cos}\varphi }{\sqrt{2}R}$

Here, ${\epsilon }_{\text{m}}$ is the peak voltage and $I_{\text{rms}}$ is the rms current.

Conclusion:

Substitute $480\text{V}$ for ${\epsilon }_{\text{m}}$, $255\Omega$ for $R$ and $-67.4°$ for $\varphi$ in the above equation.

$\begin{array}{c}{I}_{\text{rms}}=\frac{\left(480\text{V}\right)\text{cos}\left(-67.4°\right)}{\sqrt{2}\left(255\Omega \right)}\\ =0.51\text{A}\end{array}$

Therefore, the rms current in this circuit is $\underset{\_}{0.51\text{A}}$.

(e)

To determine

The average power dissipated in the circuit.

Answer to Problem 86P

The average power dissipated in the circuit is $\underset{\_}{67\text{W}}$.

Explanation of Solution

Write the expression for average power.

$P_{\text{av}}=I_{\text{rms}}^{2}R$

Here, $P_{\text{av}}$ is the average power dissipated.

Conclusion:

Substitute $0.512\text{A}$ for $I_{\text{rms}}$ and $255\Omega$ for $R$ in the above equation.

$\begin{array}{c}{P}_{\text{av}}={\left(0.512\text{A}\right)}^2\left(255\Omega \right)\\ =67\text{W}\end{array}$

Therefore, the average power dissipated in the circuit is $\underset{\_}{67\text{W}}$.

(f)

To determine

The maximum voltage across the resistor, inductor and capacitor.

Answer to Problem 86P

The maximum voltage across the resistor, inductor and capacitor are $\underset{\_}{180\text{V}}$, $\underset{\_}{150\text{V}}$ and $\underset{\_}{590\text{V}}$ respectively.

Explanation of Solution

Write the expression for the maximum voltage across the resistor.

$V_{\text{R}}=IR$

Substitute $\sqrt{2}{I}_{\text{rms}}$ for $I$ in the above equation.

$V_{\text{R}}=\sqrt{2}{I}_{\text{rms}}R$ (I)

Write the expression for the maximum voltage across the inductor.

$V_{\text{L}}=I{X}_{\text{L}}$

Here, $X_{\text{L}}$ is the reactance of the inductor.

Substitute $\sqrt{2}{I}_{\text{rms}}$ for $I$ and $\omega L$ for $X_{\text{L}}$ the above equation.

$V_{\text{L}}=\sqrt{2}{I}_{\text{rms}}\omega L$ (II)

Write the expression for the maximum voltage across the capacitor.

$V_{\text{C}}=I{X}_{\text{C}}$

Here, $X_{\text{C}}$ is the reactance of the capacitor.

Substitute $\sqrt{2}{I}_{\text{rms}}$ for $I$ and $\frac{1}{\omega C}$ for $X_{\text{C}}$ the above equation.

$V_{\text{C}}=\sqrt{2}{I}_{\text{rms}}\frac{1}{\omega C}$ (III)

Conclusion:

Substitute $0.512\text{A}$ for $I_{\text{rms}}$ and $255\Omega$ for $R$ in equation (I).

$\begin{array}{c}{V}_{\text{R}}=\sqrt{2}\left(0.512\text{A}\right)\left(255\Omega \right)\\ =180\text{V}\end{array}$

Substitute $0.512\text{A}$ for $I_{\text{rms}}$, $2\pi \left(0.50\right)\left(445\text{Hz}\right)$ for $\omega$ and $0.146\text{H}$ for $L$ in equation (II).

$\begin{array}{c}{V}_{\text{L}}=\sqrt{2}\left(0.512\text{A}\right)\left(2\pi \left(0.50\right)\left(445\text{Hz}\right)\right)\left(0.146\text{H}\right)\\ =150\text{V}\end{array}$

Substitute $0.512\text{A}$ for $I_{\text{rms}}$, $2\pi \left(0.50\right)\left(445\text{Hz}\right)$ for $\omega$ and $877\times {10}^{-9}\text{F}$ for $C$ in equation (III).

$\begin{array}{c}{V}_{\text{C}}=\sqrt{2}\left(0.512\text{A}\right)\frac{1}{\left(2\pi \left(0.50\right)\left(445\text{Hz}\right)\right)\left(877\times {10}^{-9}\text{F}\right)}\\ =590\text{V}\end{array}$

Therefore, the maximum voltage across the resistor, inductor and capacitor are $\underset{\_}{180\text{V}}$, $\underset{\_}{150\text{V}}$ and $\underset{\_}{590\text{V}}$ respectively.