Chapter 21, Problem 65P

(a)

To determine

The resonant angular frequency ${\omega }_0$.

Answer to Problem 65P

Resonant angular frequency is $750\text{rad}/\text{s}$.

Explanation of Solution

Write the equation to find the resonant frequency of a series LCR circuit.

    ${\omega }_0=\sqrt{\frac{1}{LC}}$

Here, ${\omega }_0$ is the resonant angular frequency, $L$ is the inductance, and $C$ is the capacitance.

Conclusion:

Substitute $0.800\text{H}$ for $L$ and $2.22\text{μF}$ for $C$ in the above equation to find ${\omega }_0$.

    $\begin{array}{c}{\omega }_0=\sqrt{\frac{1}{\left(0.800\text{H}\right)\left(2.22\text{μF}\left(\frac{{10}^{-6}\text{F}}{1\text{μF}}\right)\right)}}\\ =\sqrt{5.6\times {10}^5{\text{rad}}^2/{\text{s}}^2}\\ =750\text{rad}/\text{s}\end{array}$

Therefore, the resonant angular frequency is $750\text{rad}/\text{s}$.

(b)

To determine

Draw the phasor diagram at resonance.

Explanation of Solution

Write the relation between inductive and capacitive reactance at resonance.

    $X_L=X_C$

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

Write the relation between voltages across inductor and capacitor at resonance.

    $V_L=V_C$

Here, $V_L$ is the voltage across inductor and $V_C$ is the voltage across capacitor at resonance.

Write the relation between impedance and resistance at resonance.

    $Z=R$

Here, $Z$ is the impedance and $R$ is the resistance.

Draw the Phasor diagram at resonance.

Here, $V_R$ is the voltage across the resistance $R$. The phasor diagram shows that $V_L$ and $V_C$ differ by $180°$ out of phase.

(c)

To determine

The rms voltages $V_{ab},V_{bc},V_{cd},V_{bd},\text{and}{V}_{ad}$.

Answer to Problem 65P

Voltages $V_{ab},V_{bc},V_{cd},V_{bd},\text{and}{V}_{ad}$ are $440\text{V,1}\text{.1}\text{kV, 1}\text{.1}\text{kV, 0}\text{V}\text{and}440\text{V}$ respectively.

Explanation of Solution

Write the equation to find the rms current.

    $I_{rms}=\frac{{\epsilon }_{rms}}{Z}$                                                                                                                (I)

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

Write the equation to find $V_{ab}$.

  $V_{ab}=I_{rms}R$                                                                                                              (II)

Write the equation to find $V_{bc}$.

  $V_{bc}=I_{rms}{X}_L$                                                                                                           (III)

Write the equation for $X_L$.

    $X_L={\omega }_{0}L$                                                                                                               (IV)

Rewrite equation (III) by substituting equations (I) and (IV).

    $V_{bc}=\left(\frac{{\epsilon }_{rms}}{Z}\right)\left({\omega }_{0}L\right)$                                                                                                 (V)

Write the relation between $V_{bc}$ and $V_{cd}$.

    $V_{bc}=V_{cd}$                                                                                                              (VI)

Write the equation to find $V_{ad}$.

  $V_{ad}={\epsilon }_{rms}$                                                                                                           (VII)

Conclusion:

Substitute $440\text{V}$ for ${\epsilon }_{rms}$ and $250\Omega$ for $R$ in equation (I) to find $I_{rms}$.

    $\begin{array}{c}{I}_{rms}=\frac{440\text{V}}{250\Omega }\\ =1.76\text{A}\end{array}$

Substitute $1.76\text{A}$ for $I_{rms}$ and $250\Omega$ for $R$ in equation (II) to find $V_{ab}$.

  $\begin{array}{c}{V}_{ab}=\left(1.76\text{A}\right)\left(250\Omega \right)\\ =440\text{V}\end{array}$

Substitute $440\text{V}$ for ${\epsilon }_{rms}$, $750\text{rad}/\text{s}$ for ${\omega }_0$, $0.800\text{H}$ for $L$, and $250\Omega$ for $Z$ in equation (III) to find $V_{bc}$.

  $\begin{array}{c}{V}_{bc}=\left(\frac{440\text{V}}{250\Omega }\right)\left(\left(750\text{rad}/\text{s}\right)\left(0.800\text{H}\right)\right)\\ =1100\text{V}\left(\frac{\text{1}\text{kV}}{{\text{10}}^{\text{3}}\text{V}}\right)\\ =1.1\text{kV}\end{array}$

Substitute $1.1\text{kV}$ for $V_{bc}$ in equation (VI) to find $V_{cd}$.

Since the voltage across the inductor and capacitor is $180°$ out of phase, $V_{bd}$ will be zero.

Substitute $440\text{V}$ for ${\epsilon }_{rms}$ in equation (VII) to find $V_{ad}$.

  $V_{ad}=440\text{V}$

Therefore, voltages $V_{ab},V_{bc},V_{cd},V_{bd},\text{and}{V}_{ad}$ are $440\text{V,1}\text{.1}\text{kV, 1}\text{.1}\text{kV, 0}\text{V}\text{and}440\text{V}$ respectively.

(d)

To determine

The angular frequency of resonance at $R=125\Omega$.

Answer to Problem 65P

Resonant angular frequency is $750\text{rad}/\text{s}$.

Explanation of Solution

Resonant angular frequency is independent of resistance in the circuit. That means the value ${\omega }_0$ remains unchanged even the value of $R$ is changed.

Therefore, the resonant angular frequency is $750\text{rad}/\text{s}$.

(e)

To determine

New rms current with $R=125\Omega$.

Answer to Problem 65P

New rms current is $3.5\text{A}$.

Explanation of Solution

Write the equation to find $I_{rms}$.

  $I_{rms}=\frac{{\epsilon }_{rms}}{R}$

Conclusion:

Substitute $440\text{V}$ for ${\epsilon }_{rms}$ and $125\Omega$ for $R$ in the above equation to find $I_{rms}$.

    $\begin{array}{c}{I}_{rms}=\frac{440\text{V}}{125\Omega }\\ =3.5\text{A}\end{array}$

Therefore, new rms current is $3.5\text{A}$.