Chapter 33, Problem 33.81CP

To determine

The frequency will the AC source deliver a power of $250\text{W}$.

Answer to Problem 33.81CP

There are two values of frequencies at which the AC source deliver a power of $250\text{W}$ that are $58.7\text{Hz}$ and $35.9\text{Hz}$ because the circuit can be either above or below resonance.

Explanation of Solution

Given info: The rms potential difference of an AC source is $120\text{V}$.

Formula to calculate the rms current in the circuit is,

$I_{\text{rms}}=\frac{\Delta V_{\text{rms}}}{Z}$

Here,

$I_{\text{rms}}$ is the rms current in the circuit.

$\Delta V_{\text{rms}}$ is the rms source voltage.

Write the expression for the power deliver to the circuit.

$P={\left(I_{\text{rms}}\right)}^{2}R$

Substitute $\frac{\Delta V_{\text{rms}}}{Z}$ for $I_{\text{rms}}$ in above circuit.

$P={\left(\frac{\Delta V_{\text{rms}}}{Z}\right)}^{2}R$

Rearrange the equation for $Z^2$.

$Z^2=\left(\frac{{\left(\Delta V_{\text{rms}}\right)}^2}{P}\right)R$ (1)

Formula to calculate the impedance of the circuit is,

$Z=\sqrt{R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2}$

Here,

$Z$ is the impedance in the circuit.

$R$ is the resistance in the circuit.

Substitute $\omega L$ for $X_{\text{L}}$ and $\frac{1}{\omega C}$ for $X_{\text{C}}$ to find $Z$.

$Z=\sqrt{R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2}$ (2)

Substitute $\sqrt{R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2}$ for $Z$ in equation (1).

$\begin{array}{c}{\left(\sqrt{R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2}\right)}^2=\left(\frac{{\left(\Delta V_{\text{rms}}\right)}^2}{P}\right)R\\ R^2+{\left(\omega L-\frac{1}{\omega C}\right)}^2=\left(\frac{{\left(\Delta V_{\text{rms}}\right)}^2}{P}\right)R\\ {\left(\omega L-\frac{1}{\omega C}\right)}^2=\left(\frac{{\left(\Delta V_{\text{rms}}\right)}^2}{P}\right)R-R^2\end{array}$ (3)

Assume $\left(\frac{{\left(\Delta V_{\text{rms}}\right)}^2}{P}\right)R-R^2$ is equal to $A^2$.

Substitute $120\text{V}$ for $\Delta V_{\text{rms}}$, $40.0\Omega$ for $R$ and $250\text{W}$ for $P$ to find the value of $A$.

$\begin{array}{c}{A}^2=\left(\frac{{\left(120\text{V}\right)}^2}{250\text{W}}\right)\left(40.0\Omega \right)-{\left(40.0\Omega \right)}^2\\ A=\sqrt{704}\Omega \end{array}$

Substitute ${\left(\omega L-\frac{1}{\omega C}\right)}^2$ for $A^2$ in equation

Simplify the equation more,

$\begin{array}{c}\left(\omega L-\frac{1}{\omega C}\right)=\pm A\\ {\omega }^{2}LC\mp \omega CA-1=0\end{array}$

Solve for first equation,

${\omega }^{2}LC-\omega CA-1=0$

Since the angular frequency is a positive quantity. So, equation can be rewrite as,

Solve the quadratic equation.

$\begin{array}{c}{\omega }_1=\frac{-\left(-CA\right)+\sqrt{{\left(-CA\right)}^2-4LC\left(-1\right)}}{2LC}\\ =\frac{A+\sqrt{A^2+4L/C}}{2L}\end{array}$

Substitute $\sqrt{704}\Omega$ for $A$, $185\text{mH}$ for $L$ and $65.0\text{μF}$ for $C$.

$\begin{array}{l}{\omega }_1=\frac{\sqrt{704}\Omega +\sqrt{704\Omega +4\times \frac{\left(185\text{mH}\times \frac{{10}^{-3}\text{H}}{1\text{mH}}\right)}{\left(65.0\text{μF}\times \frac{{10}^{-6}\text{F}}{1\text{F}}\right)}}}{2\times 185\text{mH}\times \frac{{10}^{-3}\text{H}}{1\text{mH}}}\\ =\frac{\sqrt{704}\Omega +\sqrt{12088.61}\Omega }{0.37\text{H}}\\ =368.86\text{rad}/\text{s}\end{array}$

Formula to calculate the frequency is,

$f_1=\frac{{\omega }_1}{2\pi }$

Substitute $368.86\text{rad}/\text{s}$ for ${\omega }_1$ to find $f_1$.

$\begin{array}{c}{f}_1=\frac{368.86\text{rad}/\text{s}}{2\pi }\\ =58.7\text{Hz}\end{array}$

Solve for second equation,

${\omega }^{2}LC+\omega CA-1=0$

Since the angular frequency is a positive quantity. So, equation can be rewrite as,

Solve the quadratic equation.

$\begin{array}{c}{\omega }_2=\frac{-\left(CA\right)+\sqrt{{\left(CA\right)}^2-4LC\left(-1\right)}}{2LC}\\ =\frac{-A+\sqrt{A^2+4L/C}}{2L}\end{array}$

Substitute $\sqrt{704}\Omega$ for $A$, $185\text{mH}$ for $L$ and $65.0\text{μF}$ for $C$.

$\begin{array}{c}{\omega }_2=\frac{-\sqrt{704}\Omega +\sqrt{704\Omega +4\times \frac{\left(185\text{mH}\times \frac{{10}^{-3}\text{H}}{1\text{mH}}\right)}{\left(65.0\text{μF}\times \frac{{10}^{-6}\text{F}}{1\text{F}}\right)}}}{2\times 185\text{mH}\times \frac{{10}^{-3}\text{H}}{1\text{mH}}}\\ =\frac{-\sqrt{704}\Omega +\sqrt{12088.61}\Omega }{0.37\text{H}}\\ =225.44\text{rad}/\text{s}\end{array}$

Formula to calculate the frequency is,

$f_2=\frac{{\omega }_2}{2\pi }$

Substitute $225.44\text{rad}/\text{s}$ for ${\omega }_2$ to find $f_1$.

$\begin{array}{c}{f}_1=\frac{225.44\text{rad}/\text{s}}{2\pi }\\ =35.9\text{Hz}\end{array}$

Since the value of frequency comes out to be two that means the circuit deliver the given power two times one is below resonance and another one is above resonance.

Conclusion:

Therefore, there are two values of frequencies at which the AC source deliver a power of $250\text{W}$ that are $58.7\text{Hz}$ and $35.9\text{Hz}$ because the circuit can be either above or below resonance.