Chapter 33, Problem 33.79CP

(a)

To determine

 The angular frequency at which the power delivered to the resistor is maximum.

Answer to Problem 33.79CP

 The angular frequency at which the power delivered to the resistor is maximum is $224{\text{s}}^{-1}$.

Explanation of Solution

Given info: The maximum value of voltage is $100$. The value of the inductance is $2.00\text{H}$, the value of the capacitance is $10.0\text{μF}$ and the value of the resistance is $10.0\Omega$.

For the condition of maximum power through the resistor, the current through the resistor is to be greatest. For maximum value of the current the value of impedance of the circuit has to be least. The value of impedance is least at resonance frequency.

The expression for the resonance angular frequency is,

${\omega }_{\text{ο}}=\frac{1}{\sqrt{LC}}$

Here,

$L$ is the value of impedance.

$C$ is the value of capacitance.

Substitute $2.00\text{H}$ for $L$ and $10.0\text{μF}$ for $C$ in the above expression.

$\begin{array}{c}{\omega }_{\text{ο}}=\frac{1}{\sqrt{\left(2.00\text{H}\right)\left(10.0\text{μF}\right)}}\\ =223.6\\ \approx 224{\text{s}}^{-1}\end{array}$

Conclusion:

Therefore, the angular frequency at which the power delivered to the resistor is maximum is $224{\text{s}}^{-1}$.

(b)

To determine

 The average power delivered at resonance frequency.

Answer to Problem 33.79CP

The average power delivered at resonance frequency is $500\text{W}$.

Explanation of Solution

Given info: The maximum value of voltage is $100\text{V}$. The value of the inductance is $2.00\text{H}$, the value of the capacitance is $10.0\text{μF}$ and the value of the resistance is $10.0\Omega$.

The expression for power delivered is,

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

Here,

$\Delta V_{\text{m}}^{}$ is the value of maximum voltage.

$Z$ is the impedance.

For the condition of resonance the value of impedance is equal to the value of resistance.

Substitute $P_{\text{m}}$ for $P$ and $R$ for $Z$ in the above expression.

$\begin{array}{c}{P}_{\text{m}}=\frac{{\left(\frac{\Delta V_{\text{m}}^{}}{\sqrt{2}}\right)}^{2}R}{R^2}\\ =\frac{{\left(\Delta V_{\text{m}}^{}\right)}^2}{2R}\end{array}$

Substitute $10.0\Omega$ for $R$ and $100\text{V}$ $\Delta V_{\text{m}}^{}$ in the above expression.

$\begin{array}{c}{P}_{\text{m}}=\frac{{\left(100\text{V}\right)}^{\text{2}}}{2\left(10.0\Omega \right)}\\ =500\text{W}\end{array}$

Conclusion:

Therefore, the average power delivered at resonance frequency is $500\text{W}$.

(c)

To determine

 The two angular frequencies at which power is one half of the maximum value.

Answer to Problem 33.79CP

 The two angular frequencies at which power is one half of the maximum value are $221{\text{s}}^{-1}$ and $226{\text{s}}^{-1}$.

Explanation of Solution

Given info: The maximum value of voltage is $100\text{V}$. The value of the inductance is $2.00\text{H}$, the value of the capacitance is $10.0\text{μF}$ and the value of the resistance is $10.0\Omega$.

The expression for inductive reactance is,

$X_{\text{L}}=\omega L$

Here,

$\omega$ is the source angular frequency.

$L$ is the inductance.

The expression for capacitive reactance is,

$X_{\text{C}}=\frac{1}{\omega C}$

Here,

$C$ is the capacitance.

The expression for the impedance of the circuit is.

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

The power is half of the maximum power.

The expression of power in terms of max power is

$P=\frac{P_{\text{m}}}{2}$

Substitute $\frac{{\left(\frac{\Delta V_{\text{m}}^{}}{\sqrt{2}}\right)}^{2}R}{Z^2}$ for $P$ and $\frac{\Delta V_{\text{m}}^2}{2R}$ for $P_{\text{m}}$ in the above expression.

$\begin{array}{c}\frac{{\left(\frac{\Delta V_{\text{m}}^{}}{\sqrt{2}}\right)}^{2}R}{Z^2}=\frac{1}{2}\left(\frac{\Delta V_{\text{m}}^2}{2R}\right)\\ Z^2=2R^2\end{array}$

Substitute $\sqrt{R_{}^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2}$ for $Z$ in the above expression.

$\begin{array}{l}{R}_{}^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2=2R^2\\ \left(X_{\text{L}}-X_{\text{C}}\right)=\pm R\end{array}$ (1)

Substitute $\omega L$ for $X_{\text{L}}$ and $\frac{1}{\omega C}$ for $X_{\text{C}}$ in the above expression for positive value of resistance.

$\begin{array}{l}\left(\omega L-\frac{1}{\omega C}\right)=-R\\ {\omega }^{2}LC+\omega CR-1=0\end{array}$

The first values of $\omega$ are calculated by the expression,

$\omega =\frac{-CR\pm \sqrt{{\left(CR\right)}^2+4LC}}{2LC}$

Substitute ${\omega }_1$ for $\omega$,   $2.00\text{H}$ for $L$, $10.0\text{μF}$ for $C$ and $10.0\Omega$ for $R$ in the above expression.

$\begin{array}{c}{\omega }_1=\frac{-\left(10.0\text{μF}\right)\left(10.0\Omega \right)\pm \sqrt{{\left(\left(10.0\text{μF}\right)\left(10.0\Omega \right)\right)}^2+4\left(2.00\text{H}\right)\left(10.0\text{μF}\right)}}{2\left(2.00\text{H}\right)\left(10.0\text{μF}\right)}\\ =\frac{-\left(10.0\Omega \right)\pm \sqrt{{\left(10.0\Omega \right)}^2+4\left(2.00\text{H}\right){\left(10.0\text{μF}\right)}^{-1}}}{2\left(2.00\text{H}\right)}\\ =-2.5\pm 223.6{\text{s}}^{-1}\end{array}$

The value of the angular frequency is always positive, therefore the negative value is neglected

The first values of $\omega$ is,

$\begin{array}{c}{\omega }_1=221.1{\text{s}}^{-1}\\ \approx 221{\text{s}}^{-1}\end{array}$

Substitute $\omega L$ for $X_{\text{L}}$ and $\frac{1}{\omega C}$ for $X_{\text{C}}$ in the equation (1) for negative value of resistance.

$\begin{array}{l}\left(\omega L-\frac{1}{\omega C}\right)=R\\ {\omega }^{2}LC-\omega CR-1=0\end{array}$

The second values of $\omega$ is calculated by the expression,

${\omega }_{}=\frac{CR\pm \sqrt{{\left(CR\right)}^2+4LC}}{2LC}$

Substitute ${\omega }_2$ for $\omega$, $2.00\text{H}$ for $L$, $10.0\text{μF}$ for $C$ and $10.0\Omega$ for $R$ in the above expression.

$\begin{array}{c}{\omega }_2=\frac{\left(10.0\text{μF}\right)\left(10.0\Omega \right)\pm \sqrt{{\left(\left(10.0\text{μF}\right)\left(10.0\Omega \right)\right)}^2+4\left(2.00\text{H}\right)\left(10.0\text{μF}\right)}}{2\left(2.00\text{H}\right)\left(10.0\text{μF}\right)}\\ =\frac{\left(10.0\Omega \right)\pm \sqrt{{\left(10.0\Omega \right)}^2+4\left(2.00\text{H}\right){\left(10.0\text{μF}\right)}^{-1}}}{2\left(2.00\text{H}\right)}\\ =2.5\pm 223.6\end{array}$

The value of the angular frequency is always positive, therefore the negative value is neglected

The second value of $\omega$ is,

$\begin{array}{c}{\omega }_1=226.1{\text{s}}^{-1}\\ \approx 226{\text{s}}^{-1}\end{array}$

Conclusion:

Therefore, the two angular frequencies at which power is one half of the maximum value are $221{\text{s}}^{-1}$ and $226{\text{s}}^{-1}$.