Chapter 33, Problem 79CP

(a)

To determine

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

Answer to Problem 79CP

The angular frequency at which power delivered to the resistor is a maximum is $224\text{rad}/\text{s}$.

Explanation of Solution

Write the expression to obtain the resonance frequency in a series $RLC$ circuit.

    ${\omega }_0=\frac{1}{\sqrt{LC}}$

Here, ${\omega }_0$ is the resonance frequency, $L$ is the inductance of the inductor and $C$ is the capacitance of the capacitor.

Conclusion:

Substitute $10.0\mu \text{F}$ for $C$ and $2.00\text{H}$ for $L$ in the above equation to calculate ${\omega }_0$.

    $\begin{array}{c}{\omega }_0=\frac{1}{\sqrt{\left(2.00\text{H}\right)\left(10.0\mu \text{F}\right)}}\\ =\frac{1}{\sqrt{\left(2.00\text{H}\right)\left(10.0\mu \text{F}\times \frac{1\text{F}}{{10}^6\mu \text{F}}\right)}}\\ =224\text{rad}/\text{s}\end{array}$

Therefore, the angular frequency at which power delivered to the resistor is a maximum is $224\text{rad}/\text{s}$.

(b)

To determine

The average power delivered at the frequency which is calculates in part (a).

Answer to Problem 79CP

The average power delivered at the frequency which is calculates in part (a) is $500\text{W}$.

Explanation of Solution

Write the expression to obtain the average power delivered to the circuit at the calculated frequency.

    $P_{\text{avg}}=\frac{1}{2}\Delta V_{\max }{I}_{\max }$

Here, $P$ is the average power delivered to the circuit at the calculated frequency, $\Delta V_{\max }$ is the maximum voltage across the circuit and $I_{\max }$ is the maximum current across the circuit.

Substitute $\frac{\Delta V_{\max }}{R}$ for $I_{\max }$ in the above equation.

    $\begin{array}{c}{P}_{\text{avg}}=\frac{1}{2}\Delta V_{\max }\left(\frac{\Delta V_{\max }}{R}\right)\\ =\frac{1}{2}\frac{\Delta V_{\max }^2}{R}\end{array}$

Conclusion:

Substitute $100\text{V}$ for $\Delta V_{\max }$ and $10.0\text{Ω}$ for $R$ in the above equation to calculate $P_{\text{avg}}$.

    $\begin{array}{c}{P}_{\text{avg}}=\frac{1}{2}\frac{{\left(100\text{V}\right)}^2}{10.0\text{Ω}}\\ =\frac{1}{2}\frac{10000{\text{V}}^2}{10.0\text{Ω}}\\ =500\text{W}\end{array}$

Therefore, the average power delivered at the frequency which is calculates in part (a) is $500\text{W}$.

(c)

To determine

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

Answer to Problem 79CP

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

Explanation of Solution

Write the expression to obtain the average power delivered in the circuit.

    $P_{\text{avg}}=\frac{{\left(\Delta V_{\max }\right)}^{2}R}{2\left(R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}$

Here, $P_{\text{avg}}$ is the average power delivered in the circuit, ${\Delta }_{\max }$ is the maximum voltage in the circuit, $R$ is the resistance of the resistor, $X_{\text{L}}$ is the inductive reactance and $X_{\text{C}}$ is the capacitive reactance.

Substitute $\frac{1}{2}\frac{\Delta V_{\max }^2}{2R}$ for $P_{\text{avg}}$ in the above equation.

    $\begin{array}{c}\frac{1}{2}\frac{\Delta V_{\max }^2}{2R}=\frac{{\left(\Delta V_{\max }\right)}^{2}R}{2\left(R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}\\ \frac{1}{2R}=\frac{R}{\left(R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}\\ \left(R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)=2R^2\\ {\left(X_{\text{L}}-X_{\text{C}}\right)}^2=R^2\end{array}$

Further solve the above equation.

    $\left(X_{\text{L}}-X_{\text{C}}\right)=\pm R$

Case (i): when $\left(X_{\text{L}}-X_{\text{C}}\right)=R$

Substitute ${\omega }_{1}L$ for $X_{\text{L}}$ and $\frac{1}{{\omega }_{1}C}$ for $X_{\text{C}}$ in the above equation.

    $\begin{array}{c}\left({\omega }_{1}L-\frac{1}{{\omega }_{1}C}\right)=R\\ {\omega }_1^{2}LC-1=R{\omega }_{1}C\\ {\omega }_1^{2}LC-R{\omega }_{1}C-1=0\end{array}$                                                                                       (I)

Write expression to obtain the quadratic equation.

    $a{x}^2+bx+c=0$                                                                                                       (II)

Here, $x$ is the variable and $a,b$ and $c$ are constants.

Compare equation (I) and (II).

    $a=LC$, $b=-RC$, $c=-1$ and $x={\omega }_1$.

Write the expression to obtain the roots of the quadratic equation.

    $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$                                                                                                (III)

Substitute $LC$ for $a$, $-RC$ for $b$, $-1$ for $c$ and ${\omega }_1$ for $x$ in equation (III).

    $\begin{array}{c}{\omega }_1=\frac{RC\pm \sqrt{{\left(RC\right)}^2-4\left(LC\right)\left(-1\right)}}{2\left(LC\right)}\\ =\frac{RC\pm \sqrt{{\left(RC\right)}^2+4\left(LC\right)}}{2\left(LC\right)}\end{array}$                                                                          (IV)

Case (ii): when $\left(X_{\text{L}}-X_{\text{C}}\right)=-R$

Substitute ${\omega }_{2}L$ for $X_{\text{L}}$ and $\frac{1}{{\omega }_{2}C}$ for $X_{\text{C}}$ in the above equation.

    $\begin{array}{c}\left({\omega }_{2}L-\frac{1}{{\omega }_{2}C}\right)=-R\\ {\omega }_2^{2}LC-1=-R{\omega }_{2}C\\ {\omega }_2^{2}LC+R{\omega }_{2}C-1=0\end{array}$                                                                                   (V)

Compare equation (IV) and (II).

    $a=LC$, $b=RC$, $c=-1$ and $x={\omega }_1$.

Substitute $LC$ for $a$, $RC$ for $b$, $-1$ for $c$ and ${\omega }_2$ for $x$ in equation (V)

    $\begin{array}{c}{\omega }_2=\frac{-RC\pm \sqrt{{\left(RC\right)}^2-4\left(LC\right)\left(-1\right)}}{2\left(LC\right)}\\ =\frac{-RC\pm \sqrt{{\left(RC\right)}^2+4\left(LC\right)}}{2\left(LC\right)}\end{array}$                                                                   (IV)

Conclusion:

Substitute $10.0\text{Ω}$ for $R$, $10.0\mu \text{F}$ for $C$ and $2.00\text{H}$ for $L$ in equation (IV) to calculate ${\omega }_1$.

    $\begin{array}{c}{\omega }_1=\frac{\left(10.0\text{Ω}\right)\left(10.0\mu \text{F}\right)\pm \sqrt{{\left(\left(10.0\text{Ω}\right)\left(10.0\mu \text{F}\right)\right)}^2+4\left(\left(2.00\text{H}\right)\left(10.0\mu \text{F}\right)\right)}}{2\left(\left(2.00\text{H}\right)\left(10.0\mu \text{F}\right)\right)}\\ =\left[\frac{\left(10.0\text{Ω}\right)\left(10.0\mu \text{F}\times \frac{1\text{F}}{{10}^6\mu \text{F}}\right)\pm \sqrt{\begin{array}{l}{\left(\left(10.0\text{Ω}\right)\left(10.0\mu \text{F}\times \frac{1\text{F}}{{10}^6\mu \text{F}}\right)\right)}^2\\ +4\left(\left(2.00\text{H}\right)\left(10.0\mu \text{F}\times \frac{1\text{F}}{{10}^6\mu \text{F}}\right)\right)\end{array}}}{2\left(\left(2.00\text{H}\right)\left(10.0\mu \text{F}\times \frac{1\text{F}}{{10}^6\mu \text{F}}\right)\right)}\right]\\ =-221{\text{s}}^{-1}\\ or\end{array}$

    ${\omega }_1=226{\text{s}}^{-1}$

Consider the positive value of ${\omega }_1$.

Substitute $10.0\text{Ω}$ for $R$, $10.0\mu \text{F}$ for $C$ and $2.00\text{H}$ for $L$ in equation (VI) to calculate ${\omega }_2$.

    $\begin{array}{c}{\omega }_2=\frac{-\left(10.0\text{Ω}\right)\left(10.0\mu \text{F}\right)\pm \sqrt{{\left(\left(10.0\text{Ω}\right)\left(10.0\mu \text{F}\right)\right)}^2+4\left(\left(2.00\text{H}\right)\left(10.0\mu \text{F}\right)\right)}}{2\left(\left(2.00\text{H}\right)\left(10.0\mu \text{F}\right)\right)}\\ =\left[\frac{-\left(10.0\text{Ω}\right)\left(10.0\mu \text{F}\times \frac{1\text{F}}{{10}^6\mu \text{F}}\right)\pm \sqrt{\begin{array}{l}{\left(\left(10.0\text{Ω}\right)\left(10.0\mu \text{F}\times \frac{1\text{F}}{{10}^6\mu \text{F}}\right)\right)}^2\\ +4\left(\left(2.00\text{H}\right)\left(10.0\mu \text{F}\times \frac{1\text{F}}{{10}^6\mu \text{F}}\right)\right)\end{array}}}{2\left(\left(2.00\text{H}\right)\left(10.0\mu \text{F}\times \frac{1\text{F}}{{10}^6\mu \text{F}}\right)\right)}\right]\\ =-226{\text{s}}^{-1}\\ or\end{array}$

    ${\omega }_2=221{\text{s}}^{-1}$

Consider the positive value of ${\omega }_2$.

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