Chapter 33, Problem 77CP

(a)

To determine

The inductance of the inductor.

Answer to Problem 77CP

The inductance of the inductor is $580\mu \text{H}$.

Explanation of Solution

Write the expression to obtain the inductive reactance.

    $X_{\text{L}}=2\pi fL$

Here, $X_{\text{L}}$ is the inductive reactance, $f$ is the frequency and $L$ is the inductance of the inductor.

Write the expression to obtain the capacitive reactance.

    $X_{\text{C}}=\frac{1}{2\pi fC}$

Here, $X_{\text{C}}$ is the capacitive reactance, $f$ is the frequency and $C$ is the capacitance of the capacitor.

Write the expression to obtain the output voltage across the resistor.

    $\Delta V_{\text{out}}=\frac{\Delta V_{\text{in}}R}{\sqrt{\left(R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}}$

Here, $\Delta V_{\text{out}}$ is the output voltage, $\Delta V_{\text{in}}$ is the input voltage, $R$ is the resistance of the resistor, $X_{\text{L}}$ is the inductive reactance and $X_{\text{C}}$ is the capacitive reactance.

Re-write the above equation.

    $\frac{\Delta V_{\text{out}}}{\Delta V_{\text{in}}}=\frac{R}{\sqrt{\left(R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}}$

Substitute $\frac{1}{2}$ for $\frac{\Delta V_{\text{out}}}{\Delta V_{\text{in}}}$ in the above equation.

    $\begin{array}{l}\frac{1}{2}=\frac{R}{\sqrt{\left(R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}}\\ 4R^2=R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\\ 3R^2={\left(X_{\text{L}}-X_{\text{C}}\right)}^2\\ \sqrt{3}R=\left|\left(X_{\text{L}}-X_{\text{C}}\right)\right|\end{array}$

Further substitute $2\pi fL$ for $X_{\text{L}}$ and $\frac{1}{2\pi fC}$ for $X_{\text{C}}$ in the above equation.

    $\sqrt{3}R=\left(2\pi fL-\frac{1}{2\pi fC}\right)$                                                                                         (I)

Substitute $200\text{Hz}$ for $\omega$ in equation (I).

    $\sqrt{3}R=\left(\frac{1}{2\pi \left(200\text{Hz}\right)C}-2\pi \left(200\text{Hz}\right)L\right)$                                                           (II)

Substitute $4.00\times {10}^3\text{Hz}$ for $\omega$ in equation (I).

    $\sqrt{3}R=\left(2\pi \left(4.00\times {10}^3\text{Hz}\right)L-\frac{1}{2\pi \left(4.00\times {10}^3\text{Hz}\right)C}\right)$                                       (III)

On multiplying equation (II) by $20$ and add with the equation (III).

    $\begin{array}{l}20\sqrt{3}R+\sqrt{3}R=\left[\begin{array}{l}\frac{20}{2\pi \left(200\text{Hz}\right)C}-2\pi \left(4000\text{Hz}\right)L\\ +2\pi \left(4000\text{Hz}\right)L-\frac{1}{2\pi \left(4000\text{Hz}\right)C}\end{array}\right]\\ 20\sqrt{3}R+\sqrt{3}R=\frac{20}{2\pi \left(200\text{Hz}\right)C}+\frac{1}{2\pi \left(4000\text{Hz}\right)C}\\ C=\frac{1}{21\sqrt{3}R}\left(\frac{20}{2\pi \left(200\text{Hz}\right)}+\frac{1}{2\pi \left(4000\text{Hz}\right)C}\right)\end{array}$                                 (IV)

Conclusion:

Substitute $8.00\text{Ω}$ for $R$ in equation (IV) to calculate $C$.

    $\begin{array}{c}C=\frac{1}{21\sqrt{3}\left(8.00\text{Ω}\right)}\left(\frac{20}{2\pi \left(200\text{Hz}\right)}+\frac{1}{2\pi \left(4000\text{Hz}\right)}\right)\\ =54.6\times {10}^{-6}\text{F}\times \frac{1\mu \text{F}}{{10}^6\text{F}}\\ =54.6\mu \text{F}\end{array}$

Substitute $54.6\times {10}^{-6}\text{F}$ for $C$ and $8.00\text{Ω}$ for $R$ in equation (II) to calculate $L$.

    $\begin{array}{c}\sqrt{3}\left(8.00\Omega \right)=\left(2\pi \left(4.00\times {10}^3\text{Hz}\right)L-\frac{1}{2\pi \left(4.00\times {10}^3\text{Hz}\right)\left(54.6\times {10}^{-6}\text{F}\right)}\right)\\ 2\pi \left(4.00\times {10}^3\text{Hz}\right)L=\sqrt{3}\left(8.00\Omega \right)+\frac{1}{2\pi \left(4.00\times {10}^3\text{Hz}\right)\left(54.6\times {10}^{-6}\text{F}\right)}\\ L=580\times {10}^{-6}\text{H}\times \frac{1\mu \text{H}}{{10}^{-6}\text{H}}\\ =580\mu \text{H}\end{array}$

Therefore, the inductance of the inductor is $580\mu \text{H}$.

(b)

To determine

The capacitance of the capacitor.

Answer to Problem 77CP

The capacitance of the capacitor is $54.6\mu \text{F}$.

Explanation of Solution

Conclusion:

As the value of capacitance of the capacitor is already calculated in part (a) that is equal to $54.6\mu \text{F}$.

Therefore, the capacitance of the capacitor is $54.6\mu \text{F}$.

(c)

To determine

The maximum value of the ratio of $\Delta V_{\text{out}}/\Delta V_{\text{in}}$.

Answer to Problem 77CP

The maximum value of the ratio of $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ is $1.00$.

Explanation of Solution

The maximum value of the $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ is obtained when inductive reactance is equal to the capacitive reactance.

    $X_{\text{L}}=X_{\text{C}}$

Write the expression to obtain the ratio of $\Delta V_{\text{out}}/\Delta V_{\text{in}}$.

    $\frac{\Delta V_{\text{out}}}{\Delta V_{\text{in}}}=\frac{R}{\sqrt{\left(R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2\right)}}$

Here, $\Delta V_{\text{out}}$ is the output voltage, $\Delta V_{\text{in}}$ is the input voltage, $R$ is the resistance of the resistor, $X_{\text{L}}$ is the inductive reactance and $X_{\text{C}}$ is the capacitive reactance.

Conclusion:

Substitute $X_{\text{L}}$ for $X_{\text{C}}$ in the above equation.

    $\begin{array}{c}\frac{\Delta V_{\text{out}}}{\Delta V_{\text{in}}}=\frac{R}{\sqrt{\left(R^2+{\left(X_{\text{L}}-X_{\text{L}}\right)}^2\right)}}\\ =\frac{R}{\sqrt{\left(R^2+{\left(0\right)}^2\right)}}\\ =\frac{R}{R}\\ =1.00\end{array}$

Therefore, the maximum value of the ratio of $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ is $1.00$.

(d)

To determine

The frequency at which it has the maximum value of the ratio of $\Delta V_{\text{out}}/\Delta V_{\text{in}}$.

Answer to Problem 77CP

The frequency at which it has the maximum value of the ratio of $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ is $894\text{Hz}$.

Explanation of Solution

The maximum ratio of $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ is at the resonance condition.

Write the expression to obtain the frequency at resonance condition.

    $f_0=\frac{1}{2\pi \sqrt{LC}}$

Here, $f_0$ is the frequency, $L$ is the inductance of the inductor and $C$ is the capacitance of the capacitor.

Conclusion:

Substitute $580\times {10}^{-6}\text{H}$ for $L$ and $54.6\times {10}^{-6}\text{F}$ for $C$ in the above equation to calculate $f_0$.

    $\begin{array}{c}{f}_0=\frac{1}{2\pi \sqrt{\left(580\times {10}^{-6}\text{H}\right)\left(54.6\times {10}^{-6}\text{F}\right)}}\\ =894.3\text{Hz}\\ \approx 894\text{Hz}\end{array}$

Therefore, the frequency at which it has the maximum value of the ratio of $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ is $894\text{Hz}$.

(e)

To determine

The phase shift between $\Delta V_{\text{in}}$ and $\Delta V_{\text{out}}$ at frequency $200\text{Hz}$, at $f_0$ and at $4.00\times {10}^3\text{Hz}$.

Answer to Problem 77CP

The phase shift at $200\text{Hz}$ frequency is $-60°$ that means $\Delta V_{\text{out}}$ leads $\Delta V_{\text{in}}$, the phase shift at $f_0$ frequency is $0°$ that means $\Delta V_{\text{out}}$ is in phase with $\Delta V_{\text{in}}$ and the phase shift at $4.00\times {10}^3\text{Hz}$ frequency is $60°$ that means $\Delta V_{\text{out}}$ lags $\Delta V_{\text{in}}$.

Explanation of Solution

Case (i): At $200\text{Hz}$, $X_{\text{C}}>X_{\text{L}}$.

Write the expression to obtain the phase shift between $\Delta V_{\text{in}}$ and $\Delta V_{\text{out}}$.

    $\varphi =-{\cos }^{-1}\left(\frac{R}{Z}\right)$                                                                                                   (V)

Here, $\varphi$ is the phase shift.

Write the expression to obtain the ratio of output and input voltage.

    $\frac{\Delta V_{\text{out}}}{\Delta V_{\text{in}}}=\frac{R}{Z}$                                                                                                 (VI)

Here, $\Delta V_{\text{out}}$ is the output voltage and $\Delta V_{\text{in}}$ is the input voltage, $R$ is the resistance of the resistor and $Z$ is the impedance of the circuit.

Substitute $\frac{1}{2}$ for $\frac{\Delta V_{\text{out}}}{\Delta V_{\text{in}}}$ in the equation (VI).

    $\frac{1}{2}=\frac{R}{Z}$

Case (ii): At $f_0$ which is the resonance frequency calculated in part (d).

Substitute $1$ for $\frac{\Delta V_{\text{out}}}{\Delta V_{\text{in}}}$ in equation (VI).

    $1=\frac{R}{Z}$

Case (iii): At $4.00\times {10}^3\text{Hz}$, $X_{\text{C}}<X_{\text{L}}$.

Write the expression to obtain the phase shift between $\Delta V_{\text{in}}$ and $\Delta V_{\text{out}}$.

    $\varphi ={\cos }^{-1}\left(\frac{R}{Z}\right)$                                                                                         (VII)

Here, $\varphi$ is the phase shift.

Substitute $\frac{1}{2}$ for $\frac{\Delta V_{\text{out}}}{\Delta V_{\text{in}}}$ in the equation (VI).

    $\frac{1}{2}=\frac{R}{Z}$

Conclusion:

At $200\text{Hz}$, $X_{\text{C}}>X_{\text{L}}$.

Substitute $\frac{1}{2}$ for $\frac{R}{Z}$ in equation (V) to calculate $\varphi$.

    $\begin{array}{c}\varphi =-{\cos }^{-1}\left(\frac{1}{2}\right)\\ =-60°\end{array}$

At $f_0$.

Substitute $1$ for $\frac{R}{Z}$ in equation (VII) to calculate $\varphi$.

    $\begin{array}{c}\varphi ={\cos }^{-1}\left(1\right)\\ =0°\end{array}$

At $4.00\times {10}^3\text{Hz}$, $X_{\text{C}}<X_{\text{L}}$.

Substitute $\frac{1}{2}$ for $\frac{R}{Z}$ in equation (VII) to calculate $\varphi$.

    $\begin{array}{c}\varphi ={\cos }^{-1}\left(\frac{1}{2}\right)\\ =60°\end{array}$

Therefore, the phase shift at $200\text{Hz}$ frequency is $-60°$ that means $\Delta V_{\text{out}}$ leads $\Delta V_{\text{in}}$, the phase shift at $f_0$ frequency is $0°$ that means $\Delta V_{\text{out}}$ is in phase with $\Delta V_{\text{in}}$ and the phase shift at $4.00\times {10}^3\text{Hz}$ frequency is $60°$ that means $\Delta V_{\text{out}}$ lags $\Delta V_{\text{in}}$.

(f)

To determine

The average power transferred to the speaker at $200\text{Hz}$, at $f_0$ and at $4.00\times {10}^3\text{Hz}$ frequencies.

Answer to Problem 77CP

The average power transferred to the speaker at $200\text{Hz}$ and at $4.00\times {10}^3\text{Hz}$ frequencies is $1.56\text{W}$ and the average power transferred to the speaker at $f_0$ frequency is $6.25\text{W}$.

Explanation of Solution

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

    $P_{\text{avg}}=\frac{1}{2}\left(\frac{{\left(\Delta V_{\text{in}}\right)}^2\cos \varphi }{\sqrt{R^2+{\left(X_{\text{L}}-X_{\text{C}}\right)}^2}}\right)$                                                                                    (VIII)

Here, $P_{\text{avg}}$ is the average power in the circuit, $\Delta V_{\text{in}}$ is the input voltage, $R$ is the resistance in the circuit, $X_{\text{L}}$ is the inductive reactance and $X_{\text{C}}$ is the capacitive reactance.

Conclusion:

Case (i): At frequency $200\text{Hz}$ and at $4.00\times {10}^3\text{Hz}$.

Substitute $10.0\text{V}$ for $\Delta V_{\text{in}}$, $\sqrt{3}R$ for $\left(X_{\text{L}}-X_{\text{C}}\right)$ and $60°$ for $\varphi$ in equation (VIII) to calculate $P_{\text{avg}}$.

    $P_{\text{avg}}=\frac{1}{2}\left(\frac{{\left(10.0\text{V}\right)}^2\cos 60°}{\sqrt{R^2+{\left(\sqrt{3}R\right)}^2}}\right)$

Further substitute $8.00\text{Ω}$ for $R$ in the above equation.

    $\begin{array}{c}{P}_{\text{avg}}=\frac{1}{2}\left(\frac{{\left(10.0\text{V}\right)}^2\cos 60°}{\sqrt{{\left(8.00\text{Ω}\right)}^2+{\left(\sqrt{3}\left(8.00\text{Ω}\right)\right)}^2}}\right)\\ =\frac{1}{2}\left(\frac{100{\text{V}}^2\left(\frac{1}{2}\right)}{\sqrt{{\left(8.00\text{Ω}\right)}^2+{\left(\sqrt{3}\left(8.00\text{Ω}\right)\right)}^2}}\right)\\ =1.56\text{W}\end{array}$

Case (ii): At frequency $f_0$.

Substitute $10.0\text{V}$ for $\Delta V_{\text{in}}$, $X_{\text{L}}$ for $X_{\text{C}}$, $8.00\text{Ω}$ for $R$ and $0°$ for $\varphi$ in equation (VIII) to calculate $P_{\text{avg}}$.

    $\begin{array}{c}{P}_{\text{avg}}=\frac{1}{2}\left(\frac{{\left(10.0\text{V}\right)}^2\cos 0°}{\sqrt{{\left(8.00\text{Ω}\right)}^2+{\left(X_{\text{L}}-X_{\text{L}}\right)}^2}}\right)\\ =\frac{1}{2}\left(\frac{100{\text{V}}^2}{\sqrt{{\left(8.00\text{Ω}\right)}^2}}\right)\\ =6.25\text{W}\end{array}$

Therefore, the average power transferred to the speaker at $200\text{Hz}$ and at $4.00\times {10}^3\text{Hz}$ frequencies is $1.56\text{W}$ and the average power transferred to the speaker at $f_0$ frequency is $6.25\text{W}$.

(g)

To determine

The quality factor while treating the filter as a resonance circuit.

Answer to Problem 77CP

The quality factor while treating the filter as a resonance circuit is $0.408$.

Explanation of Solution

Write the expression to obtain the quality factor while treating the filter as a resonance circuit.

    $Q=\frac{\left(2\pi f_0\right)L}{R}$

Here, $Q$ is the quality factor, $f_0$ is the resonance frequency, $L$ is the inductance of the inductor and $R$ is the resistor of the circuit.

Conclusion:

Substitute $894\text{Hz}$ for $f_0$, $580\mu \text{F}$ for $L$ and $8.00\text{Ω}$ for $R$ in the above equation.

    $\begin{array}{c}Q=\frac{\left(2\pi \left(894\text{Hz}\right)\right)\left(580\mu \text{F}\right)}{8.00\text{Ω}}\\ =\frac{\left(5617.16\text{rad}/\text{s}\right)\left(580\mu \text{F}\times \frac{1\text{F}}{{10}^6\mu \text{F}}\right)}{8.00\text{Ω}}\\ =0.408\end{array}$

Therefore, the quality factor while treating the filter as a resonance circuit is $0.408$.