Chapter 32, Problem 49CP

The resistor in Figure P32.49 represents the midrange speaker in a three-speaker system. Assume its resistance to be constant at 8.00 Ω. The source represents an audio amplifier producing signals of uniform amplitude ΔV_max = 10.0 V at all audio frequencies. The inductor and capacitor are to function as a band-pass filter with $\Delta V_{\text{out}}/\Delta V_{\text{in}}=\frac{1}{2}$ at 200 Hz and at 4.00 × 10³ Hz. Determine the required values of (a) L and (b) C. Find (c) the maximum value of the ratio ΔV_out/ΔV_in; (d) the frequency f_o at which the ratio has its maximum value; (e) the phase shift between Δv_in and Δv_out at 200 Hz, at f_o, and at 4.00 × 10³ Hz; and (f) the average power transferred to the speaker at 200 Hz, at f₀, and at 4.00 × 10³ Hz. (g) Recognizing that the diagram represents an RLC circuit driven by an AC source, find its quality factor.

Figure P32.49

To determine

The required value of inductance $L$.

Answer to Problem 49CP

The required value of inductance $L$ is $580\text{μH}$.

Explanation of Solution

Given info: The value of resistance is $8.00\Omega$, maximum potential difference is $10.0\text{V}$, ratio of $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ at $200\text{Hz}$ and at $4.00\times {10}^3\text{Hz}$ is $\frac{1}{2}$.

Formula to calculate the output potential difference is,

$\Delta V_{\text{out}}=IR$ (1)

Here,

$I$ is the current flowing in the circuit.

$R$ is the resistance in the circuit.

Formula to calculate the input potential difference is,

$\Delta V_{\text{in}}=IZ$ (2)

Here,

$I$ is the current flowing in the circuit.

$Z$ is the impedance in the circuit.

Divide equation (1) and equation (1).

$\begin{array}{c}\frac{\Delta V_{\text{out}}}{\Delta V_{\text{in}}}=\frac{IR}{IZ}\\ \frac{\Delta V_{\text{out}}}{\Delta V_{\text{in}}}=\frac{R}{Z}\end{array}$ (3)

Formula to calculate the inductive reactance of the circuit is,

$X_{\text{L}}=2\pi fL$

Here,

$X_{\text{L}}$ is the inductive reactance of the circuit.

$f$ is the frequency of the source.

$L$ is the inductance of the inductor.

Formula to calculate the inductive reactance of the circuit is,

$X_{\text{C}}=\frac{1}{2\pi fC}$

Here,

$X_{\text{C}}$ is the inductive reactance of the circuit.

$C$ is the capacitance of the capacitor.

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 $2\pi fL$ for $X_{\text{L}}$ and $\frac{1}{2\pi fC}$ for $X_{\text{C}}$ to find $Z$.

$Z=\sqrt{R^2+{\left(2\pi fL-\frac{1}{2\pi fC}\right)}^2}$ (4)

At low frequency that is $200\text{Hz}$, the impedance of the circuit is and the value must satisfy $X_{\text{L}}-X_{\text{C}}<0$.

Substitute $8.00\Omega$ for $R$ and $200\text{Hz}$ for $f$ in equation (4).

$\begin{array}{c}Z=\sqrt{{\left(8.00\Omega \right)}^2+{\left(\left(2\pi \times 200\text{Hz}\right)L-\frac{1}{\left(2\pi \times 200\text{Hz}\right)C}\right)}^2}\\ =\sqrt{64{\Omega }^2+{\left(\left(400\text{Hz}\right)L-\frac{1}{\left(400\text{Hz}\right)C}\right)}^2}\end{array}$

Substitute $8.00\Omega$ for $R$, $\sqrt{64{\Omega }^2+{\left(\left(400\text{Hz}\right)L-\frac{1}{\left(400\text{Hz}\right)C}\right)}^2}$ for $Z$ and $\frac{1}{2}$ for $\frac{\Delta V_{\text{out}}}{\Delta V_{\text{in}}}$ in equation (3).

$\begin{array}{c}\frac{8.00\Omega }{\sqrt{64{\Omega }^2+{\left(\left(400\text{Hz}\right)L-\frac{1}{\left(400\text{Hz}\right)C}\right)}^2}}=\frac{1}{2}\\ {\left(\frac{8.00\Omega }{\sqrt{64{\Omega }^2+{\left(\left(400\text{Hz}\right)L-\frac{1}{\left(400\text{Hz}\right)C}\right)}^2}}\right)}^2={\left(\frac{1}{2}\right)}^2\\ \frac{64{\Omega }^2}{\left(64{\Omega }^2+{\left(\left(400\text{Hz}\right)L-\frac{1}{\left(400\text{Hz}\right)C}\right)}^2\right)}=\frac{1}{4}\\ 64{\Omega }^2+{\left(\left(400\text{Hz}\right)L-\frac{1}{\left(400\text{Hz}\right)C}\right)}^2=256{\Omega }^2\end{array}$

Solve the equation further,

$\begin{array}{c}{\left(\left(400\pi \text{Hz}\right)L-\frac{1}{\left(400\pi \text{Hz}\right)C}\right)}^2=256{\Omega }^2-64{\Omega }^2\\ \left(400\pi \text{Hz}\right)L-\frac{1}{\left(400\pi \text{Hz}\right)C}=\sqrt{192}\Omega \\ \left(400\pi \text{Hz}\right)L-\frac{1}{\left(400\pi \text{Hz}\right)C}=-13.9\Omega \end{array}$

Divide the equation by $20$ both side.

$\left(20\pi \text{Hz}\right)L-\frac{1}{\left(8000\pi \text{Hz}\right)C}=-0.695\Omega$ (5)

At high frequency that is $4.00\times {10}^3\text{Hz}$, the impedance of the circuit and the value must satisfy $X_{\text{L}}-X_{\text{C}}>0$.

Substitute $8.00\Omega$ for $R$ and $4.00\times {10}^3\text{Hz}$ for $f$ in equation (4).

$\begin{array}{c}Z=\sqrt{{\left(8.00\Omega \right)}^2+{\left(\left(2\pi \times 4.00\times {10}^3\text{Hz}\right)L-\frac{1}{\left(2\pi \times 4.00\times {10}^3\text{Hz}\right)C}\right)}^2}\\ =\sqrt{64{\Omega }^2+{\left(\left(8000\pi \text{Hz}\right)L-\frac{1}{\left(8000\pi \text{Hz}\right)C}\right)}^2}\end{array}$

Substitute $8.00\Omega$ for $R$, $\sqrt{64{\Omega }^2+{\left(\left(8000\pi \text{Hz}\right)L-\frac{1}{\left(8000\pi \text{Hz}\right)C}\right)}^2}$ for $Z$ and $\frac{1}{2}$ for $\frac{\Delta V_{\text{out}}}{\Delta V_{\text{in}}}$ in equation (3).

$\begin{array}{c}\frac{8.00\Omega }{\sqrt{64{\Omega }^2+{\left(\left(8000\pi \text{Hz}\right)L-\frac{1}{\left(8000\pi \text{Hz}\right)C}\right)}^2}}=\frac{1}{2}\\ {\left(\frac{8.00\Omega }{\sqrt{64{\Omega }^2+{\left(\left(8000\pi \text{Hz}\right)L-\frac{1}{\left(8000\pi \text{Hz}\right)C}\right)}^2}}\right)}^2={\left(\frac{1}{2}\right)}^2\\ \frac{64{\Omega }^2}{\left(64{\Omega }^2+{\left(\left(8000\pi \text{Hz}\right)L-\frac{1}{\left(8000\pi \text{Hz}\right)C}\right)}^2\right)}=\frac{1}{4}\\ 64{\Omega }^2+{\left(\left(8000\pi \text{Hz}\right)L-\frac{1}{\left(8000\pi \text{Hz}\right)C}\right)}^2=256{\Omega }^2\end{array}$

Solve the equation further,

$\begin{array}{c}{\left(\left(8000\pi \text{Hz}\right)L-\frac{1}{\left(8000\pi \text{Hz}\right)C}\right)}^2=256{\Omega }^2-64{\Omega }^2\\ \left(8000\pi \text{Hz}\right)L-\frac{1}{\left(8000\pi \text{Hz}\right)C}=\sqrt{192}\Omega \\ \left(8000\pi \text{Hz}\right)L-\frac{1}{\left(8000\pi \text{Hz}\right)C}=13.9\Omega \end{array}$ (6)

Subtract the equation (5) and equation (6) to find the value of $L$.

$\begin{array}{c}\left[\begin{array}{l}\left(8000\pi \text{Hz}\right)L-\\ \left(20\pi \text{Hz}\right)L\end{array}\right]=13.9\Omega -0.695\Omega \\ L=580\times {10}^{-6}\text{H}\times \frac{{10}^6\text{μH}}{1\text{H}}\\ =580\text{μH}\end{array}$

Conclusion:

Therefore, the required value of inductance $L$ is $580\text{μH}$.

To determine

The required value of capacitance $C$.

Answer to Problem 49CP

The required value of capacitance $C$ is $54.6\text{μF}$.

Explanation of Solution

Given info: The value of resistance is $8.00\Omega$, maximum potential difference is $10.0\text{V}$, ratio of $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ at $200\text{Hz}$ and at $4.00\times {10}^3\text{Hz}$ is $\frac{1}{2}$.

The equation (6) is given as,

$\left(8000\pi \text{Hz}\right)L-\frac{1}{\left(8000\pi \text{Hz}\right)C}=13.9\Omega$

Substitute $580\text{μH}$ for $L$ in above equation to find $C$.

$\begin{array}{c}\left(1579043.5\text{Hz}\right)\left(580\text{μH}\times \frac{{10}^{-6}\text{H}}{1\text{μH}}\right)-\frac{1}{C}=17411.9\text{Hz}\cdot \Omega \\ C=54.6\times {10}^{-6}\text{F}\frac{{\text{10}}^6\text{μF}}{1\text{F}}\\ C=54.6\text{μF}\end{array}$

Conclusion:

Therefore, the required value of capacitance $C$ is $54.6\text{μF}$.

To determine

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

Answer to Problem 49CP

The maximum value of the ratio $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ is $1$.

Explanation of Solution

Given info: The value of resistance is $8.00\Omega$, maximum potential difference is $10.0\text{V}$, ratio of $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ at $200\text{Hz}$ and at $4.00\times {10}^3\text{Hz}$ is $\frac{1}{2}$.

The value $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ will be maximum when the circuit is in resonance condition.

At resonance condition, $X_{\text{L}}=X_{\text{C}}$ so the impedance of the circuit will becomes equal to the resistance in the circuit that is $Z=R$.

Substitute $R$ for $Z$ in equation (3).

$\begin{array}{l}{\left(\frac{\Delta V_{\text{out}}}{\Delta V_{\text{in}}}\right)}_{\max }=\frac{R}{R}\\ {\left(\frac{\Delta V_{\text{out}}}{\Delta V_{\text{in}}}\right)}_{\max }=1\end{array}$

Conclusion:

Therefore, the maximum value of the ratio $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ is $1$.

To determine

The frequency $f_0$ at which the ratio $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ has maximum value.

Answer to Problem 49CP

The frequency $f_0$ at which the ratio $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ has maximum value is $894\text{Hz}$.

Explanation of Solution

Given info: The value of resistance is $8.00\Omega$, maximum potential difference is $10.0\text{V}$, ratio of $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ at $200\text{Hz}$ and at $4.00\times {10}^3\text{Hz}$ is $\frac{1}{2}$.

Since the ratio $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ will be maximum at resonance point.

Formula to calculate the resonance frequency is,

$f_0=\frac{1}{2\pi \sqrt{LC}}$

Substitute $580\text{μH}$ for $L$ and $54.6\text{μC}$ for $C$ to find $f_0$.

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

Conclusion:

Therefore, the frequency $f_0$ at which the ratio $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ has maximum value is $894\text{Hz}$.

To determine

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

Answer to Problem 49CP

The phase shift between $\Delta v_{\text{in}}$ and $\Delta v_{\text{out}}$ at $200\text{Hz}$ is $60°$, at $f_0$ is $0°$ and at $4.00\times {10}^3\text{Hz}$ is $60°$.

Explanation of Solution

Given info: The value of resistance is $8.00\Omega$, maximum potential difference is $10.0\text{V}$, ratio of $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ at $200\text{Hz}$ and at $4.00\times {10}^3\text{Hz}$ is $\frac{1}{2}$.

Formula to calculate the phase shift between $\Delta v_{\text{in}}$ and $\Delta v_{\text{out}}$ is,

${\varphi }_1={\cos }^{-1}\left(\frac{R}{Z}\right)$

At $200\text{Hz}$ the reactance of the capacitor is greater than the reactance of the inductor that is $X_{\text{C}}>X_{\text{L}}$ hence $\Delta v_{\text{out}}$ leads $\Delta v_{\text{in}}$ and the value of $\left(\frac{R}{Z}\right)$ is $\frac{1}{2}$ from equation (3).

Substitute $\frac{1}{2}$ for $\left(\frac{R}{Z}\right)$ to find ${\varphi }_1$.

$\begin{array}{c}{\varphi }_1={\cos }^{-1}\left(\frac{1}{2}\right)\\ =60°\end{array}$

At $f_0$ the reactance of the capacitor is equal to the reactance of the inductor that is $X_{\text{C}}=X_{\text{L}}$ hence $\Delta v_{\text{out}}$ is in phase with $\Delta v_{\text{in}}$ and the impedance is equal to the resistance.

Substitute $R$ for $Z$ to find ${\varphi }_1$.

$\begin{array}{c}{\varphi }_1={\cos }^{-1}\left(\frac{R}{R}\right)\\ ={\cos }^{-1}\left(1\right)\\ =0°\end{array}$

At $4.00\times {10}^3\text{Hz}$ the reactance of the capacitor is less than the reactance of the inductor that is $X_{\text{C}}<X_{\text{L}}$ hence $\Delta v_{\text{out}}$ lags $\Delta v_{\text{in}}$ and the value of $\left(\frac{R}{Z}\right)$ is $\frac{1}{2}$ from equation (3).

Substitute $\frac{1}{2}$ for $\left(\frac{R}{Z}\right)$ to find ${\varphi }_1$.

$\begin{array}{c}{\varphi }_1={\cos }^{-1}\left(\frac{1}{2}\right)\\ =60°\end{array}$

Conclusion:

Therefore, the phase shift between $\Delta v_{\text{in}}$ and $\Delta v_{\text{out}}$ at $200\text{Hz}$ is $60°$, at $f_0$ is $0°$ and at $4.00\times {10}^3\text{Hz}$ is $60°$.

To determine

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

Answer to Problem 49CP

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

Explanation of Solution

Given info: The value of resistance is $8.00\Omega$, maximum potential difference is $10.0\text{V}$, ratio of $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ at $200\text{Hz}$ and at $4.00\times {10}^3\text{Hz}$ is $\frac{1}{2}$.

Formula to calculate the rms output voltage is,

${\left(\Delta v_{out}\right)}_{\text{rms}}=\frac{\left(\Delta v_{out}\right)}{\sqrt{2}}$

Formula to calculate the power deliver to the speaker is,

$P=\frac{{\left(\Delta v_{\text{out}}\right)}_{\text{rms}}^2}{R}$

Substitute $\frac{\left(\Delta v_{out}\right)}{\sqrt{2}}$ for ${\left(\Delta v_{out}\right)}_{\text{rms}}$ in above equation.

$\begin{array}{c}P=\frac{{\left(\frac{\left(\Delta v_{out}\right)}{\sqrt{2}}\right)}_{\text{rms}}^2}{R}\\ =\frac{{\left(\Delta v_{out}\right)}^2}{2R}\end{array}$ (7)

For low frequency $200\text{Hz}$ and high frequency $4.00\times {10}^3\text{Hz}$, the value of ratio $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ is.

$\begin{array}{l}\frac{\Delta v_{\text{out}}}{\Delta v_{\text{in}}}=\frac{1}{2}\\ \Delta v_{\text{out}}=\frac{1}{2}\Delta v_{\text{in}}\end{array}$

Substitute $\frac{1}{2}\Delta v_{\text{in}}$ for $\Delta v_{\text{out}}$ in equation (7).

$\begin{array}{c}P=\frac{{\left(\frac{1}{2}\Delta v_{\text{in}}\right)}^2}{2R}\\ =\frac{1}{8}\left(\frac{{\left(\Delta v_{\text{in}}\right)}^2}{R}\right)\end{array}$

Substitute $10.0\text{V}$ for $\Delta v_{\text{in}}$ and $8.00\Omega$ for $R$ to find $P$.

$\begin{array}{c}P=\frac{1}{8}\left(\frac{{\left(10.0\text{V}\right)}^2}{8.00\Omega }\right)\\ =1.56\text{W}\end{array}$

For resonance frequency $f_0$ the value of ratio $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ is 1.

$\begin{array}{l}\frac{\Delta v_{\text{out}}}{\Delta v_{\text{in}}}=1\\ \Delta v_{\text{out}}=\Delta v_{\text{in}}\end{array}$

Substitute $\Delta v_{\text{in}}$ for $\Delta v_{\text{out}}$ in equation (7).

$\begin{array}{c}P=\frac{{\left(\Delta v_{\text{in}}\right)}^2}{2R}\\ =\left(\frac{{\left(\Delta v_{\text{in}}\right)}^2}{2R}\right)\end{array}$

Substitute $10.0\text{V}$ for $\Delta v_{\text{in}}$ and $8.00\Omega$ for $R$ to find $P$.

$\begin{array}{c}P=\left(\frac{{\left(10.0\text{V}\right)}^2}{2\times 8.00\Omega }\right)\\ =6.25\text{W}\end{array}$

Conclusion:

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

To determine

The quality factor of the circuit.

Answer to Problem 49CP

The quality factor of the circuit is $0.408$.

Explanation of Solution

Given info: The value of resistance is $8.00\Omega$, maximum potential difference is $10.0\text{V}$, ratio of $\Delta V_{\text{out}}/\Delta V_{\text{in}}$ at $200\text{Hz}$ and at $4.00\times {10}^3\text{Hz}$ is $\frac{1}{2}$.

Formula to calculate the quality factor is,

$Q=\frac{2\pi f_{0}L}{R}$

Substitute $894\text{Hz}$ for $f_0$, $580\text{μH}$ for $L$ and $8.00\Omega$ for $R$.

$\begin{array}{c}Q=\frac{2\pi \times 894\text{Hz}\times 580\text{μH}\times \frac{{10}^{-6}\text{H}}{1\text{μH}}}{8.00\Omega }\\ =0.408\end{array}$

Conclusion:

Therefore, the quality factor of the circuit is $0.408$.