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 Δ V max = 10.0 V at all audio frequencies. The inductor and capacitor are to function as a band-pass filter with Δ V out / Δ V in = 1 2 at 200 Hz and at 4.00 × 10 3 Hz. Determine the required values of (a) L and (b) C . Find (c) the maximum value of the ratio Δ V out /Δ V in ; (d) the frequency f o at which the ratio has its maximum value; (e) the phase shift between Δ v in and Δ v out at 200 Hz, at f o , and at 4.00 × 10 3 Hz; and (f) the average power transferred to the speaker at 200 Hz, at f 0 , and at 4.00 × 10 3 Hz. (g) Recognizing that the diagram represents an RLC circuit driven by an AC source, find its quality factor. Figure P32.49

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Answer 1

Textbook Question

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 Δ V out / Δ V in = 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

Chapter 32, Problem 49CP, The resistor in Figure P32.49 represents the midrange speaker in a three-speaker system. Assume its

(a)

Expert Solution

To determine

The required value of inductance L .

Answer to Problem 49CP

The required value of inductance L is 580 μH .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the output potential difference is,

ΔVout=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,

ΔVin=IZ (2)

Here,

I is the current flowing in the circuit.

Z is the impedance in the circuit.

Divide equation (1) and equation (1).

ΔVoutΔVin=IRIZΔVoutΔVin=RZ (3)

Formula to calculate the inductive reactance of the circuit is,

XL=2πfL

Here,

XL 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,

XC=12πfC

Here,

XC is the inductive reactance of the circuit.

C is the capacitance of the capacitor.

Formula to calculate the impedance of the circuit is,

Z=R2+(XL−XC)2

Here,

Z is the impedance in the circuit.

R is the resistance in the circuit.

Substitute 2πfL for XL and 12πfC for XC to find Z .

Z=R2+(2πfL−12πfC)2 (4)

At low frequency that is 200 Hz , the impedance of the circuit is and the value must satisfy XL−XC<0 .

Substitute 8.00 Ω for R and 200 Hz for f in equation (4).

Z=(8.00 Ω)2+((2π×200 Hz)L−1(2π×200 Hz)C)2=64 Ω2+((400 Hz)L−1(400 Hz)C)2

Substitute 8.00 Ω for R , 64 Ω2+((400 Hz)L−1(400 Hz)C)2 for Z and 12 for ΔVoutΔVin in equation (3).

8.00 Ω64 Ω2+((400 Hz)L−1(400 Hz)C)2=12(8.00 Ω64 Ω2+((400 Hz)L−1(400 Hz)C)2)2=(12)264 Ω2(64 Ω2+((400 Hz)L−1(400 Hz)C)2)=1464 Ω2+((400 Hz)L−1(400 Hz)C)2=256 Ω2

Solve the equation further,

((400π Hz)L−1(400π Hz)C)2=256 Ω2−64 Ω2(400π Hz)L−1(400π Hz)C=192 Ω(400π Hz)L−1(400π Hz)C=−13.9 Ω

Divide the equation by 20 both side.

(20π Hz)L−1(8000π Hz)C=−0.695 Ω (5)

At high frequency that is 4.00×103 Hz , the impedance of the circuit and the value must satisfy XL−XC>0 .

Substitute 8.00 Ω for R and 4.00×103 Hz for f in equation (4).

Z=(8.00 Ω)2+((2π×4.00×103 Hz)L−1(2π×4.00×103 Hz)C)2=64 Ω2+((8000π Hz)L−1(8000π Hz)C)2

Substitute 8.00 Ω for R , 64 Ω2+((8000π Hz)L−1(8000π Hz)C)2 for Z and 12 for ΔVoutΔVin in equation (3).

8.00 Ω64 Ω2+((8000π Hz)L−1(8000π Hz)C)2=12(8.00 Ω64 Ω2+((8000π Hz)L−1(8000π Hz)C)2)2=(12)264 Ω2(64 Ω2+((8000π Hz)L−1(8000π Hz)C)2)=1464 Ω2+((8000π Hz)L−1(8000π Hz)C)2=256 Ω2

Solve the equation further,

((8000π Hz)L−1(8000π Hz)C)2=256 Ω2−64 Ω2(8000π Hz)L−1(8000π Hz)C=192 Ω(8000π Hz)L−1(8000π Hz)C=13.9 Ω (6)

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

[(8000π Hz)L−(20π Hz)L]=13.9 Ω−0.695 ΩL=580×10−6 H×106 μH1 H=580 μH

Conclusion:

Therefore, the required value of inductance L is 580 μH .

(b)

Expert Solution

To determine

The required value of capacitance C .

Answer to Problem 49CP

The required value of capacitance C is 54.6 μF .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

The equation (6) is given as,

(8000π Hz)L−1(8000π Hz)C=13.9 Ω

Substitute 580 μH for L in above equation to find C .

(1579043.5 Hz)(580 μH×10−6 H1 μH)−1C=17411.9 Hz⋅ΩC=54.6×10−6 F 106 μF1 FC=54.6 μF

Conclusion:

Therefore, the required value of capacitance C is 54.6 μF .

(c)

Expert Solution

To determine

The maximum value of the ratio ΔVout/ΔVin .

Answer to Problem 49CP

The maximum value of the ratio ΔVout/ΔVin is 1 .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

The value ΔVout/ΔVin will be maximum when the circuit is in resonance condition.

At resonance condition, XL=XC 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).

(ΔVoutΔVin)max=RR(ΔVoutΔVin)max=1

Conclusion:

Therefore, the maximum value of the ratio ΔVout/ΔVin is 1 .

(d)

Expert Solution

To determine

The frequency f0 at which the ratio ΔVout/ΔVin has maximum value.

Answer to Problem 49CP

The frequency f0 at which the ratio ΔVout/ΔVin has maximum value is 894 Hz .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Since the ratio ΔVout/ΔVin will be maximum at resonance point.

Formula to calculate the resonance frequency is,

f0=12πLC

Substitute 580 μH for L and 54.6 μC for C to find f0 .

f0=12π(580 μH×10−6 H1 μH)(54.6 μC×10−6 C1 μC)=894 Hz

Conclusion:

Therefore, the frequency f0 at which the ratio ΔVout/ΔVin has maximum value is 894 Hz .

(e)

Expert Solution

To determine

The phase shift between Δvin and Δvout at 200 Hz , f0 and 4.00×103 Hz

Answer to Problem 49CP

The phase shift between Δvin and Δvout at 200 Hz is 60° , at f0 is 0° and at 4.00×103 Hz is 60° .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the phase shift between Δvin and Δvout is,

ϕ1=cos−1(RZ)

At 200 Hz the reactance of the capacitor is greater than the reactance of the inductor that is XC>XL hence Δvout leads Δvin and the value of (RZ) is 12 from equation (3).

Substitute 12 for (RZ) to find ϕ1 .

ϕ1=cos−1(12)=60°

At f0 the reactance of the capacitor is equal to the reactance of the inductor that is XC=XL hence Δvout is in phase with Δvin and the impedance is equal to the resistance.

Substitute R for Z to find ϕ1 .

ϕ1=cos−1(RR)=cos−1(1)=0°

At 4.00×103 Hz the reactance of the capacitor is less than the reactance of the inductor that is XC<XL hence Δvout lags Δvin and the value of (RZ) is 12 from equation (3).

Substitute 12 for (RZ) to find ϕ1 .

ϕ1=cos−1(12)=60°

Conclusion:

Therefore, the phase shift between Δvin and Δvout at 200 Hz is 60° , at f0 is 0° and at 4.00×103 Hz is 60° .

(f)

Expert Solution

To determine

The average power transferred to the speaker at 200 Hz , f0 and 4.00×103 Hz

Answer to Problem 49CP

The average power transferred to the speaker at 200 Hz and 4.00×103 Hz is 1.56 W and at f0 is 6.25 W .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the rms output voltage is,

(Δvout)rms=(Δvout)2

Formula to calculate the power deliver to the speaker is,

P=(Δvout)rms2R

Substitute (Δvout)2 for (Δvout)rms in above equation.

P=((Δvout)2)rms2R=(Δvout)22R (7)

For low frequency 200 Hz and high frequency 4.00×103 Hz , the value of ratio ΔVout/ΔVin is.

ΔvoutΔvin=12Δvout=12Δvin

Substitute 12Δvin for Δvout in equation (7).

P=(12Δvin)22R=18((Δvin)2R)

Substitute 10.0 V for Δvin and 8.00 Ω for R to find P .

P=18((10.0 V)28.00 Ω)=1.56 W

For resonance frequency f0 the value of ratio ΔVout/ΔVin is 1.

ΔvoutΔvin=1Δvout=Δvin

Substitute Δvin for Δvout in equation (7).

P=(Δvin)22R=((Δvin)22R)

Substitute 10.0 V for Δvin and 8.00 Ω for R to find P .

P=((10.0 V)22×8.00 Ω)=6.25 W

Conclusion:

Therefore, the average power transferred to the speaker at 200 Hz and 4.00×103 Hz is 1.56 W and at f0 is 6.25 W .

(g)

Expert Solution

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 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the quality factor is,

Q=2πf0LR

Substitute 894 Hz for f0 , 580 μH for L and 8.00 Ω for R .

Q=2π×894 Hz×580 μH×10−6 H1 μH8.00 Ω=0.408

Conclusion:

Therefore, the quality factor of the circuit is 0.408 .

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A simple series circuit consists of a 150 Ω resistor, a 27.0 V battery, a switch, and a 2.00 pF parallel-plate capacitor (initially uncharged) with plates 5.0 mm apart. The switch is closed at t =0s . Part A Part complete Part B Part complete Part C Find the electric flux at t =0.50ns. Express your answer in volt-meters. View Available Hint(s)for Part C Activate to select the appropriates template from the following choices. Operate up and down arrow for selection and press enter to choose the input value typeActivate to select the appropriates symbol from the following choices. Operate up and down arrow for selection and press enter to choose the input value type nothing V⋅m Part D Find the displacement current at t =0.50ns.

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Answer 2

Textbook Question

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 Δ V out / Δ V in = 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

Chapter 32, Problem 49CP, The resistor in Figure P32.49 represents the midrange speaker in a three-speaker system. Assume its

(a)

Expert Solution

To determine

The required value of inductance L .

Answer to Problem 49CP

The required value of inductance L is 580 μH .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the output potential difference is,

ΔVout=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,

ΔVin=IZ (2)

Here,

I is the current flowing in the circuit.

Z is the impedance in the circuit.

Divide equation (1) and equation (1).

ΔVoutΔVin=IRIZΔVoutΔVin=RZ (3)

Formula to calculate the inductive reactance of the circuit is,

XL=2πfL

Here,

XL 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,

XC=12πfC

Here,

XC is the inductive reactance of the circuit.

C is the capacitance of the capacitor.

Formula to calculate the impedance of the circuit is,

Z=R2+(XL−XC)2

Here,

Z is the impedance in the circuit.

R is the resistance in the circuit.

Substitute 2πfL for XL and 12πfC for XC to find Z .

Z=R2+(2πfL−12πfC)2 (4)

At low frequency that is 200 Hz , the impedance of the circuit is and the value must satisfy XL−XC<0 .

Substitute 8.00 Ω for R and 200 Hz for f in equation (4).

Z=(8.00 Ω)2+((2π×200 Hz)L−1(2π×200 Hz)C)2=64 Ω2+((400 Hz)L−1(400 Hz)C)2

Substitute 8.00 Ω for R , 64 Ω2+((400 Hz)L−1(400 Hz)C)2 for Z and 12 for ΔVoutΔVin in equation (3).

8.00 Ω64 Ω2+((400 Hz)L−1(400 Hz)C)2=12(8.00 Ω64 Ω2+((400 Hz)L−1(400 Hz)C)2)2=(12)264 Ω2(64 Ω2+((400 Hz)L−1(400 Hz)C)2)=1464 Ω2+((400 Hz)L−1(400 Hz)C)2=256 Ω2

Solve the equation further,

((400π Hz)L−1(400π Hz)C)2=256 Ω2−64 Ω2(400π Hz)L−1(400π Hz)C=192 Ω(400π Hz)L−1(400π Hz)C=−13.9 Ω

Divide the equation by 20 both side.

(20π Hz)L−1(8000π Hz)C=−0.695 Ω (5)

At high frequency that is 4.00×103 Hz , the impedance of the circuit and the value must satisfy XL−XC>0 .

Substitute 8.00 Ω for R and 4.00×103 Hz for f in equation (4).

Z=(8.00 Ω)2+((2π×4.00×103 Hz)L−1(2π×4.00×103 Hz)C)2=64 Ω2+((8000π Hz)L−1(8000π Hz)C)2

Substitute 8.00 Ω for R , 64 Ω2+((8000π Hz)L−1(8000π Hz)C)2 for Z and 12 for ΔVoutΔVin in equation (3).

8.00 Ω64 Ω2+((8000π Hz)L−1(8000π Hz)C)2=12(8.00 Ω64 Ω2+((8000π Hz)L−1(8000π Hz)C)2)2=(12)264 Ω2(64 Ω2+((8000π Hz)L−1(8000π Hz)C)2)=1464 Ω2+((8000π Hz)L−1(8000π Hz)C)2=256 Ω2

Solve the equation further,

((8000π Hz)L−1(8000π Hz)C)2=256 Ω2−64 Ω2(8000π Hz)L−1(8000π Hz)C=192 Ω(8000π Hz)L−1(8000π Hz)C=13.9 Ω (6)

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

[(8000π Hz)L−(20π Hz)L]=13.9 Ω−0.695 ΩL=580×10−6 H×106 μH1 H=580 μH

Conclusion:

Therefore, the required value of inductance L is 580 μH .

(b)

Expert Solution

To determine

The required value of capacitance C .

Answer to Problem 49CP

The required value of capacitance C is 54.6 μF .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

The equation (6) is given as,

(8000π Hz)L−1(8000π Hz)C=13.9 Ω

Substitute 580 μH for L in above equation to find C .

(1579043.5 Hz)(580 μH×10−6 H1 μH)−1C=17411.9 Hz⋅ΩC=54.6×10−6 F 106 μF1 FC=54.6 μF

Conclusion:

Therefore, the required value of capacitance C is 54.6 μF .

(c)

Expert Solution

To determine

The maximum value of the ratio ΔVout/ΔVin .

Answer to Problem 49CP

The maximum value of the ratio ΔVout/ΔVin is 1 .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

The value ΔVout/ΔVin will be maximum when the circuit is in resonance condition.

At resonance condition, XL=XC 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).

(ΔVoutΔVin)max=RR(ΔVoutΔVin)max=1

Conclusion:

Therefore, the maximum value of the ratio ΔVout/ΔVin is 1 .

(d)

Expert Solution

To determine

The frequency f0 at which the ratio ΔVout/ΔVin has maximum value.

Answer to Problem 49CP

The frequency f0 at which the ratio ΔVout/ΔVin has maximum value is 894 Hz .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Since the ratio ΔVout/ΔVin will be maximum at resonance point.

Formula to calculate the resonance frequency is,

f0=12πLC

Substitute 580 μH for L and 54.6 μC for C to find f0 .

f0=12π(580 μH×10−6 H1 μH)(54.6 μC×10−6 C1 μC)=894 Hz

Conclusion:

Therefore, the frequency f0 at which the ratio ΔVout/ΔVin has maximum value is 894 Hz .

(e)

Expert Solution

To determine

The phase shift between Δvin and Δvout at 200 Hz , f0 and 4.00×103 Hz

Answer to Problem 49CP

The phase shift between Δvin and Δvout at 200 Hz is 60° , at f0 is 0° and at 4.00×103 Hz is 60° .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the phase shift between Δvin and Δvout is,

ϕ1=cos−1(RZ)

At 200 Hz the reactance of the capacitor is greater than the reactance of the inductor that is XC>XL hence Δvout leads Δvin and the value of (RZ) is 12 from equation (3).

Substitute 12 for (RZ) to find ϕ1 .

ϕ1=cos−1(12)=60°

At f0 the reactance of the capacitor is equal to the reactance of the inductor that is XC=XL hence Δvout is in phase with Δvin and the impedance is equal to the resistance.

Substitute R for Z to find ϕ1 .

ϕ1=cos−1(RR)=cos−1(1)=0°

At 4.00×103 Hz the reactance of the capacitor is less than the reactance of the inductor that is XC<XL hence Δvout lags Δvin and the value of (RZ) is 12 from equation (3).

Substitute 12 for (RZ) to find ϕ1 .

ϕ1=cos−1(12)=60°

Conclusion:

Therefore, the phase shift between Δvin and Δvout at 200 Hz is 60° , at f0 is 0° and at 4.00×103 Hz is 60° .

(f)

Expert Solution

To determine

The average power transferred to the speaker at 200 Hz , f0 and 4.00×103 Hz

Answer to Problem 49CP

The average power transferred to the speaker at 200 Hz and 4.00×103 Hz is 1.56 W and at f0 is 6.25 W .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the rms output voltage is,

(Δvout)rms=(Δvout)2

Formula to calculate the power deliver to the speaker is,

P=(Δvout)rms2R

Substitute (Δvout)2 for (Δvout)rms in above equation.

P=((Δvout)2)rms2R=(Δvout)22R (7)

For low frequency 200 Hz and high frequency 4.00×103 Hz , the value of ratio ΔVout/ΔVin is.

ΔvoutΔvin=12Δvout=12Δvin

Substitute 12Δvin for Δvout in equation (7).

P=(12Δvin)22R=18((Δvin)2R)

Substitute 10.0 V for Δvin and 8.00 Ω for R to find P .

P=18((10.0 V)28.00 Ω)=1.56 W

For resonance frequency f0 the value of ratio ΔVout/ΔVin is 1.

ΔvoutΔvin=1Δvout=Δvin

Substitute Δvin for Δvout in equation (7).

P=(Δvin)22R=((Δvin)22R)

Substitute 10.0 V for Δvin and 8.00 Ω for R to find P .

P=((10.0 V)22×8.00 Ω)=6.25 W

Conclusion:

Therefore, the average power transferred to the speaker at 200 Hz and 4.00×103 Hz is 1.56 W and at f0 is 6.25 W .

(g)

Expert Solution

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 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the quality factor is,

Q=2πf0LR

Substitute 894 Hz for f0 , 580 μH for L and 8.00 Ω for R .

Q=2π×894 Hz×580 μH×10−6 H1 μH8.00 Ω=0.408

Conclusion:

Therefore, the quality factor of the circuit is 0.408 .

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Answer 3

Textbook Question

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 Δ V out / Δ V in = 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

Chapter 32, Problem 49CP, The resistor in Figure P32.49 represents the midrange speaker in a three-speaker system. Assume its

(a)

Expert Solution

To determine

The required value of inductance L .

Answer to Problem 49CP

The required value of inductance L is 580 μH .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the output potential difference is,

ΔVout=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,

ΔVin=IZ (2)

Here,

I is the current flowing in the circuit.

Z is the impedance in the circuit.

Divide equation (1) and equation (1).

ΔVoutΔVin=IRIZΔVoutΔVin=RZ (3)

Formula to calculate the inductive reactance of the circuit is,

XL=2πfL

Here,

XL 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,

XC=12πfC

Here,

XC is the inductive reactance of the circuit.

C is the capacitance of the capacitor.

Formula to calculate the impedance of the circuit is,

Z=R2+(XL−XC)2

Here,

Z is the impedance in the circuit.

R is the resistance in the circuit.

Substitute 2πfL for XL and 12πfC for XC to find Z .

Z=R2+(2πfL−12πfC)2 (4)

At low frequency that is 200 Hz , the impedance of the circuit is and the value must satisfy XL−XC<0 .

Substitute 8.00 Ω for R and 200 Hz for f in equation (4).

Z=(8.00 Ω)2+((2π×200 Hz)L−1(2π×200 Hz)C)2=64 Ω2+((400 Hz)L−1(400 Hz)C)2

Substitute 8.00 Ω for R , 64 Ω2+((400 Hz)L−1(400 Hz)C)2 for Z and 12 for ΔVoutΔVin in equation (3).

8.00 Ω64 Ω2+((400 Hz)L−1(400 Hz)C)2=12(8.00 Ω64 Ω2+((400 Hz)L−1(400 Hz)C)2)2=(12)264 Ω2(64 Ω2+((400 Hz)L−1(400 Hz)C)2)=1464 Ω2+((400 Hz)L−1(400 Hz)C)2=256 Ω2

Solve the equation further,

((400π Hz)L−1(400π Hz)C)2=256 Ω2−64 Ω2(400π Hz)L−1(400π Hz)C=192 Ω(400π Hz)L−1(400π Hz)C=−13.9 Ω

Divide the equation by 20 both side.

(20π Hz)L−1(8000π Hz)C=−0.695 Ω (5)

At high frequency that is 4.00×103 Hz , the impedance of the circuit and the value must satisfy XL−XC>0 .

Substitute 8.00 Ω for R and 4.00×103 Hz for f in equation (4).

Z=(8.00 Ω)2+((2π×4.00×103 Hz)L−1(2π×4.00×103 Hz)C)2=64 Ω2+((8000π Hz)L−1(8000π Hz)C)2

Substitute 8.00 Ω for R , 64 Ω2+((8000π Hz)L−1(8000π Hz)C)2 for Z and 12 for ΔVoutΔVin in equation (3).

8.00 Ω64 Ω2+((8000π Hz)L−1(8000π Hz)C)2=12(8.00 Ω64 Ω2+((8000π Hz)L−1(8000π Hz)C)2)2=(12)264 Ω2(64 Ω2+((8000π Hz)L−1(8000π Hz)C)2)=1464 Ω2+((8000π Hz)L−1(8000π Hz)C)2=256 Ω2

Solve the equation further,

((8000π Hz)L−1(8000π Hz)C)2=256 Ω2−64 Ω2(8000π Hz)L−1(8000π Hz)C=192 Ω(8000π Hz)L−1(8000π Hz)C=13.9 Ω (6)

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

[(8000π Hz)L−(20π Hz)L]=13.9 Ω−0.695 ΩL=580×10−6 H×106 μH1 H=580 μH

Conclusion:

Therefore, the required value of inductance L is 580 μH .

(b)

Expert Solution

To determine

The required value of capacitance C .

Answer to Problem 49CP

The required value of capacitance C is 54.6 μF .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

The equation (6) is given as,

(8000π Hz)L−1(8000π Hz)C=13.9 Ω

Substitute 580 μH for L in above equation to find C .

(1579043.5 Hz)(580 μH×10−6 H1 μH)−1C=17411.9 Hz⋅ΩC=54.6×10−6 F 106 μF1 FC=54.6 μF

Conclusion:

Therefore, the required value of capacitance C is 54.6 μF .

(c)

Expert Solution

To determine

The maximum value of the ratio ΔVout/ΔVin .

Answer to Problem 49CP

The maximum value of the ratio ΔVout/ΔVin is 1 .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

The value ΔVout/ΔVin will be maximum when the circuit is in resonance condition.

At resonance condition, XL=XC 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).

(ΔVoutΔVin)max=RR(ΔVoutΔVin)max=1

Conclusion:

Therefore, the maximum value of the ratio ΔVout/ΔVin is 1 .

(d)

Expert Solution

To determine

The frequency f0 at which the ratio ΔVout/ΔVin has maximum value.

Answer to Problem 49CP

The frequency f0 at which the ratio ΔVout/ΔVin has maximum value is 894 Hz .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Since the ratio ΔVout/ΔVin will be maximum at resonance point.

Formula to calculate the resonance frequency is,

f0=12πLC

Substitute 580 μH for L and 54.6 μC for C to find f0 .

f0=12π(580 μH×10−6 H1 μH)(54.6 μC×10−6 C1 μC)=894 Hz

Conclusion:

Therefore, the frequency f0 at which the ratio ΔVout/ΔVin has maximum value is 894 Hz .

(e)

Expert Solution

To determine

The phase shift between Δvin and Δvout at 200 Hz , f0 and 4.00×103 Hz

Answer to Problem 49CP

The phase shift between Δvin and Δvout at 200 Hz is 60° , at f0 is 0° and at 4.00×103 Hz is 60° .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the phase shift between Δvin and Δvout is,

ϕ1=cos−1(RZ)

At 200 Hz the reactance of the capacitor is greater than the reactance of the inductor that is XC>XL hence Δvout leads Δvin and the value of (RZ) is 12 from equation (3).

Substitute 12 for (RZ) to find ϕ1 .

ϕ1=cos−1(12)=60°

At f0 the reactance of the capacitor is equal to the reactance of the inductor that is XC=XL hence Δvout is in phase with Δvin and the impedance is equal to the resistance.

Substitute R for Z to find ϕ1 .

ϕ1=cos−1(RR)=cos−1(1)=0°

At 4.00×103 Hz the reactance of the capacitor is less than the reactance of the inductor that is XC<XL hence Δvout lags Δvin and the value of (RZ) is 12 from equation (3).

Substitute 12 for (RZ) to find ϕ1 .

ϕ1=cos−1(12)=60°

Conclusion:

Therefore, the phase shift between Δvin and Δvout at 200 Hz is 60° , at f0 is 0° and at 4.00×103 Hz is 60° .

(f)

Expert Solution

To determine

The average power transferred to the speaker at 200 Hz , f0 and 4.00×103 Hz

Answer to Problem 49CP

The average power transferred to the speaker at 200 Hz and 4.00×103 Hz is 1.56 W and at f0 is 6.25 W .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the rms output voltage is,

(Δvout)rms=(Δvout)2

Formula to calculate the power deliver to the speaker is,

P=(Δvout)rms2R

Substitute (Δvout)2 for (Δvout)rms in above equation.

P=((Δvout)2)rms2R=(Δvout)22R (7)

For low frequency 200 Hz and high frequency 4.00×103 Hz , the value of ratio ΔVout/ΔVin is.

ΔvoutΔvin=12Δvout=12Δvin

Substitute 12Δvin for Δvout in equation (7).

P=(12Δvin)22R=18((Δvin)2R)

Substitute 10.0 V for Δvin and 8.00 Ω for R to find P .

P=18((10.0 V)28.00 Ω)=1.56 W

For resonance frequency f0 the value of ratio ΔVout/ΔVin is 1.

ΔvoutΔvin=1Δvout=Δvin

Substitute Δvin for Δvout in equation (7).

P=(Δvin)22R=((Δvin)22R)

Substitute 10.0 V for Δvin and 8.00 Ω for R to find P .

P=((10.0 V)22×8.00 Ω)=6.25 W

Conclusion:

Therefore, the average power transferred to the speaker at 200 Hz and 4.00×103 Hz is 1.56 W and at f0 is 6.25 W .

(g)

Expert Solution

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 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the quality factor is,

Q=2πf0LR

Substitute 894 Hz for f0 , 580 μH for L and 8.00 Ω for R .

Q=2π×894 Hz×580 μH×10−6 H1 μH8.00 Ω=0.408

Conclusion:

Therefore, the quality factor of the circuit is 0.408 .

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Answer 4

Textbook Question

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 Δ V out / Δ V in = 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

Chapter 32, Problem 49CP, The resistor in Figure P32.49 represents the midrange speaker in a three-speaker system. Assume its

(a)

Expert Solution

To determine

The required value of inductance L .

Answer to Problem 49CP

The required value of inductance L is 580 μH .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the output potential difference is,

ΔVout=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,

ΔVin=IZ (2)

Here,

I is the current flowing in the circuit.

Z is the impedance in the circuit.

Divide equation (1) and equation (1).

ΔVoutΔVin=IRIZΔVoutΔVin=RZ (3)

Formula to calculate the inductive reactance of the circuit is,

XL=2πfL

Here,

XL 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,

XC=12πfC

Here,

XC is the inductive reactance of the circuit.

C is the capacitance of the capacitor.

Formula to calculate the impedance of the circuit is,

Z=R2+(XL−XC)2

Here,

Z is the impedance in the circuit.

R is the resistance in the circuit.

Substitute 2πfL for XL and 12πfC for XC to find Z .

Z=R2+(2πfL−12πfC)2 (4)

At low frequency that is 200 Hz , the impedance of the circuit is and the value must satisfy XL−XC<0 .

Substitute 8.00 Ω for R and 200 Hz for f in equation (4).

Z=(8.00 Ω)2+((2π×200 Hz)L−1(2π×200 Hz)C)2=64 Ω2+((400 Hz)L−1(400 Hz)C)2

Substitute 8.00 Ω for R , 64 Ω2+((400 Hz)L−1(400 Hz)C)2 for Z and 12 for ΔVoutΔVin in equation (3).

8.00 Ω64 Ω2+((400 Hz)L−1(400 Hz)C)2=12(8.00 Ω64 Ω2+((400 Hz)L−1(400 Hz)C)2)2=(12)264 Ω2(64 Ω2+((400 Hz)L−1(400 Hz)C)2)=1464 Ω2+((400 Hz)L−1(400 Hz)C)2=256 Ω2

Solve the equation further,

((400π Hz)L−1(400π Hz)C)2=256 Ω2−64 Ω2(400π Hz)L−1(400π Hz)C=192 Ω(400π Hz)L−1(400π Hz)C=−13.9 Ω

Divide the equation by 20 both side.

(20π Hz)L−1(8000π Hz)C=−0.695 Ω (5)

At high frequency that is 4.00×103 Hz , the impedance of the circuit and the value must satisfy XL−XC>0 .

Substitute 8.00 Ω for R and 4.00×103 Hz for f in equation (4).

Z=(8.00 Ω)2+((2π×4.00×103 Hz)L−1(2π×4.00×103 Hz)C)2=64 Ω2+((8000π Hz)L−1(8000π Hz)C)2

Substitute 8.00 Ω for R , 64 Ω2+((8000π Hz)L−1(8000π Hz)C)2 for Z and 12 for ΔVoutΔVin in equation (3).

8.00 Ω64 Ω2+((8000π Hz)L−1(8000π Hz)C)2=12(8.00 Ω64 Ω2+((8000π Hz)L−1(8000π Hz)C)2)2=(12)264 Ω2(64 Ω2+((8000π Hz)L−1(8000π Hz)C)2)=1464 Ω2+((8000π Hz)L−1(8000π Hz)C)2=256 Ω2

Solve the equation further,

((8000π Hz)L−1(8000π Hz)C)2=256 Ω2−64 Ω2(8000π Hz)L−1(8000π Hz)C=192 Ω(8000π Hz)L−1(8000π Hz)C=13.9 Ω (6)

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

[(8000π Hz)L−(20π Hz)L]=13.9 Ω−0.695 ΩL=580×10−6 H×106 μH1 H=580 μH

Conclusion:

Therefore, the required value of inductance L is 580 μH .

(b)

Expert Solution

To determine

The required value of capacitance C .

Answer to Problem 49CP

The required value of capacitance C is 54.6 μF .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

The equation (6) is given as,

(8000π Hz)L−1(8000π Hz)C=13.9 Ω

Substitute 580 μH for L in above equation to find C .

(1579043.5 Hz)(580 μH×10−6 H1 μH)−1C=17411.9 Hz⋅ΩC=54.6×10−6 F 106 μF1 FC=54.6 μF

Conclusion:

Therefore, the required value of capacitance C is 54.6 μF .

(c)

Expert Solution

To determine

The maximum value of the ratio ΔVout/ΔVin .

Answer to Problem 49CP

The maximum value of the ratio ΔVout/ΔVin is 1 .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

The value ΔVout/ΔVin will be maximum when the circuit is in resonance condition.

At resonance condition, XL=XC 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).

(ΔVoutΔVin)max=RR(ΔVoutΔVin)max=1

Conclusion:

Therefore, the maximum value of the ratio ΔVout/ΔVin is 1 .

(d)

Expert Solution

To determine

The frequency f0 at which the ratio ΔVout/ΔVin has maximum value.

Answer to Problem 49CP

The frequency f0 at which the ratio ΔVout/ΔVin has maximum value is 894 Hz .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Since the ratio ΔVout/ΔVin will be maximum at resonance point.

Formula to calculate the resonance frequency is,

f0=12πLC

Substitute 580 μH for L and 54.6 μC for C to find f0 .

f0=12π(580 μH×10−6 H1 μH)(54.6 μC×10−6 C1 μC)=894 Hz

Conclusion:

Therefore, the frequency f0 at which the ratio ΔVout/ΔVin has maximum value is 894 Hz .

(e)

Expert Solution

To determine

The phase shift between Δvin and Δvout at 200 Hz , f0 and 4.00×103 Hz

Answer to Problem 49CP

The phase shift between Δvin and Δvout at 200 Hz is 60° , at f0 is 0° and at 4.00×103 Hz is 60° .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the phase shift between Δvin and Δvout is,

ϕ1=cos−1(RZ)

At 200 Hz the reactance of the capacitor is greater than the reactance of the inductor that is XC>XL hence Δvout leads Δvin and the value of (RZ) is 12 from equation (3).

Substitute 12 for (RZ) to find ϕ1 .

ϕ1=cos−1(12)=60°

At f0 the reactance of the capacitor is equal to the reactance of the inductor that is XC=XL hence Δvout is in phase with Δvin and the impedance is equal to the resistance.

Substitute R for Z to find ϕ1 .

ϕ1=cos−1(RR)=cos−1(1)=0°

At 4.00×103 Hz the reactance of the capacitor is less than the reactance of the inductor that is XC<XL hence Δvout lags Δvin and the value of (RZ) is 12 from equation (3).

Substitute 12 for (RZ) to find ϕ1 .

ϕ1=cos−1(12)=60°

Conclusion:

Therefore, the phase shift between Δvin and Δvout at 200 Hz is 60° , at f0 is 0° and at 4.00×103 Hz is 60° .

(f)

Expert Solution

To determine

The average power transferred to the speaker at 200 Hz , f0 and 4.00×103 Hz

Answer to Problem 49CP

The average power transferred to the speaker at 200 Hz and 4.00×103 Hz is 1.56 W and at f0 is 6.25 W .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the rms output voltage is,

(Δvout)rms=(Δvout)2

Formula to calculate the power deliver to the speaker is,

P=(Δvout)rms2R

Substitute (Δvout)2 for (Δvout)rms in above equation.

P=((Δvout)2)rms2R=(Δvout)22R (7)

For low frequency 200 Hz and high frequency 4.00×103 Hz , the value of ratio ΔVout/ΔVin is.

ΔvoutΔvin=12Δvout=12Δvin

Substitute 12Δvin for Δvout in equation (7).

P=(12Δvin)22R=18((Δvin)2R)

Substitute 10.0 V for Δvin and 8.00 Ω for R to find P .

P=18((10.0 V)28.00 Ω)=1.56 W

For resonance frequency f0 the value of ratio ΔVout/ΔVin is 1.

ΔvoutΔvin=1Δvout=Δvin

Substitute Δvin for Δvout in equation (7).

P=(Δvin)22R=((Δvin)22R)

Substitute 10.0 V for Δvin and 8.00 Ω for R to find P .

P=((10.0 V)22×8.00 Ω)=6.25 W

Conclusion:

Therefore, the average power transferred to the speaker at 200 Hz and 4.00×103 Hz is 1.56 W and at f0 is 6.25 W .

(g)

Expert Solution

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 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the quality factor is,

Q=2πf0LR

Substitute 894 Hz for f0 , 580 μH for L and 8.00 Ω for R .

Q=2π×894 Hz×580 μH×10−6 H1 μH8.00 Ω=0.408

Conclusion:

Therefore, the quality factor of the circuit is 0.408 .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The required value of inductance L .

Answer to Problem 49CP

The required value of inductance L is 580 μH .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the output potential difference is,

ΔVout=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,

ΔVin=IZ (2)

Here,

I is the current flowing in the circuit.

Z is the impedance in the circuit.

Divide equation (1) and equation (1).

ΔVoutΔVin=IRIZΔVoutΔVin=RZ (3)

Formula to calculate the inductive reactance of the circuit is,

XL=2πfL

Here,

XL 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,

XC=12πfC

Here,

XC is the inductive reactance of the circuit.

C is the capacitance of the capacitor.

Formula to calculate the impedance of the circuit is,

Z=R2+(XL−XC)2

Here,

Z is the impedance in the circuit.

R is the resistance in the circuit.

Substitute 2πfL for XL and 12πfC for XC to find Z .

Z=R2+(2πfL−12πfC)2 (4)

At low frequency that is 200 Hz , the impedance of the circuit is and the value must satisfy XL−XC<0 .

Substitute 8.00 Ω for R and 200 Hz for f in equation (4).

Z=(8.00 Ω)2+((2π×200 Hz)L−1(2π×200 Hz)C)2=64 Ω2+((400 Hz)L−1(400 Hz)C)2

Substitute 8.00 Ω for R , 64 Ω2+((400 Hz)L−1(400 Hz)C)2 for Z and 12 for ΔVoutΔVin in equation (3).

8.00 Ω64 Ω2+((400 Hz)L−1(400 Hz)C)2=12(8.00 Ω64 Ω2+((400 Hz)L−1(400 Hz)C)2)2=(12)264 Ω2(64 Ω2+((400 Hz)L−1(400 Hz)C)2)=1464 Ω2+((400 Hz)L−1(400 Hz)C)2=256 Ω2

Solve the equation further,

((400π Hz)L−1(400π Hz)C)2=256 Ω2−64 Ω2(400π Hz)L−1(400π Hz)C=192 Ω(400π Hz)L−1(400π Hz)C=−13.9 Ω

Divide the equation by 20 both side.

(20π Hz)L−1(8000π Hz)C=−0.695 Ω (5)

At high frequency that is 4.00×103 Hz , the impedance of the circuit and the value must satisfy XL−XC>0 .

Substitute 8.00 Ω for R and 4.00×103 Hz for f in equation (4).

Z=(8.00 Ω)2+((2π×4.00×103 Hz)L−1(2π×4.00×103 Hz)C)2=64 Ω2+((8000π Hz)L−1(8000π Hz)C)2

Substitute 8.00 Ω for R , 64 Ω2+((8000π Hz)L−1(8000π Hz)C)2 for Z and 12 for ΔVoutΔVin in equation (3).

8.00 Ω64 Ω2+((8000π Hz)L−1(8000π Hz)C)2=12(8.00 Ω64 Ω2+((8000π Hz)L−1(8000π Hz)C)2)2=(12)264 Ω2(64 Ω2+((8000π Hz)L−1(8000π Hz)C)2)=1464 Ω2+((8000π Hz)L−1(8000π Hz)C)2=256 Ω2

Solve the equation further,

((8000π Hz)L−1(8000π Hz)C)2=256 Ω2−64 Ω2(8000π Hz)L−1(8000π Hz)C=192 Ω(8000π Hz)L−1(8000π Hz)C=13.9 Ω (6)

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

[(8000π Hz)L−(20π Hz)L]=13.9 Ω−0.695 ΩL=580×10−6 H×106 μH1 H=580 μH

Conclusion:

Therefore, the required value of inductance L is 580 μH .

(b)

Expert Solution

To determine

The required value of capacitance C .

Answer to Problem 49CP

The required value of capacitance C is 54.6 μF .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

The equation (6) is given as,

(8000π Hz)L−1(8000π Hz)C=13.9 Ω

Substitute 580 μH for L in above equation to find C .

(1579043.5 Hz)(580 μH×10−6 H1 μH)−1C=17411.9 Hz⋅ΩC=54.6×10−6 F 106 μF1 FC=54.6 μF

Conclusion:

Therefore, the required value of capacitance C is 54.6 μF .

(c)

Expert Solution

To determine

The maximum value of the ratio ΔVout/ΔVin .

Answer to Problem 49CP

The maximum value of the ratio ΔVout/ΔVin is 1 .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

The value ΔVout/ΔVin will be maximum when the circuit is in resonance condition.

At resonance condition, XL=XC 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).

(ΔVoutΔVin)max=RR(ΔVoutΔVin)max=1

Conclusion:

Therefore, the maximum value of the ratio ΔVout/ΔVin is 1 .

(d)

Expert Solution

To determine

The frequency f0 at which the ratio ΔVout/ΔVin has maximum value.

Answer to Problem 49CP

The frequency f0 at which the ratio ΔVout/ΔVin has maximum value is 894 Hz .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Since the ratio ΔVout/ΔVin will be maximum at resonance point.

Formula to calculate the resonance frequency is,

f0=12πLC

Substitute 580 μH for L and 54.6 μC for C to find f0 .

f0=12π(580 μH×10−6 H1 μH)(54.6 μC×10−6 C1 μC)=894 Hz

Conclusion:

Therefore, the frequency f0 at which the ratio ΔVout/ΔVin has maximum value is 894 Hz .

(e)

Expert Solution

To determine

The phase shift between Δvin and Δvout at 200 Hz , f0 and 4.00×103 Hz

Answer to Problem 49CP

The phase shift between Δvin and Δvout at 200 Hz is 60° , at f0 is 0° and at 4.00×103 Hz is 60° .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the phase shift between Δvin and Δvout is,

ϕ1=cos−1(RZ)

At 200 Hz the reactance of the capacitor is greater than the reactance of the inductor that is XC>XL hence Δvout leads Δvin and the value of (RZ) is 12 from equation (3).

Substitute 12 for (RZ) to find ϕ1 .

ϕ1=cos−1(12)=60°

At f0 the reactance of the capacitor is equal to the reactance of the inductor that is XC=XL hence Δvout is in phase with Δvin and the impedance is equal to the resistance.

Substitute R for Z to find ϕ1 .

ϕ1=cos−1(RR)=cos−1(1)=0°

At 4.00×103 Hz the reactance of the capacitor is less than the reactance of the inductor that is XC<XL hence Δvout lags Δvin and the value of (RZ) is 12 from equation (3).

Substitute 12 for (RZ) to find ϕ1 .

ϕ1=cos−1(12)=60°

Conclusion:

Therefore, the phase shift between Δvin and Δvout at 200 Hz is 60° , at f0 is 0° and at 4.00×103 Hz is 60° .

(f)

Expert Solution

To determine

The average power transferred to the speaker at 200 Hz , f0 and 4.00×103 Hz

Answer to Problem 49CP

The average power transferred to the speaker at 200 Hz and 4.00×103 Hz is 1.56 W and at f0 is 6.25 W .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the rms output voltage is,

(Δvout)rms=(Δvout)2

Formula to calculate the power deliver to the speaker is,

P=(Δvout)rms2R

Substitute (Δvout)2 for (Δvout)rms in above equation.

P=((Δvout)2)rms2R=(Δvout)22R (7)

For low frequency 200 Hz and high frequency 4.00×103 Hz , the value of ratio ΔVout/ΔVin is.

ΔvoutΔvin=12Δvout=12Δvin

Substitute 12Δvin for Δvout in equation (7).

P=(12Δvin)22R=18((Δvin)2R)

Substitute 10.0 V for Δvin and 8.00 Ω for R to find P .

P=18((10.0 V)28.00 Ω)=1.56 W

For resonance frequency f0 the value of ratio ΔVout/ΔVin is 1.

ΔvoutΔvin=1Δvout=Δvin

Substitute Δvin for Δvout in equation (7).

P=(Δvin)22R=((Δvin)22R)

Substitute 10.0 V for Δvin and 8.00 Ω for R to find P .

P=((10.0 V)22×8.00 Ω)=6.25 W

Conclusion:

Therefore, the average power transferred to the speaker at 200 Hz and 4.00×103 Hz is 1.56 W and at f0 is 6.25 W .

(g)

Expert Solution

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 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the quality factor is,

Q=2πf0LR

Substitute 894 Hz for f0 , 580 μH for L and 8.00 Ω for R .

Q=2π×894 Hz×580 μH×10−6 H1 μH8.00 Ω=0.408

Conclusion:

Therefore, the quality factor of the circuit is 0.408 .

Answer 8

(a)

Expert Solution

To determine

The required value of inductance L .

Answer to Problem 49CP

The required value of inductance L is 580 μH .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the output potential difference is,

ΔVout=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,

ΔVin=IZ (2)

Here,

I is the current flowing in the circuit.

Z is the impedance in the circuit.

Divide equation (1) and equation (1).

ΔVoutΔVin=IRIZΔVoutΔVin=RZ (3)

Formula to calculate the inductive reactance of the circuit is,

XL=2πfL

Here,

XL 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,

XC=12πfC

Here,

XC is the inductive reactance of the circuit.

C is the capacitance of the capacitor.

Formula to calculate the impedance of the circuit is,

Z=R2+(XL−XC)2

Here,

Z is the impedance in the circuit.

R is the resistance in the circuit.

Substitute 2πfL for XL and 12πfC for XC to find Z .

Z=R2+(2πfL−12πfC)2 (4)

At low frequency that is 200 Hz , the impedance of the circuit is and the value must satisfy XL−XC<0 .

Substitute 8.00 Ω for R and 200 Hz for f in equation (4).

Z=(8.00 Ω)2+((2π×200 Hz)L−1(2π×200 Hz)C)2=64 Ω2+((400 Hz)L−1(400 Hz)C)2

Substitute 8.00 Ω for R , 64 Ω2+((400 Hz)L−1(400 Hz)C)2 for Z and 12 for ΔVoutΔVin in equation (3).

8.00 Ω64 Ω2+((400 Hz)L−1(400 Hz)C)2=12(8.00 Ω64 Ω2+((400 Hz)L−1(400 Hz)C)2)2=(12)264 Ω2(64 Ω2+((400 Hz)L−1(400 Hz)C)2)=1464 Ω2+((400 Hz)L−1(400 Hz)C)2=256 Ω2

Solve the equation further,

((400π Hz)L−1(400π Hz)C)2=256 Ω2−64 Ω2(400π Hz)L−1(400π Hz)C=192 Ω(400π Hz)L−1(400π Hz)C=−13.9 Ω

Divide the equation by 20 both side.

(20π Hz)L−1(8000π Hz)C=−0.695 Ω (5)

At high frequency that is 4.00×103 Hz , the impedance of the circuit and the value must satisfy XL−XC>0 .

Substitute 8.00 Ω for R and 4.00×103 Hz for f in equation (4).

Z=(8.00 Ω)2+((2π×4.00×103 Hz)L−1(2π×4.00×103 Hz)C)2=64 Ω2+((8000π Hz)L−1(8000π Hz)C)2

Substitute 8.00 Ω for R , 64 Ω2+((8000π Hz)L−1(8000π Hz)C)2 for Z and 12 for ΔVoutΔVin in equation (3).

8.00 Ω64 Ω2+((8000π Hz)L−1(8000π Hz)C)2=12(8.00 Ω64 Ω2+((8000π Hz)L−1(8000π Hz)C)2)2=(12)264 Ω2(64 Ω2+((8000π Hz)L−1(8000π Hz)C)2)=1464 Ω2+((8000π Hz)L−1(8000π Hz)C)2=256 Ω2

Solve the equation further,

((8000π Hz)L−1(8000π Hz)C)2=256 Ω2−64 Ω2(8000π Hz)L−1(8000π Hz)C=192 Ω(8000π Hz)L−1(8000π Hz)C=13.9 Ω (6)

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

[(8000π Hz)L−(20π Hz)L]=13.9 Ω−0.695 ΩL=580×10−6 H×106 μH1 H=580 μH

Conclusion:

Therefore, the required value of inductance L is 580 μH .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The required value of inductance L .

Answer 13

Answer to Problem 49CP

The required value of inductance L is 580 μH .

Answer 14

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the output potential difference is,

ΔVout=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,

ΔVin=IZ (2)

Here,

I is the current flowing in the circuit.

Z is the impedance in the circuit.

Divide equation (1) and equation (1).

ΔVoutΔVin=IRIZΔVoutΔVin=RZ (3)

Formula to calculate the inductive reactance of the circuit is,

XL=2πfL

Here,

XL 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,

XC=12πfC

Here,

XC is the inductive reactance of the circuit.

C is the capacitance of the capacitor.

Formula to calculate the impedance of the circuit is,

Z=R2+(XL−XC)2

Here,

Z is the impedance in the circuit.

R is the resistance in the circuit.

Substitute 2πfL for XL and 12πfC for XC to find Z .

Z=R2+(2πfL−12πfC)2 (4)

At low frequency that is 200 Hz , the impedance of the circuit is and the value must satisfy XL−XC<0 .

Substitute 8.00 Ω for R and 200 Hz for f in equation (4).

Z=(8.00 Ω)2+((2π×200 Hz)L−1(2π×200 Hz)C)2=64 Ω2+((400 Hz)L−1(400 Hz)C)2

Substitute 8.00 Ω for R , 64 Ω2+((400 Hz)L−1(400 Hz)C)2 for Z and 12 for ΔVoutΔVin in equation (3).

8.00 Ω64 Ω2+((400 Hz)L−1(400 Hz)C)2=12(8.00 Ω64 Ω2+((400 Hz)L−1(400 Hz)C)2)2=(12)264 Ω2(64 Ω2+((400 Hz)L−1(400 Hz)C)2)=1464 Ω2+((400 Hz)L−1(400 Hz)C)2=256 Ω2

Solve the equation further,

((400π Hz)L−1(400π Hz)C)2=256 Ω2−64 Ω2(400π Hz)L−1(400π Hz)C=192 Ω(400π Hz)L−1(400π Hz)C=−13.9 Ω

Divide the equation by 20 both side.

(20π Hz)L−1(8000π Hz)C=−0.695 Ω (5)

At high frequency that is 4.00×103 Hz , the impedance of the circuit and the value must satisfy XL−XC>0 .

Substitute 8.00 Ω for R and 4.00×103 Hz for f in equation (4).

Z=(8.00 Ω)2+((2π×4.00×103 Hz)L−1(2π×4.00×103 Hz)C)2=64 Ω2+((8000π Hz)L−1(8000π Hz)C)2

Substitute 8.00 Ω for R , 64 Ω2+((8000π Hz)L−1(8000π Hz)C)2 for Z and 12 for ΔVoutΔVin in equation (3).

8.00 Ω64 Ω2+((8000π Hz)L−1(8000π Hz)C)2=12(8.00 Ω64 Ω2+((8000π Hz)L−1(8000π Hz)C)2)2=(12)264 Ω2(64 Ω2+((8000π Hz)L−1(8000π Hz)C)2)=1464 Ω2+((8000π Hz)L−1(8000π Hz)C)2=256 Ω2

Solve the equation further,

((8000π Hz)L−1(8000π Hz)C)2=256 Ω2−64 Ω2(8000π Hz)L−1(8000π Hz)C=192 Ω(8000π Hz)L−1(8000π Hz)C=13.9 Ω (6)

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

[(8000π Hz)L−(20π Hz)L]=13.9 Ω−0.695 ΩL=580×10−6 H×106 μH1 H=580 μH

Conclusion:

Therefore, the required value of inductance L is 580 μH .

Answer 15

(b)

Expert Solution

To determine

The required value of capacitance C .

Answer to Problem 49CP

The required value of capacitance C is 54.6 μF .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

The equation (6) is given as,

(8000π Hz)L−1(8000π Hz)C=13.9 Ω

Substitute 580 μH for L in above equation to find C .

(1579043.5 Hz)(580 μH×10−6 H1 μH)−1C=17411.9 Hz⋅ΩC=54.6×10−6 F 106 μF1 FC=54.6 μF

Conclusion:

Therefore, the required value of capacitance C is 54.6 μF .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The required value of capacitance C .

Answer 20

Answer to Problem 49CP

The required value of capacitance C is 54.6 μF .

Answer 21

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

The equation (6) is given as,

(8000π Hz)L−1(8000π Hz)C=13.9 Ω

Substitute 580 μH for L in above equation to find C .

(1579043.5 Hz)(580 μH×10−6 H1 μH)−1C=17411.9 Hz⋅ΩC=54.6×10−6 F 106 μF1 FC=54.6 μF

Conclusion:

Therefore, the required value of capacitance C is 54.6 μF .

Answer 22

(c)

Expert Solution

To determine

The maximum value of the ratio ΔVout/ΔVin .

Answer to Problem 49CP

The maximum value of the ratio ΔVout/ΔVin is 1 .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

The value ΔVout/ΔVin will be maximum when the circuit is in resonance condition.

At resonance condition, XL=XC 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).

(ΔVoutΔVin)max=RR(ΔVoutΔVin)max=1

Conclusion:

Therefore, the maximum value of the ratio ΔVout/ΔVin is 1 .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The maximum value of the ratio ΔVout/ΔVin .

Answer 27

Answer to Problem 49CP

The maximum value of the ratio ΔVout/ΔVin is 1 .

Answer 28

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

The value ΔVout/ΔVin will be maximum when the circuit is in resonance condition.

At resonance condition, XL=XC 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).

(ΔVoutΔVin)max=RR(ΔVoutΔVin)max=1

Conclusion:

Therefore, the maximum value of the ratio ΔVout/ΔVin is 1 .

Answer 29

(d)

Expert Solution

To determine

The frequency f0 at which the ratio ΔVout/ΔVin has maximum value.

Answer to Problem 49CP

The frequency f0 at which the ratio ΔVout/ΔVin has maximum value is 894 Hz .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Since the ratio ΔVout/ΔVin will be maximum at resonance point.

Formula to calculate the resonance frequency is,

f0=12πLC

Substitute 580 μH for L and 54.6 μC for C to find f0 .

f0=12π(580 μH×10−6 H1 μH)(54.6 μC×10−6 C1 μC)=894 Hz

Conclusion:

Therefore, the frequency f0 at which the ratio ΔVout/ΔVin has maximum value is 894 Hz .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The frequency f0 at which the ratio ΔVout/ΔVin has maximum value.

Answer 34

Answer to Problem 49CP

The frequency f0 at which the ratio ΔVout/ΔVin has maximum value is 894 Hz .

Answer 35

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Since the ratio ΔVout/ΔVin will be maximum at resonance point.

Formula to calculate the resonance frequency is,

f0=12πLC

Substitute 580 μH for L and 54.6 μC for C to find f0 .

f0=12π(580 μH×10−6 H1 μH)(54.6 μC×10−6 C1 μC)=894 Hz

Conclusion:

Therefore, the frequency f0 at which the ratio ΔVout/ΔVin has maximum value is 894 Hz .

Answer 36

(e)

Expert Solution

To determine

The phase shift between Δvin and Δvout at 200 Hz , f0 and 4.00×103 Hz

Answer to Problem 49CP

The phase shift between Δvin and Δvout at 200 Hz is 60° , at f0 is 0° and at 4.00×103 Hz is 60° .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the phase shift between Δvin and Δvout is,

ϕ1=cos−1(RZ)

At 200 Hz the reactance of the capacitor is greater than the reactance of the inductor that is XC>XL hence Δvout leads Δvin and the value of (RZ) is 12 from equation (3).

Substitute 12 for (RZ) to find ϕ1 .

ϕ1=cos−1(12)=60°

At f0 the reactance of the capacitor is equal to the reactance of the inductor that is XC=XL hence Δvout is in phase with Δvin and the impedance is equal to the resistance.

Substitute R for Z to find ϕ1 .

ϕ1=cos−1(RR)=cos−1(1)=0°

At 4.00×103 Hz the reactance of the capacitor is less than the reactance of the inductor that is XC<XL hence Δvout lags Δvin and the value of (RZ) is 12 from equation (3).

Substitute 12 for (RZ) to find ϕ1 .

ϕ1=cos−1(12)=60°

Conclusion:

Therefore, the phase shift between Δvin and Δvout at 200 Hz is 60° , at f0 is 0° and at 4.00×103 Hz is 60° .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

To determine

The phase shift between Δvin and Δvout at 200 Hz , f0 and 4.00×103 Hz

Answer 41

Answer to Problem 49CP

The phase shift between Δvin and Δvout at 200 Hz is 60° , at f0 is 0° and at 4.00×103 Hz is 60° .

Answer 42

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the phase shift between Δvin and Δvout is,

ϕ1=cos−1(RZ)

At 200 Hz the reactance of the capacitor is greater than the reactance of the inductor that is XC>XL hence Δvout leads Δvin and the value of (RZ) is 12 from equation (3).

Substitute 12 for (RZ) to find ϕ1 .

ϕ1=cos−1(12)=60°

At f0 the reactance of the capacitor is equal to the reactance of the inductor that is XC=XL hence Δvout is in phase with Δvin and the impedance is equal to the resistance.

Substitute R for Z to find ϕ1 .

ϕ1=cos−1(RR)=cos−1(1)=0°

At 4.00×103 Hz the reactance of the capacitor is less than the reactance of the inductor that is XC<XL hence Δvout lags Δvin and the value of (RZ) is 12 from equation (3).

Substitute 12 for (RZ) to find ϕ1 .

ϕ1=cos−1(12)=60°

Conclusion:

Therefore, the phase shift between Δvin and Δvout at 200 Hz is 60° , at f0 is 0° and at 4.00×103 Hz is 60° .

Answer 43

(f)

Expert Solution

To determine

The average power transferred to the speaker at 200 Hz , f0 and 4.00×103 Hz

Answer to Problem 49CP

The average power transferred to the speaker at 200 Hz and 4.00×103 Hz is 1.56 W and at f0 is 6.25 W .

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the rms output voltage is,

(Δvout)rms=(Δvout)2

Formula to calculate the power deliver to the speaker is,

P=(Δvout)rms2R

Substitute (Δvout)2 for (Δvout)rms in above equation.

P=((Δvout)2)rms2R=(Δvout)22R (7)

For low frequency 200 Hz and high frequency 4.00×103 Hz , the value of ratio ΔVout/ΔVin is.

ΔvoutΔvin=12Δvout=12Δvin

Substitute 12Δvin for Δvout in equation (7).

P=(12Δvin)22R=18((Δvin)2R)

Substitute 10.0 V for Δvin and 8.00 Ω for R to find P .

P=18((10.0 V)28.00 Ω)=1.56 W

For resonance frequency f0 the value of ratio ΔVout/ΔVin is 1.

ΔvoutΔvin=1Δvout=Δvin

Substitute Δvin for Δvout in equation (7).

P=(Δvin)22R=((Δvin)22R)

Substitute 10.0 V for Δvin and 8.00 Ω for R to find P .

P=((10.0 V)22×8.00 Ω)=6.25 W

Conclusion:

Therefore, the average power transferred to the speaker at 200 Hz and 4.00×103 Hz is 1.56 W and at f0 is 6.25 W .

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

To determine

The average power transferred to the speaker at 200 Hz , f0 and 4.00×103 Hz

Answer 48

Answer to Problem 49CP

The average power transferred to the speaker at 200 Hz and 4.00×103 Hz is 1.56 W and at f0 is 6.25 W .

Answer 49

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the rms output voltage is,

(Δvout)rms=(Δvout)2

Formula to calculate the power deliver to the speaker is,

P=(Δvout)rms2R

Substitute (Δvout)2 for (Δvout)rms in above equation.

P=((Δvout)2)rms2R=(Δvout)22R (7)

For low frequency 200 Hz and high frequency 4.00×103 Hz , the value of ratio ΔVout/ΔVin is.

ΔvoutΔvin=12Δvout=12Δvin

Substitute 12Δvin for Δvout in equation (7).

P=(12Δvin)22R=18((Δvin)2R)

Substitute 10.0 V for Δvin and 8.00 Ω for R to find P .

P=18((10.0 V)28.00 Ω)=1.56 W

For resonance frequency f0 the value of ratio ΔVout/ΔVin is 1.

ΔvoutΔvin=1Δvout=Δvin

Substitute Δvin for Δvout in equation (7).

P=(Δvin)22R=((Δvin)22R)

Substitute 10.0 V for Δvin and 8.00 Ω for R to find P .

P=((10.0 V)22×8.00 Ω)=6.25 W

Conclusion:

Therefore, the average power transferred to the speaker at 200 Hz and 4.00×103 Hz is 1.56 W and at f0 is 6.25 W .

Answer 50

(g)

Expert Solution

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 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the quality factor is,

Q=2πf0LR

Substitute 894 Hz for f0 , 580 μH for L and 8.00 Ω for R .

Q=2π×894 Hz×580 μH×10−6 H1 μH8.00 Ω=0.408

Conclusion:

Therefore, the quality factor of the circuit is 0.408 .

Answer 51

(g)

Expert Solution

Answer 52

(g)

Expert Solution

Answer 53

Expert Solution

Answer 54

To determine

The quality factor of the circuit.

Answer 55

Answer to Problem 49CP

The quality factor of the circuit is 0.408 .

Answer 56

Explanation of Solution

Given info: The value of resistance is 8.00 Ω , maximum potential difference is 10.0 V , ratio of ΔVout/ΔVin at 200 Hz and at 4.00×103 Hz is 12 .

Formula to calculate the quality factor is,

Q=2πf0LR

Substitute 894 Hz for f0 , 580 μH for L and 8.00 Ω for R .

Q=2π×894 Hz×580 μH×10−6 H1 μH8.00 Ω=0.408

Conclusion:

Therefore, the quality factor of the circuit is 0.408 .

Answer 57

Ch. 32.2 - Consider the voltage phasor in Figure 32.4, shown...Ch. 32.3 - Consider the AC circuit in Figure 32.8. The...Ch. 32.4 - Consider the AC circuit in Figure 32.11. The...Ch. 32.4 - Consider the AC circuit in Figure 32.12. The...Ch. 32.5 - Label each part of Figure 32.16, (a), (b), and...Ch. 32.6 - Prob. 32.6QQ Ch. 32.7 - Prob. 32.7QQ Ch. 32 - (a) What is the resistance of a lightbulb that...Ch. 32 - A certain lightbulb is rated at 60.0 W when...Ch. 32 - The current in the circuit shown in Figure P32.3...

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