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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 ΔVmax = 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 ΔVoutVin; (d) the frequency fo at which the ratio has its maximum value; (e) the phase shift between Δvin and Δvout at 200 Hz, at fo, 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
Check Mark
To determine

The required value of inductance L .

Answer to Problem 49CP

The required value of inductance L is 580μH .

Explanation of Solution

Given info: The value of resistance is 8.00Ω , maximum potential difference is 10.0V , ratio of ΔVout/ΔVin at 200Hz and at 4.00×103Hz 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+(XLXC)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+(2πfL12πfC)2 (4)

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

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

Z=(8.00Ω)2+((2π×200Hz)L1(2π×200Hz)C)2=64Ω2+((400Hz)L1(400Hz)C)2

Substitute 8.00Ω for R , 64Ω2+((400Hz)L1(400Hz)C)2 for Z and 12 for ΔVoutΔVin in equation (3).

8.00Ω64Ω2+((400Hz)L1(400Hz)C)2=12(8.00Ω64Ω2+((400Hz)L1(400Hz)C)2)2=(12)264Ω2(64Ω2+((400Hz)L1(400Hz)C)2)=1464Ω2+((400Hz)L1(400Hz)C)2=256Ω2

Solve the equation further,

((400πHz)L1(400πHz)C)2=256Ω264Ω2(400πHz)L1(400πHz)C=192Ω(400πHz)L1(400πHz)C=13.9Ω

Divide the equation by 20 both side.

(20πHz)L1(8000πHz)C=0.695Ω (5)

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

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

Z=(8.00Ω)2+((2π×4.00×103Hz)L1(2π×4.00×103Hz)C)2=64Ω2+((8000πHz)L1(8000πHz)C)2

Substitute 8.00Ω for R , 64Ω2+((8000πHz)L1(8000πHz)C)2 for Z and 12 for ΔVoutΔVin in equation (3).

8.00Ω64Ω2+((8000πHz)L1(8000πHz)C)2=12(8.00Ω64Ω2+((8000πHz)L1(8000πHz)C)2)2=(12)264Ω2(64Ω2+((8000πHz)L1(8000πHz)C)2)=1464Ω2+((8000πHz)L1(8000πHz)C)2=256Ω2

Solve the equation further,

((8000πHz)L1(8000πHz)C)2=256Ω264Ω2(8000πHz)L1(8000πHz)C=192Ω(8000πHz)L1(8000πHz)C=13.9Ω (6)

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

[(8000πHz)L(20πHz)L]=13.9Ω0.695ΩL=580×106H×106μH1H=580μH

Conclusion:

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

(b)

Expert Solution
Check Mark
To determine

The required value of capacitance C .

Answer to Problem 49CP

The required value of capacitance C is 54.6μF .

Explanation of Solution

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

The equation (6) is given as,

(8000πHz)L1(8000πHz)C=13.9Ω

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

(1579043.5Hz)(580μH×106H1μH)1C=17411.9HzΩC=54.6×106F106μF1FC=54.6μF

Conclusion:

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

(c)

Expert Solution
Check Mark
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 8.00Ω , maximum potential difference is 10.0V , ratio of ΔVout/ΔVin at 200Hz and at 4.00×103Hz 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
Check Mark
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 894Hz .

Explanation of Solution

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

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

Formula to calculate the resonance frequency is,

f0=12πLC

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

f0=12π(580μH×106H1μH)(54.6μC×106C1μC)=894Hz

Conclusion:

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

(e)

Expert Solution
Check Mark
To determine

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

Answer to Problem 49CP

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

Explanation of Solution

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

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

ϕ1=cos1(RZ)

At 200Hz 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=cos1(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=cos1(RR)=cos1(1)=0°

At 4.00×103Hz 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=cos1(12)=60°

Conclusion:

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

(f)

Expert Solution
Check Mark
To determine

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

Answer to Problem 49CP

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

Explanation of Solution

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

ΔvoutΔvin=12Δvout=12Δvin

Substitute 12Δvin for Δvout in equation (7).

P=(12Δvin)22R=18((Δvin)2R)

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

P=18((10.0V)28.00Ω)=1.56W

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

P=((10.0V)22×8.00Ω)=6.25W

Conclusion:

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

(g)

Expert Solution
Check Mark
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 8.00Ω , maximum potential difference is 10.0V , ratio of ΔVout/ΔVin at 200Hz and at 4.00×103Hz is 12 .

Formula to calculate the quality factor is,

Q=2πf0LR

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

Q=2π×894Hz×580μH×106H1μH8.00Ω=0.408

Conclusion:

Therefore, the quality factor of the circuit is 0.408 .

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Chapter 32 Solutions

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term

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