Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
9th Edition
ISBN: 9781305932302
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 33, Problem 77CP

(a)

To determine

The inductance of the inductor.

(a)

Expert Solution
Check Mark

Answer to Problem 77CP

The inductance of the inductor is 580μH.

Explanation of Solution

Write the expression to obtain the inductive reactance.

    XL=2πfL

Here, XL is the inductive reactance, f is the frequency and L is the inductance of the inductor.

Write the expression to obtain the capacitive reactance.

    XC=12πfC

Here, XC 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.

    ΔVout=ΔVinR(R2+(XLXC)2)

Here, ΔVout is the output voltage, ΔVin is the input voltage, R is the resistance of the resistor, XL is the inductive reactance and XC is the capacitive reactance.

Re-write the above equation.

    ΔVoutΔVin=R(R2+(XLXC)2)

Substitute 12 for ΔVoutΔVin in the above equation.

    12=R(R2+(XLXC)2)4R2=R2+(XLXC)23R2=(XLXC)23R=|(XLXC)|

Further substitute 2πfL for XL and 12πfC for XC in the above equation.

    3R=(2πfL12πfC)                                                                                         (I)

Substitute 200Hz for ω in equation (I).

    3R=(12π(200Hz)C2π(200Hz)L)                                                           (II)

Substitute 4.00×103Hz for ω in equation (I).

    3R=(2π(4.00×103Hz)L12π(4.00×103Hz)C)                                       (III)

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

    203R+3R=[202π(200Hz)C2π(4000Hz)L+2π(4000Hz)L12π(4000Hz)C]203R+3R=202π(200Hz)C+12π(4000Hz)CC=1213R(202π(200Hz)+12π(4000Hz)C)                                 (IV)

Conclusion:

Substitute 8.00Ω for R in equation (IV) to calculate C.

    C=1213(8.00Ω)(202π(200Hz)+12π(4000Hz))=54.6×106F×1μF106F=54.6μF

Substitute 54.6×106F for C and 8.00Ω for R in equation (II) to calculate L.

    3(8.00Ω)=(2π(4.00×103Hz)L12π(4.00×103Hz)(54.6×106F))2π(4.00×103Hz)L=3(8.00Ω)+12π(4.00×103Hz)(54.6×106F)L=580×106H×1μH106H=580μH

Therefore, the inductance of the inductor is 580μH.

(b)

To determine

The capacitance of the capacitor.

(b)

Expert Solution
Check Mark

Answer to Problem 77CP

The capacitance of the capacitor is 54.6μF.

Explanation of Solution

Conclusion:

As the value of capacitance of the capacitor is already calculated in part (a) that is equal to 54.6μF.

Therefore, the capacitance of the capacitor is 54.6μF.

(c)

To determine

The maximum value of the ratio of ΔVout/ΔVin.

(c)

Expert Solution
Check Mark

Answer to Problem 77CP

The maximum value of the ratio of ΔVout/ΔVin is 1.00.

Explanation of Solution

The maximum value of the ΔVout/ΔVin is obtained when inductive reactance is equal to the capacitive reactance.

    XL=XC

Write the expression to obtain the ratio of ΔVout/ΔVin.

    ΔVoutΔVin=R(R2+(XLXC)2)

Here, ΔVout is the output voltage, ΔVin is the input voltage, R is the resistance of the resistor, XL is the inductive reactance and XC is the capacitive reactance.

Conclusion:

Substitute XL for XC in the above equation.

    ΔVoutΔVin=R(R2+(XLXL)2)=R(R2+(0)2)=RR=1.00

Therefore, the maximum value of the ratio of ΔVout/ΔVin is 1.00.

(d)

To determine

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

(d)

Expert Solution
Check Mark

Answer to Problem 77CP

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

Explanation of Solution

The maximum ratio of ΔVout/ΔVin is at the resonance condition.

Write the expression to obtain the frequency at resonance condition.

    f0=12πLC

Here, f0 is the frequency, L is the inductance of the inductor and C is the capacitance of the capacitor.

Conclusion:

Substitute 580×106H for L and 54.6×106F for C in the above equation to calculate f0.

    f0=12π(580×106H)(54.6×106F)=894.3Hz894Hz

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

(e)

To determine

The phase shift between ΔVin and ΔVout at frequency 200Hz, at f0 and at 4.00×103Hz.

(e)

Expert Solution
Check Mark

Answer to Problem 77CP

The phase shift at 200Hz frequency is 60° that means ΔVout leads ΔVin, the phase shift at f0 frequency is 0° that means ΔVout is in phase with ΔVin and the phase shift at 4.00×103Hz frequency is 60° that means ΔVout lags ΔVin.

Explanation of Solution

Case (i): At 200Hz, XC>XL.

Write the expression to obtain the phase shift between ΔVin and ΔVout.

    ϕ=cos1(RZ)                                                                                                   (V)

Here, ϕ is the phase shift.

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

    ΔVoutΔVin=RZ                                                                                                 (VI)

Here, ΔVout is the output voltage and ΔVin is the input voltage, R is the resistance of the resistor and Z is the impedance of the circuit.

Substitute 12 for ΔVoutΔVin in the equation (VI).

    12=RZ

Case (ii): At f0 which is the resonance frequency calculated in part (d).

Substitute 1 for ΔVoutΔVin in equation (VI).

    1=RZ

Case (iii): At 4.00×103Hz, XC<XL.

Write the expression to obtain the phase shift between ΔVin and ΔVout.

    ϕ=cos1(RZ)                                                                                         (VII)

Here, ϕ is the phase shift.

Substitute 12 for ΔVoutΔVin in the equation (VI).

    12=RZ

Conclusion:

At 200Hz, XC>XL.

Substitute 12 for RZ in equation (V) to calculate ϕ.

    ϕ=cos1(12)=60°

At f0.

Substitute 1 for RZ in equation (VII) to calculate ϕ.

    ϕ=cos1(1)=0°

At 4.00×103Hz, XC<XL.

Substitute 12 for RZ in equation (VII) to calculate ϕ.

    ϕ=cos1(12)=60°

Therefore, the phase shift at 200Hz frequency is 60° that means ΔVout leads ΔVin, the phase shift at f0 frequency is 0° that means ΔVout is in phase with ΔVin and the phase shift at 4.00×103Hz frequency is 60° that means ΔVout lags ΔVin.

(f)

To determine

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

(f)

Expert Solution
Check Mark

Answer to Problem 77CP

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

Explanation of Solution

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

    Pavg=12((ΔVin)2cosϕR2+(XLXC)2)                                                                                    (VIII)

Here, Pavg is the average power in the circuit, ΔVin is the input voltage, R is the resistance in the circuit, XL is the inductive reactance and XC is the capacitive reactance.

Conclusion:

Case (i): At frequency 200Hz and at 4.00×103Hz.

Substitute 10.0V for ΔVin, 3R for (XLXC) and 60° for ϕ in equation (VIII) to calculate Pavg.

    Pavg=12((10.0V)2cos60°R2+(3R)2)

Further substitute 8.00Ω for R in the above equation.

    Pavg=12((10.0V)2cos60°(8.00Ω)2+(3(8.00Ω))2)=12(100V2(12)(8.00Ω)2+(3(8.00Ω))2)=1.56W

Case (ii): At frequency f0.

Substitute 10.0V for ΔVin, XL for XC, 8.00Ω for R and 0° for ϕ in equation (VIII) to calculate Pavg.

    Pavg=12((10.0V)2cos0°(8.00Ω)2+(XLXL)2)=12(100V2(8.00Ω)2)=6.25W

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

(g)

To determine

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

(g)

Expert Solution
Check Mark

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=(2πf0)LR

Here, Q is the quality factor, f0 is the resonance frequency, L is the inductance of the inductor and R is the resistor of the circuit.

Conclusion:

Substitute 894Hz for f0, 580μF for L and 8.00Ω for R in the above equation.

    Q=(2π(894Hz))(580μF)8.00Ω=(5617.16rad/s)(580μF×1F106μF)8.00Ω=0.408

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

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Chapter 33 Solutions

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term

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