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 79CP

(a)

To determine

The angular frequency at which power delivered to the resistor is a maximum.

(a)

Expert Solution
Check Mark

Answer to Problem 79CP

The angular frequency at which power delivered to the resistor is a maximum is 224rad/s.

Explanation of Solution

Write the expression to obtain the resonance frequency in a series RLC circuit.

    ω0=1LC

Here, ω0 is the resonance frequency, L is the inductance of the inductor and C is the capacitance of the capacitor.

Conclusion:

Substitute 10.0μF for C and 2.00H for L in the above equation to calculate ω0.

    ω0=1(2.00H)(10.0μF)=1(2.00H)(10.0μF×1F106μF)=224rad/s

Therefore, the angular frequency at which power delivered to the resistor is a maximum is 224rad/s.

(b)

To determine

The average power delivered at the frequency which is calculates in part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 79CP

The average power delivered at the frequency which is calculates in part (a) is 500W.

Explanation of Solution

Write the expression to obtain the average power delivered to the circuit at the calculated frequency.

    Pavg=12ΔVmaxImax

Here, P is the average power delivered to the circuit at the calculated frequency, ΔVmax is the maximum voltage across the circuit and Imax is the maximum current across the circuit.

Substitute ΔVmaxR for Imax in the above equation.

    Pavg=12ΔVmax(ΔVmaxR)=12ΔVmax2R

Conclusion:

Substitute 100V for ΔVmax and 10.0Ω for R in the above equation to calculate Pavg.

    Pavg=12(100V)210.0Ω=1210000V210.0Ω=500W

Therefore, the average power delivered at the frequency which is calculates in part (a) is 500W.

(c)

To determine

The two angular frequencies at which power delivered is one-half the maximum value.

(c)

Expert Solution
Check Mark

Answer to Problem 79CP

The two angular frequencies at which power delivered is one-half the maximum value are 226s1 and 221s1.

Explanation of Solution

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

    Pavg=(ΔVmax)2R2(R2+(XLXC)2)

Here, Pavg is the average power delivered in the circuit, Δmax is the maximum voltage in the circuit, R is the resistance of the resistor, XL is the inductive reactance and XC is the capacitive reactance.

Substitute 12ΔVmax22R for Pavg in the above equation.

    12ΔVmax22R=(ΔVmax)2R2(R2+(XLXC)2)12R=R(R2+(XLXC)2)(R2+(XLXC)2)=2R2(XLXC)2=R2

Further solve the above equation.

    (XLXC)=±R

Case (i): when (XLXC)=R

Substitute ω1L for XL and 1ω1C for XC in the above equation.

    (ω1L1ω1C)=Rω12LC1=Rω1Cω12LCRω1C1=0                                                                                       (I)

Write expression to obtain the quadratic equation.

    ax2+bx+c=0                                                                                                       (II)

Here, x is the variable and a,b and c are constants.

Compare equation (I) and (II).

    a=LC, b=RC, c=1 and x=ω1.

Write the expression to obtain the roots of the quadratic equation.

    x=b±b24ac2a                                                                                                (III)

Substitute LC for a, RC for b, 1 for c and ω1 for x in equation (III).

    ω1=RC±(RC)24(LC)(1)2(LC)=RC±(RC)2+4(LC)2(LC)                                                                          (IV)

Case (ii): when (XLXC)=R

Substitute ω2L for XL and 1ω2C for XC in the above equation.

    (ω2L1ω2C)=Rω22LC1=Rω2Cω22LC+Rω2C1=0                                                                                   (V)

Compare equation (IV) and (II).

    a=LC, b=RC, c=1 and x=ω1.

Substitute LC for a, RC for b, 1 for c and ω2 for x in equation (V)

    ω2=RC±(RC)24(LC)(1)2(LC)=RC±(RC)2+4(LC)2(LC)                                                                   (IV)

Conclusion:

Substitute 10.0Ω for R, 10.0μF for C and 2.00H for L in equation (IV) to calculate ω1.

    ω1=(10.0Ω)(10.0μF)±((10.0Ω)(10.0μF))2+4((2.00H)(10.0μF))2((2.00H)(10.0μF))=[(10.0Ω)(10.0μF×1F106μF)±((10.0Ω)(10.0μF×1F106μF))2+4((2.00H)(10.0μF×1F106μF))2((2.00H)(10.0μF×1F106μF))]=221s1or

    ω1=226s1

Consider the positive value of ω1.

Substitute 10.0Ω for R, 10.0μF for C and 2.00H for L in equation (VI) to calculate ω2.

    ω2=(10.0Ω)(10.0μF)±((10.0Ω)(10.0μF))2+4((2.00H)(10.0μF))2((2.00H)(10.0μF))=[(10.0Ω)(10.0μF×1F106μF)±((10.0Ω)(10.0μF×1F106μF))2+4((2.00H)(10.0μF×1F106μF))2((2.00H)(10.0μF×1F106μF))]=226s1or

    ω2=221s1

Consider the positive value of ω2.

Therefore, the two angular frequencies at which power delivered is one-half the maximum value are 226s1 and 221s1.

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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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