Physics for Scientists and Engineers, Volume 1, Chapters 1-22
Physics for Scientists and Engineers, Volume 1, Chapters 1-22
8th Edition
ISBN: 9781439048382
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 33, Problem 33.81CP
To determine

The frequency will the AC source deliver a power of 250W .

Expert Solution & Answer
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Answer to Problem 33.81CP

There are two values of frequencies at which the AC source deliver a power of 250W that are 58.7Hz and 35.9Hz because the circuit can be either above or below resonance.

Explanation of Solution

Given info: The rms potential difference of an AC source is 120V .

Formula to calculate the rms current in the circuit is,

Irms=ΔVrmsZ

Here,

Irms is the rms current in the circuit.

ΔVrms is the rms source voltage.

Write the expression for the power deliver to the circuit.

P=(Irms)2R

Substitute ΔVrmsZ for Irms in above circuit.

P=(ΔVrmsZ)2R

Rearrange the equation for Z2 .

Z2=((ΔVrms)2P)R (1)

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 ωL for XL and 1ωC for XC to find Z .

Z=R2+(ωL1ωC)2 (2)

Substitute R2+(ωL1ωC)2 for Z in equation (1).

(R2+(ωL1ωC)2)2=((ΔVrms)2P)RR2+(ωL1ωC)2=((ΔVrms)2P)R(ωL1ωC)2=((ΔVrms)2P)RR2 (3)

Assume ((ΔVrms)2P)RR2 is equal to A2 .

Substitute 120V for ΔVrms , 40.0Ω for R and 250W for P to find the value of A .

A2=((120V)2250W)(40.0Ω)(40.0Ω)2A=704Ω

Substitute (ωL1ωC)2 for A2 in equation

Simplify the equation more,

(ωL1ωC)=±Aω2LCωCA1=0

Solve for first equation,

ω2LCωCA1=0

Since the angular frequency is a positive quantity. So, equation can be rewrite as,

Solve the quadratic equation.

ω1=(CA)+(CA)24LC(1)2LC=A+A2+4L/C2L

Substitute 704Ω for A , 185mH for L and 65.0μF for C .

ω1=704Ω+704Ω+4×(185mH×103H1mH)(65.0μF×106F1F)2×185mH×103H1mH=704Ω+12088.61Ω0.37H=368.86rad/s

Formula to calculate the frequency is,

f1=ω12π

Substitute 368.86rad/s for ω1 to find f1 .

f1=368.86rad/s2π=58.7Hz

Solve for second equation,

ω2LC+ωCA1=0

Since the angular frequency is a positive quantity. So, equation can be rewrite as,

Solve the quadratic equation.

ω2=(CA)+(CA)24LC(1)2LC=A+A2+4L/C2L

Substitute 704Ω for A , 185mH for L and 65.0μF for C .

ω2=704Ω+704Ω+4×(185mH×103H1mH)(65.0μF×106F1F)2×185mH×103H1mH=704Ω+12088.61Ω0.37H=225.44rad/s

Formula to calculate the frequency is,

f2=ω22π

Substitute 225.44rad/s for ω2 to find f1 .

f1=225.44rad/s2π=35.9Hz

Since the value of frequency comes out to be two that means the circuit deliver the given power two times one is below resonance and another one is above resonance.

Conclusion:

Therefore, there are two values of frequencies at which the AC source deliver a power of 250W that are 58.7Hz and 35.9Hz because the circuit can be either above or below resonance.

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Chapter 33 Solutions

Physics for Scientists and Engineers, Volume 1, Chapters 1-22

Ch. 33 - Prob. 33.4OQCh. 33 - Prob. 33.5OQCh. 33 - A sinusoidally varying potential difference has...Ch. 33 - A series RLCcircuit contains a 20.0- resistor, a...Ch. 33 - A resistor, a capacitor, and an inductor are...Ch. 33 - (a) Why does a capacitor act as a short circuit at...Ch. 33 - What is the plia.se angle in a series RLC circuit...Ch. 33 - Prob. 33.11OQCh. 33 - A 6.00-V battery is connected across the primary...Ch. 33 - Do AC ammeters and voltmeters read (a)...Ch. 33 - (a) Explain how the quality factor is related to...Ch. 33 - (a) Explain how the mnemonic ELI the ICE man can...Ch. 33 - Why is the sum of the maximum voltages across each...Ch. 33 - (a) Does the phase angle in an RLC series circuit...Ch. 33 - Prob. 33.5CQCh. 33 - As shown in Figure CQ33.6, a person pulls a vacuum...Ch. 33 - Prob. 33.7CQCh. 33 - Will a transformer operate if a battery is used...Ch. 33 - Prob. 33.9CQCh. 33 - Prob. 33.10CQCh. 33 - When an AC source is connected across a 12.0-...Ch. 33 - (a) What is the resistance of a lightbulb that...Ch. 33 - An AC power supply produces a maximum voltage Vmax...Ch. 33 - A certain lightbulb is rated at 60.0 W when...Ch. 33 - The current in the circuit shown in Figure P32.3...Ch. 33 - In the AC circuit shown in Figure P32.3, R = 70.0 ...Ch. 33 - An audio amplifier, represented by the AC I source...Ch. 33 - Figure P32.4 shows three lightbulbs connected to a...Ch. 33 - An inductor has a .54.0- reactance when connected...Ch. 33 - In a purely inductive AC circuit as shown in...Ch. 33 - Prob. 33.11PCh. 33 - An inductor is connected to an AC power supply...Ch. 33 - An AC source has an output rms voltage of 78.0 V...Ch. 33 - A 20.0-mH inductor is connected to a North...Ch. 33 - Review. 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In an RLC series circuit that includes a...Ch. 33 - Prob. 33.34PCh. 33 - A series RLC circuit has a resistance of 45.0 and...Ch. 33 - An AC voltage of the form = 100 sin 1 000t, where...Ch. 33 - A series RLC circuit has a resistance of 22.0 and...Ch. 33 - An AC voltage of the form v = 90.0 sin 350t, where...Ch. 33 - ln a certain series RLC circuit, Irms = 9.00 A,...Ch. 33 - Prob. 33.40PCh. 33 - Prob. 33.41PCh. 33 - A series RLC circuit has components with the...Ch. 33 - An RLC circuit is used in a radio to tune into an...Ch. 33 - The LC circuit of a radar transmitter oscillates...Ch. 33 - A 10.0- resistor, 10.0-mH inductor, and 100-F...Ch. 33 - A resistor R, inductor L, and capacitor C are...Ch. 33 - Review. A radar transmitter contains an LC circuit...Ch. 33 - A step-down transformer is used for recharging the...Ch. 33 - The primary coil of a transformer has N1 = 350...Ch. 33 - A transmission line that has a resistance per unit...Ch. 33 - In the transformer shown in Figure P33.51, the...Ch. 33 - A person is working near the secondary of a...Ch. 33 - The RC high-pass filter shown in Figure P33.53 has...Ch. 33 - Consider the RC high-pass filler circuit shown in...Ch. 33 - Prob. 33.55PCh. 33 - Consider the Filter circuit shown in Figure...Ch. 33 - A step-up transformer is designed to have an...Ch. 33 - Prob. 33.58APCh. 33 - Review. The voltage phasor diagram for a certain...Ch. 33 - Prob. 33.60APCh. 33 - Energy is to be transmitted over a pair of copper...Ch. 33 - Energy is to be transmitted over a pair of copper...Ch. 33 - A 400- resistor, an inductor, and a capacitor are...Ch. 33 - Show that the rms value for the sawtooth voltage...Ch. 33 - A transformer may be used to provide maximum power...Ch. 33 - A capacitor, a coil, and two resistors of equal...Ch. 33 - Marie Cornu, a physicist at the Polytechnic...Ch. 33 - A series RLC circuit has resonance angular...Ch. 33 - Review. One insulated conductor from a household...Ch. 33 - (a) Sketch a graph of the phase angle for an RLC...Ch. 33 - In Figure P33.71, find the rms current delivered...Ch. 33 - Review. In the circuit shown in Figure P32.44,...Ch. 33 - Prob. 33.73APCh. 33 - A series RLC circuit is operating at 2.00 103 Hz....Ch. 33 - A series RLC circuit consists of an 8.00-...Ch. 33 - A series RLC circuit in which R = l.00 , L = 1.00...Ch. 33 - The resistor in Figure P32.49 represents the...Ch. 33 - An 80.0- resistor and a 200-mH inductor are...Ch. 33 - Prob. 33.79CPCh. 33 - P33.80a shows a parallel RLC circuit. The...Ch. 33 - Prob. 33.81CP
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