Physics for Scientists and Engineers
Physics for Scientists and Engineers
10th Edition
ISBN: 9781337553278
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 26, Problem 34AP

Lightbulb A is marked “25 W 120 V,” and lightbulb B is marked “100 W 120 V.” These labels mean that each lightbulb has its respective power delivered to it when it is connected to a constant 120-V source. (a) Find the resistance of each lightbulb. (b) During what time interval does 1.00 C pass into lightbulb A? (c) Is this charge different upon its exit versus its entry into the lightbulb? Explain. (d) In what time interval does 1.00 J pass into lightbulb A? (e) By what mechanisms does this energy enter and exit the lightbulb? Explain. (f) Find the cost of running lightbulb A continuously for 30.0 days, assuming the electric company sells its product at $0.110 per kWh.

(a)

Expert Solution
Check Mark
To determine
The resistance of each light bulb.

Answer to Problem 34AP

The resistance of lightbulb A is 576Ω and resistance of lightbulb B is 144Ω .

Explanation of Solution

Given information: Power of light bulb A is 25W , and voltage across light bulb A is 120V , power of light bulb B is 100W , and voltage across light bulb B is 120V , constant voltage source is 120V .

Explanation:

Formula to calculate the resistance of lightbulb A.

PA=V2RARA=V2PA (1)

Here,

PA is the power of light bulb A.

RA is the resistance of lightbulb A.

V is the voltage across light bulb A.

Substitute 120V for V , 25W for PA in equation (1) to find RA ,

RA=(120V)225W=576Ω

Thus, the resistance of lightbulb A is 576Ω .

Formula to calculate the resistance of lightbulb B.

PB=V2RBRB=V2PB (2)

Here,

PB is the power of light bulb B.

RB is the resistance of lightbulb B.

V is the voltage across light bulb B.

Substitute 120V for V , 100W for PB in equation (2) to find RB ,

RB=(120V)2100W=144Ω

Thus, the resistance of lightbulb B is 144Ω .

Conclusion:

Therefore, the resistance of lightbulb A is 576Ω and resistance of lightbulb B is 144Ω .

(b)

Expert Solution
Check Mark
To determine
The time interval through which 1.00C pass into light bulb A.

Answer to Problem 34AP

The time interval through which 1.00C pass into light bulb A is 4.808s .

Explanation of Solution

Given information: Power of light bulb A is 25W , and voltage across light bulb A is 120V , power of light bulb B is 100W , and voltage across light bulb B is 120V , constant voltage source is 120V , charge across that passes through light bulb A is 1.00C .

Explanation:

Formula to calculate the current flowing in th light bulb A.

IA=VRA (3)

Here,

IA is the current flowing in th light bulb A.

Substitute 120V for V , 576Ω for RA in equation (4) to find IA ,

IA=120V576Ω=0.208A

Thus, the current flowing in th light bulb A is 0.208A .

Formula to calculate the time interval through which 1.00C pass into light bulb A.

IA=Q1.00Ct1.00Ct1.00C=Q1.00CIA (4)

Here,

t1.00C is the time interval through which 1.00C pass into light bulb A.

Q1.00C is the charge across that passes through light bulb A.

Substitute 1.00C for Q1.00C , 0.208A for IA in equation (4) to find t1.00C ,

t1.00C=1.00C0.208A=4.8076s4.808s

Thus, the time interval through which 1.00C pass into light bulb A is 4.807s .

Conclusion:

Therefore, the time interval through which 1.00C pass into light bulb A is 4.807s .

(c)

Expert Solution
Check Mark
To determine
The reason that this charge is different upon its exit versus its entry into the light bulb or not.

Answer to Problem 34AP

This charge is not different upon its exit versus its entry into the light bulb because the current is charged over the time and is the same everywhere on the series circuit.

Explanation of Solution

Given information: Power of light bulb A is 25W , and voltage across light bulb A is 120V , power of light bulb B is 100W , and voltage across light bulb B is 120V , constant voltage source is 120V , charge across that passes through light bulb A is 1.00C .

Explanation:

No, the existing charge is the same amount as the entering charge into the light bulb because the current is charged over the time and is the same everywhere on the series circuit.

Thus, this charge is not different upon its exit versus its entry into the light bulb because the current is charged over the time and is the same everywhere on the series circuit.

Conclusion:

Therefore, this charge is not different upon its exit versus its entry into the light bulb because the current is charged over the time and is the same everywhere on the series circuit.

(d)

Expert Solution
Check Mark
To determine
The time interval through which 1.00J pass into light bulb A.

Answer to Problem 34AP

The time interval through which 1.00J pass into light bulb A is 0.04s .

Explanation of Solution

Given information: Power of light bulb A is 25W , and voltage across light bulb A is 120V , power of light bulb B is 100W , and voltage across light bulb B is 120V , constant voltage source is 120V , energy that passes through light bulb A is 1.00J .

Explanation:

Formula to calculate the time interval through which 1.00J pass into light bulb A.

EA=PA×t1.00Jt1.00J=EAPA (5)

Here,

t1.00J is the time interval through which 1.00J pass into light bulb A.

EA is the energy that passes through light bulb A

Substitute 1.00J for EA , 25W for PA in equation (5) to find t1.00J ,

t1.00J=1.00J25W=0.04s

Thus, the time interval through which 1.00J pass into light bulb A is 0.04s .

Conclusion:

Therefore, the time interval through which 1.00J pass into light bulb A is 0.04s .

(e)

Expert Solution
Check Mark
To determine
The mechanism through which this energy enter and exit the light bulb.

Answer to Problem 34AP

The mechanism through which this energy enter and exit the light bulb is that the bulb is connected to the electrical energy source by wires, usually by a wire like copper, which mainly carries the current of electrons into and out the bulb.

Explanation of Solution

Given information: Power of light bulb A is 25W , and voltage across light bulb A is 120V , power of light bulb B is 100W , and voltage across light bulb B is 120V , constant voltage source is 120V , energy that passes through light bulb A is 1.00J .

Explanation:

In this mechanism, the bulb is connected to the electrical energy source by wires, usually by a wire like copper, which mainly carries the current of electrons into and out the bulb.

Thus, the mechanism through which this energy enter and exit the light bulb is that the bulb is connected to the electrical energy source by wires, usually by a wire like copper, which mainly carries the current of electrons into and out the bulb.

Conclusion:

Therefore, the mechanism through which this energy enter and exit the light bulb is that the bulb is connected to the electrical energy source by wires, usually by a wire like copper, which mainly carries the current of electrons into and out the bulb.

(f)

Expert Solution
Check Mark
To determine
The cost of running light bulb A continuously for 30.0days .

Answer to Problem 34AP

The cost of running light bulb A continuously for 30.0days is $1.98 .

Explanation of Solution

Given information: Power of light bulb A is 25W , and voltage across light bulb A is 120V , power of light bulb B is 100W , and voltage across light bulb B is 120V , constant voltage source is 120V , energy that passes through light bulb A is 1.00J , running time of light bulb A is 30.0days , selling cost at which electric company sells product is $0.110/kWh .

Explanation:

Write the expression for the energy for light bulb A works continuously for 30.0days .

E30.0days=P×t (6)

Here,

E30.0days is the energy for light bulb A works continuously for 30.0days .

P is the power of light bulb A.

t is the running time of light bulb A.

Substitute 25W for P , 30.0days for t in equation (6) to find E30.0days ,

E30.0days=25W×(30.0days×24h1day)=18000Wh=18kWh

Thus, the energy for light bulb A works continuously for 30.0days is 18kWh .

Formula to calculate the cost of running light bulb A continuously for 30.0days .

CR=E30.0days×CS (7)

Here,

CR is the cost of running light bulb A continuously for 30.0days .

CS is the selling cost at which electric company sells product.

Substitute 18kWh for E30.0days , $0.110/kWh for CS in equation (7) to find CR ,

CR=18kWh×$0.110/kWh=$1.98

Thus, the cost of running light bulb A continuously for 30.0days is $1.98 .

Conclusion:

Therefore, the cost of running light bulb A continuously for 30.0days is $1.98 .

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Students have asked these similar questions
Lightbulb A is marked “25 W 120 V,” and lightbulb B is marked “100 W 120 V.” These labels mean that each lightbulb has its respective power delivered to it when it is connected to a constant 120-V source. (a) Find the resistance of each lightbulb. (b) During what time interval does 1.00 C pass into lightbulb A? (c) Is this charge different upon its exit versus its entry into the lightbulb? Explain. (d) In what time interval does 1.00 J pass into lightbulb A? (e) By what mechanisms does this energy enter and exit the lightbulb? Explain. (f) Find the cost of running lightbulb A continuously for 30.0 days, assuming the electric company sells its product at $0.110 per kWh.
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Lightbulb A is marked "25 W 120 V," and lightbulb B is marked "100 W 120 V." These labels mean that each lightbulb has its respective power delivered to it when it is connected to a constant 120 V source. (a) Find the resistance of each lightbulb. (b) During what time interval does 1.00 C pass into lightbulb A? (c) Is this charge different upon its exit versus its entry into the lightbulb? Explain. (d) In what time interval does 1.00 J pass into lightbulb A? (e) By what mechanisms does this energy enter and exit the lightbulb? Explain. (f) Find the cost of running lightbulb A continuously for 30.0 days, assuming the electric company sells its product at $0.110 per kWh.

Chapter 26 Solutions

Physics for Scientists and Engineers

Ch. 26 - The quantity of charge q (in coulombs) that has...Ch. 26 - A Van de Graaff generator (see Problem 24)...Ch. 26 - An electric current in a conductor varies with...Ch. 26 - Prob. 10PCh. 26 - An electric heater carries a current of 13.5 A...Ch. 26 - You are working at a company that manufactures...Ch. 26 - Prob. 13PCh. 26 - Prob. 14PCh. 26 - Prob. 15PCh. 26 - Prob. 16PCh. 26 - Prob. 17PCh. 26 - Prob. 18PCh. 26 - An aluminum wire with a diameter of 0.100 mm has a...Ch. 26 - Plethysmographs are devices used for measuring...Ch. 26 - At what temperature will aluminum have a...Ch. 26 - You are working in a laboratory that studies the...Ch. 26 - Assume that global lightning on the Earth...Ch. 26 - The Van de Graaff generator, diagrammed in Figure...Ch. 26 - A 100-W lightbulb connected to a 120-V source...Ch. 26 - The potential difference across a resting neuron...Ch. 26 - The cost of energy delivered to residences by...Ch. 26 - Residential building codes typically require the...Ch. 26 - Assuming the cost of energy from the electric...Ch. 26 - An 11.0-W energy-efficient fluorescent lightbulb...Ch. 26 - A 500-W heating coil designed to operate from 110...Ch. 26 - Why is the following situation impossible? A...Ch. 26 - Make an order-of-magnitude estimate of the cost of...Ch. 26 - Lightbulb A is marked 25 W 120 V, and lightbulb B...Ch. 26 - One wire in a high-voltage transmission line...Ch. 26 - You are working with an oceanographer who is...Ch. 26 - A charge Q is placed on a capacitor of capacitance...Ch. 26 - An experiment is conducted to measure the...Ch. 26 - Prob. 39APCh. 26 - Prob. 40APCh. 26 - Review. An office worker uses an immersion heater...Ch. 26 - The strain in a wire can be monitored and computed...Ch. 26 - A close analogy exists between the flow of energy...Ch. 26 - The dielectric material between the plates of a...Ch. 26 - Review. A parallel-plate capacitor consists of...Ch. 26 - Prob. 46APCh. 26 - Why is the following situation impossible? An...Ch. 26 - Prob. 48CPCh. 26 - A spherical shell with inner radius ra and outer...Ch. 26 - Material with uniform resistivity is formed into...
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