College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 17, Problem 68AP

A man wishes to vacuum his car with a canister vacuum cleaner marked 535 W at 120. V. The car is parked far from the building, so he uses an extension cord 15.0 m long to plug the cleaner into a 120.-V source. Assume the cleaner has constant resistance. (a) If the resistance of each of the two conductors of the extension cord is 0.900 Ω, what is the actual power delivered to the cleaner? (b) If, instead, the power is to be at least 525 W, what must be the diameter of each of two identical copper conductors in the cord the young man buys? (c) Repeat part (b) if the power is to be at least 532 W. Suggestion: A symbolic solution can simplify the calculations.

(a)

Expert Solution
Check Mark
To determine
The actual power delivered to vacuum cleaner.

Answer to Problem 68AP

The actual power delivered to vacuum cleaner is 470W

Explanation of Solution

Given Info: A vacuum cleaner utilizes 535W power when connected to 120V supply. The extension cord from vacuum cleaner which used to clean the car consists of two conducting wires. The total length of extension cord is 15.0m . The resistance of each conductor is 0.900Ω .

Explanation:

Formula to calculate the total current in the circuit is,

I=ΔVRtotal (I)

  • I is the current in the circuit,
  • ΔV is the voltage of power supply,
  • Rtotal is the total resistance in the path of current,

Formula to calculate the total resistance in the path of current is,

Rtotal=Rvaccum+2Rc (II)

  • Rvaccum is the resistance offered by the vacuum cleaner,
  • Rc is the resistance offered by each conductor in the extension cord,

Formula to calculate the resistance offered by vacuum cleaner is,

Rvaccum=(ΔV)2P (III)

  • P is the rated power of vacuum cleaner,

Substitute equation (III) in equation (II) and rewrite equation (II).

Rtotal=(ΔV)2P+2Rc (IV)

Substitute equation (IV) in equation (I) and rewrite equation (I).

I=ΔV((ΔV)2P+2Rc)

Substitute 120V for ΔV , 535W for P and 0.900Ω for Rc in the above equation to find I .

I=120V(((120V)2535W)+2(0.900Ω))=4.18A

The total current in the circuit is 4.18A

Conclusion

Formula to calculate the actual power delivered to vacuum cleaner is,

Pdelivered=I2Rvaccum (V)

  • Pdelivered is the actual power delivered to vacuum cleaner

Formula to calculate the resistance offered by vacuum cleaner is,

Rvaccum=(ΔV)2P

Use (ΔV)2/P the above equation for Rvaccum in the above equation to rewrite P .

Pdelivered=I2(ΔV)2P

Substitute 4.18A for I , 120V for ΔV and 535W for P in the above equation to find P .

P=(4.18A)2((120V)2535W)=470.28W470W

Therefore, the actual power delivered to vacuum cleaner is 470W

(b)

Expert Solution
Check Mark
To determine
The diameter of each of identical copper wire in the extension cord when minimum power is 525W .

Answer to Problem 68AP

The diameter of each of identical copper wire in the extension cord is 1.59×103m

Explanation of Solution

Given Info: A vacuum cleaner utilizes 535W power when connected to 120V supply Vacuum cleaner utilizes at least 525W power. The extension cord from vacuum cleaner which used to clean the car consists of two conducting wires. The total length of extension cord is 15.0m .

Explanation:

Formula to calculate the actual minimum power delivered to vacuum cleaner is,

Pmin=I2Rvaccum

  • Pmin is the minimum acceptable power delivered to vacuum cleaner,

Use (ΔV/Rtotal) for I and ((ΔV)2/P) for Rvaccum to rewrite Pdelivered .

Pmin=(ΔVRtotal)2((ΔV)2P)

Rewrite the above equation in terms of Rtotal

Rtotal=(ΔV)2PminP (VI)

Equate (IV) and equation (VI)

(ΔV)2P+2Rc=(ΔV)2PminP

Use Rc,max for Rc in the above relation to rewrite it.

  • Rc,max is the maximum value of resistance of wire when power delivered to vacuum cleaner is minimum.

Rc,max=12[(ΔV)2PminP(ΔV)2P]Rc,max=(ΔV)22[1PminP1P]

Substitute 120V for ΔV , 525W for Pmin and 535W for P in the above equation to find Rc,max .

Rc,max=(120V)22[1(525W)(535W)1(535W)]=0.1276Ω0.128Ω

The maximum resistance of copper wire is 0.128Ω .

Conclusion:

Formula to calculate the diameter of copper wire is,

d=4ρLπRc,max

  • d is the diameter of copper wire,
  • L is the length of each copper wire,
  • ρ is the resistivity of copper,

Substitute 1.7×108Ω.m for ρ , 15.0m for L , 3.14 for π and 0.128Ω for Rc,max in the above equation to find d .

d=4(1.7×108Ω.m)(15.0m)(3.14)(0.128Ω)=1.59×103m=1.59mm

Therefore, the diameter of each of identical copper wire in the extension cord is 1.59mm

(c)

Expert Solution
Check Mark
To determine
The diameter of each of identical copper wire in the extension cord when minimum power is 532W .

Answer to Problem 68AP

The diameter of each of identical copper wire in the extension cord is 2.93mm .

Explanation of Solution

Formula to find the maximum resistance of copper wire is,

Rc,max=(ΔV)22[1PminP1P]

Substitute 120V for ΔV , 532W for Pmin and 535W for P in the above equation to find Rc,max .

Rc,max=(120V)22[1(532W)(535W)1(535W)]=0.03789Ω0.0379Ω

Formula to calculate the diameter of copper wire is,

d=4ρLπRc,max

Substitute 1.7×108Ω.m for ρ , 15.0m for L 3.14 for π and 0.0379Ω for Rc,max in the above equation to find d .

d=4(1.7×108Ω.m)(15.0m)(3.14)(0.0379Ω)=2.93×103m=2.93mm

Conclusion:

Therefore, the diameter of each of identical copper wire in the extension cord is 2.93mm .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A 660 W electric heater is designed to operate from a 120 V source. If the source voltage drops to 80.0 V, what will be the power dissipated by the same heater? (Assume the resistance of the heater is constant.) a. 440 W b. 660 W c. 293 W d. 73.3 W
When the switch S is open, the voltmeter V of the battery reads 3.05 V. When the switch is closed, the voltmeter reading drops to 2.88 V, and the ammeter A reads 1.57 A.  Find the (a) emf, (b) internal resistance r of the battery, and (c) circuit resistance R. Assume that the two meters are ideal, so they don’t affect the circuit.
There is a 120V 40 W light bulb, which statement is INCORRECT? Why?         the rate of energy dissipation is 40 W       the current flowing through is 0.33 A       the capacitance is 3.0 Farad       the light bulb will function well when plugged into a 120 V socket

Chapter 17 Solutions

College Physics

Ch. 17 - We have seen that an electric field must exist...Ch. 17 - A 12-V battery is connected across a device with...Ch. 17 - Prob. 3CQCh. 17 - In an analogy between traffic flow and electrical...Ch. 17 - Two copper wires A and B have the same length and...Ch. 17 - Two lightbulbs are each connected to a voltage of...Ch. 17 - Newspaper articles often have statements such as...Ch. 17 - There is an old admonition given to experimenters...Ch. 17 - What could happen to the drift velocity of the...Ch. 17 - Power P0 = I0 V0 is delivered to a resistor of...Ch. 17 - When is more power delivered to a lightbulb,...Ch. 17 - Prob. 1PCh. 17 - A copper wire has a circular cross section with a...Ch. 17 - In the Bohr model of the hydrogen atom, an...Ch. 17 - A typical lightning bolt may last for 0.200 s and...Ch. 17 - Prob. 5PCh. 17 - Prob. 6PCh. 17 - A 2.0 102-km-long high-voltage transmission line...Ch. 17 - An aluminum wire having a cross-sectional area of...Ch. 17 - An iron wire has a cross-sectional area of 5.00 ...Ch. 17 - Prob. 10PCh. 17 - Prob. 11PCh. 17 - Germanium is a semiconducting metal with a...Ch. 17 - Prob. 13PCh. 17 - Prob. 14PCh. 17 - Nichrome wire of cross-sectional radius 0.791 mm...Ch. 17 - Prob. 16PCh. 17 - A potential difference of 12 V is found to produce...Ch. 17 - The current supplied by a battery in a portable...Ch. 17 - A wire 50.0 m long and 2.00 mm in diameter is...Ch. 17 - Prob. 20PCh. 17 - Prob. 21PCh. 17 - The human body can exhibit a wide range of...Ch. 17 - Starting from Ohms law, show that E = J, where E...Ch. 17 - Prob. 24PCh. 17 - Prob. 25PCh. 17 - Prob. 26PCh. 17 - Prob. 27PCh. 17 - At what temperature will aluminum have a...Ch. 17 - At 20.0C, the carbon resistor in an electric...Ch. 17 - Prob. 30PCh. 17 - Prob. 31PCh. 17 - An engineer needs a resistor with a zero overall...Ch. 17 - In one form of plethysmograph (a device for...Ch. 17 - Prob. 34PCh. 17 - A 5.00-V power supply provides a maximum current...Ch. 17 - If electrical energy costs 0.12 per kilowatt-hour,...Ch. 17 - Residential building codes typically require the...Ch. 17 - A portable coffee heater supplies a potential...Ch. 17 - The heating element of a coffeemaker operates at...Ch. 17 - A typical cell phone consumes an average of about...Ch. 17 - Lightbulb A is marked 25.0 W 120. V, and lightbulb...Ch. 17 - Prob. 42PCh. 17 - A copper cable is designed to carry a current of...Ch. 17 - Batteries are rated in terms of ampere-hours (A ...Ch. 17 - The potential difference across a resting neuron...Ch. 17 - The cost of electricity varies widely throughout...Ch. 17 - An electric utility company supplies a customers...Ch. 17 - An office worker uses an immersion heater to warm...Ch. 17 - Two wires A and B made of the same material and...Ch. 17 - Prob. 50PCh. 17 - If a battery is rated at 60.0 A h, how much total...Ch. 17 - A car owner forgets to turn off the headlights of...Ch. 17 - Prob. 53APCh. 17 - A given copper wire has a resistance of 5.00 at...Ch. 17 - Prob. 55APCh. 17 - Birds resting on high-voltage power lines are a...Ch. 17 - Prob. 58APCh. 17 - You are cooking breakfast for yourself and a...Ch. 17 - The current in a conductor varies in time as shown...Ch. 17 - A 120.-V motor has mechanical power output of 2.50...Ch. 17 - Prob. 62APCh. 17 - A length of metal wire has a radius of 5.00 103 m...Ch. 17 - In a certain stereo system, each speaker has a...Ch. 17 - A resistor is constructed by forming a material of...Ch. 17 - When a straight wire is heated, its resistance...Ch. 17 - An x-ray tube used for cancer therapy operates at...Ch. 17 - A man wishes to vacuum his car with a canister...
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics Volume 2
Physics
ISBN:9781938168161
Author:OpenStax
Publisher:OpenStax
Ohm's law Explained; Author: ALL ABOUT ELECTRONICS;https://www.youtube.com/watch?v=PV8CMZZKrB4;License: Standard YouTube License, CC-BY