Does This Material Follow Ohm's Law?

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Homework Help Overview

The discussion revolves around the application of Ohm's Law in relation to a material's behavior under varying voltage and current conditions, as well as calculations involving power and energy consumption in electrical circuits.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to determine if a material follows Ohm's Law based on given voltage and current values, while also exploring calculations related to power usage and energy expenditure.
  • Some participants question the assumptions made regarding the relationship between resistance, voltage, and current, and seek clarification on the correct approach to the energy calculation.
  • There is a discussion about the interpretation of energy units and the relevance of joules in the context of kilowatt-hours.

Discussion Status

Participants are engaged in clarifying concepts and calculations, with some guidance provided on how to approach the problems. There is an acknowledgment of the original poster's correct answer to one part of the problem, but no consensus has been reached on the other questions.

Contextual Notes

Participants are encouraged to show their work and reasoning, and there is a note that the thread has been moved to a specific sub-forum for homework help.

physicsphobic
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urgently needed help on intro physics...

This is my first time posting on here so I hope someone will answer my questions.

Here they are:

1. A scientist testing a material discovers that when a voltage of 10V is placed across a material a current of 10A flows. When a voltage of 20V is placed across it, however, a current of 15A flows through it. Does this material obey Ohm's law? Give reason for your answer.

2.a. How much power does a light bulb connected to a 120V outlet use if if draws 0.5A of current? My answer is: 120V *0.5A= 60W
b. One kilowatt-hour is a measure of energy which is equal to 3,600,000J. How many hours would you have to leave on the light bulb from part a in order to expend one kilowatt-hour of energy?

c. If one kilowatt-hour costs $.10, how long would you have to leave the light bulb from part a on in order to spend $1.00 on electricity?

3. The electric charge of an electron is -1.6x10^-19C. What is the force exerted between two electrons separated by one meter? (Be sure to note whether the force is attractive or repulsive).

Thanks for any help given to this newbie!

Joanne
 
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Welcome to PF!

Hi Joanne! Welcome to PF! :wink:

Your answer to 2a is correct.

For the others, show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 


OK, since nobody wants to give me the answers I'll try to give what I think are the answers and, hopefully someone will be kind enough to give me the correct answers.

Since 2a is correct, here's what I think about 2b:
according to my textbook I use this formula..energy=power * time.
plugging the 3,600,000J in for energy and 60W in for power I get: 3600000=60W * time
leaving time on one side, time=60W/3600000=60,000 hours? Is this right because my algebras a little rusty.

2c has me stumped so any help is greatly appreciated.
The first question I think it does obey ohms law because the greater the resistance, the less the current and if the resistance is doubled the current will be half what it would be otherwise. Right??
 
Hi Joanne! :smile:

Sorry, but on this forum we don't give you the answers. :wink:
physicsphobic said:
Since 2a is correct, here's what I think about 2b:
according to my textbook I use this formula..energy=power * time.
plugging the 3,600,000J in for energy and 60W in for power I get: 3600000=60W * time
leaving time on one side, time=60W/3600000=60,000 hours? Is this right because my algebras a little rusty.

Why are you using 3,600,000?

The question asks for 1 kWhr, (that's the same power as a kW heater for an hour) … joules have nothing to do with it.

Try again. :smile:
The first question I think it does obey ohms law because the greater the resistance, the less the current and if the resistance is doubled the current will be half what it would be otherwise. Right??

Ohm's law mentions the amount of the resistance … how much is the resistance in this case? :smile:
 


tiny-tim said:
Sorry, but on this forum we don't give you the answers. :wink:

We also have forums specifically for getting help with homework / coursework exercises. They're sub-forums of "Homework & Coursework Questions" which is near the top of the list of forums. I've moved this thread to "Introductory Physics."
 


figured out most of the problems myself so thanks anyway, tiny-tim for your "help".
 
physicsphobic said:
figured out most of the problems myself so thanks anyway, tiny-tim for your "help".

you're welcome :wink: … we always think our "help" is most successful when the OP ends up answering the question him/herself :biggrin:
 

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