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ProfuselyQuarky

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How has your textbook explained it?

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It is quite confusing in the book as they have used different formula for different question. But I know that P=I^2R is used to find the heatHow has your textbook explained it?

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Dale

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ProfuselyQuarky

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Okay, well, do you know what each variable stands for?It is quite confusing in the book as they have used different formula for different question. But I know that P=I^2R is used to find the heat

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I think that P=IV is mainly used to find the power of the whole circuit whereas P=I^2R is used to find how much energy is given out by the component per second.

But not sure if I am right.

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Okay, well, do you know what each variable stands for?

I for current

R resistance

I understand the fact that it is simple substitution and the formulas are literally same. But when it come to practice questions it is used differently in different context

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Dale

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OK, that is a good start.I think that P=IV is mainly used to find the power of the whole circuit whereas P=I^2R is used to find how much energy is given out by the component per second.

But not sure if I am right.

What is different about P=I^2 R that makes it better for individual components than P=IV? Or what do components have that might lead you to use P=I^2R.

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Well we are looking at difficulty of current going through due to resistivity.OK, that is a good start.

What is different about P=I^2 R that makes it better for individual components than P=IV?

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Dale

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Right. So would it be possible to use P=I^2 R if you don't know the resistance?Well we are looking at difficulty of current going through due to resistivity.

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ProfuselyQuarky

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Not really It just depends on what you're looking for--and the only way to know what you're looking for would be to know what each part means. Perhaps this picture will help: click hereBut when it come to practice questions it is used differently in different context

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Well it is possible if you know the voltageRight. So would it be possible to use P=I^2 R if you don't know the resistance?

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ProfuselyQuarky

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That's correct! See, you can use different equations (in this case, Ohm's Law) to find the missing parts you need. So where's the trouble?Well it is possible if you know the voltage

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Dale

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Yes, exactly. There are only two formulas here: P=IV and V=IR. You have two equations in four variables. You can always substitute one equation into the other to eliminate any one variable.Well it is possible if you know the voltage

That is all those equations do. You are then left with one equation in three variables. Use the one that fits.

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Yes, exactly. There are only two formulas here: P=IV and V=IR. You have two equations in four variables. You can always substitute one equation into the other to eliminate any one variable.

That is all those ewuayions do. You are then left with one equation in three variables. Use the one that fits.

Yes, exactly. There are only two formulas here: P=IV and V=IR. You have two equations in four variables. You can always substitute one equation into the other to eliminate any one variable.

That is all those ewuayions do. You are then left with one equation in three variables. Use the one that fits.

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Dale

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What are the knowns and what are the unknowns? Which formula fits?If you look ag question 3c you can not use all the formulas. Only one will give you the right answer.

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ProfuselyQuarky

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And based on that, which formula for power matches what you know?What are the knowns and what are the unknowns?

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P=v`2/rAnd based on that, which formula for power matches what you know?

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This gives me the wrong answerP=v`2/r

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ProfuselyQuarky

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What numbers did you use?This gives me the wrong answer

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Well I can use the P=v^2/R formula but it gives me the wrong answerWhat are the knowns and what are the unknowns? Which formula fits?

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P=6^2/9What numbers did you use?

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ProfuselyQuarky

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Why did you use "6" for volts? The picture's not that clear, but doesn't it read "3"?P=6^2/9

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russ_watters

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When? Always, always use the right formulas!

In addition to what others have said about the variables provided in the problem, it also matters what the problem is asking you to calculate the power of. If it asks you for the power used by a light bulb, it may be different than if they asked you to find the power loss by the wires going to the light bulb.

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