Understanding Power Formulas; When to use Them

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The discussion centers on the three power formulas: P=IV, P=I^2R, and P=V^2/R, and their appropriate applications. P=IV is primarily used for calculating the total power in a circuit, while P=I^2R is better suited for determining energy dissipation in individual components. Participants highlight the importance of understanding the variables involved and the context of the problem to select the correct formula. The conversation emphasizes that knowing the knowns and unknowns in a problem can guide the choice of formula. Ultimately, using the right formula is crucial for obtaining accurate results in electrical calculations.
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There are three Formulas for Power, ( P=IV, P=I^2R, P=V^2/R). Can someone tell me when to use the right forumlas?
 
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How has your textbook explained it?
 
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ProfuselyQuarky said:
How has your textbook explained it?
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
 
What are your thoughts? What is different about the three equations and when might those differences be important?
 
Gajan1234 said:
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
Okay, well, do you know what each variable stands for?
 
Dale said:
What are your thoughts? What is different about the three equations and when might those differences be important?

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.
 
ProfuselyQuarky said:
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
 
Gajan1234 said:
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.
OK, that is a good start.

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.
 
Dale said:
OK, that is a good start.

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

Well we are looking at difficulty of current going through due to resistivity.
 
  • #10
Gajan1234 said:
Well we are looking at difficulty of current going through due to resistivity.
Right. So would it be possible to use P=I^2 R if you don't know the resistance?
 
  • #11
Gajan1234 said:
But when it come to practice questions it is used differently in different context
Not really :smile: 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 here
 
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  • #12
Dale said:
Right. So would it be possible to use P=I^2 R if you don't know the resistance?

Well it is possible if you know the voltage
 
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  • #13
Gajan1234 said:
Well it is possible if you know the voltage
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?
 
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  • #14
Gajan1234 said:
Well it is possible if you know the voltage
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 anyone variable.

That is all those equations do. You are then left with one equation in three variables. Use the one that fits.
 
  • #15
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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Dale said:
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 anyone variable.

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

Dale said:
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 anyone variable.

That is all those ewuayions do. You are then left with one equation in three variables. Use the one that fits.
 
  • #16
Gajan1234 said:
If you look ag question 3c you can not use all the formulas. Only one will give you the right answer.
What are the knowns and what are the unknowns? Which formula fits?
 
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  • #17
Dale said:
What are the knowns and what are the unknowns?
And based on that, which formula for power matches what you know?
 
  • #18
ProfuselyQuarky said:
And based on that, which formula for power matches what you know?
P=v`2/r
 
  • #19
Gajan1234 said:
P=v`2/r
This gives me the wrong answer
 
  • #20
Gajan1234 said:
This gives me the wrong answer
What numbers did you use?
 
  • #21
Dale said:
What are the knowns and what are the unknowns? Which formula fits?
Well I can use the P=v^2/R formula but it gives me the wrong answer
 
  • #22
ProfuselyQuarky said:
What numbers did you use?
P=6^2/9
 
  • #23
Gajan1234 said:
P=6^2/9
Why did you use "6" for volts? The picture's not that clear, but doesn't it read "3"?
 
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  • #24
My suggestion is to use P = IV. Even though you get the same results this equation is more general. For instance for an inductor you can still use P=IV but not the other ones.
 
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  • #25
Gajan1234 said:
There are three Formulas for Power, ( P=IV, P=I^2R, P=V^2/R). Can someone tell me when to use the right forumlas?
When? Always, always use the right formulas! :-p

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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