When to use P=I^2R, P=VI, P=V^2/R?

[FaroukYasser]

When to use P=I^2R, P=VI, P=V^2/R??

Hi,
I know that P = I^2R is to find heat loss. but can we use P = V^2/R or P = IV to find the heat loss? simply when do we use which?? (or can we use all anytime?)

The question said to find 1) the power GENERATED by a dc voltage source and to find the 2) Power dissipated in the power source
Should I multiply the voltage difference of the voltage source by the current passing through it to get the power generated and multiply current squared by its internal resistance to get the heat dissipated through it?
Thanks a lot

 

[adjacent]

Hi,
I know that P = I^2R is to find heat loss. but can we use P = V^2/R or P = IV to find the heat loss? simply when do we use which?? (or can we use all anytime?)

$P=\frac{V^2}{R}$ is derived from $V=IR$ and $P=IV$
You get $P=\frac{V^2}{R}$ because $P=IV=\frac{V}{R}V$

I solved the equation $V=IR$ for $I$ and plugged it into that Power equation $P=IV$
Do the same for $P=I^2R$

For the first equation if you don't know V , find it.Then you can use P=IV.
You can do the same for the second equation too.
You can use anyone of these anytime.

The question said to find 1) the power GENERATED by a dc voltage source and to find the 2) Power dissipated in the power source
Should I multiply the voltage difference of the voltage source by the current passing through it to get the power generated and multiply current squared by its internal resistance to get the heat dissipated through it?
Thanks a lot

Yes you are correct.
Note that heat is not the same thing as power.

 

[FaroukYasser]

For the first equation if you don't know V , find it.Then you can use P=IV.
You can do the same for the second equation too.
You can use anyone of these anytime.

I don't understand this part would you be kind enough to elaborate? Thanks for the reply

 

[adjacent]

I don't understand this part would you be kind enough to elaborate? Thanks for the reply

The first equation in your first post is this:$P=I^2R$
Here you have no variable for $V$.
So you can use $V=IR$ and plug in I and R from the first equation to it and find V.
Then you can use $P=IV$.

This is the same as using $P=I^2R$ as I mentioned in my previous post.

Then you have $P=\frac{V^2}{R}$
There you can again use $V=IR$ to find the current and then you can use $P=IV$

 

[FaroukYasser]

The first equation in your first post is this:$P=I^2R$
Here you have no variable for $V$.
So you can use $V=IR$ and plug in I and R from the first equation to it and find V.
Then you can use $P=IV$.

This is the same as using $P=I^2R$ as I mentioned in my previous post.

Then you have $P=\frac{V^2}{R}$
There you can again use $V=IR$ to find the current and then you can use $P=IV$

Ohhhhhhhhh So I guess it is kind of related to each others, You just sort of save a step in the middle Thanks a TON )

 

[enorbet]

Yes they are interdependent. Functionally, it matters a great deal in understanding and measuring what occurs in a system, since it may be considerably easier to get 2 functions but difficult to get the third. For example, because voltage is a differential it is very easy to measure that difference from one part of a system to another. Resistance is a little bit harder since some circuits create an effective resistance because of interaction. In AC circuits or some pulsed DC circuits, we must add impedance, frequency dependent resistance. Current is theoretically very easy because we can insert a known small resistance and measure the voltage drop and translate that as current flow. It is harder to measure directly. Also some circuits are difficult or even dangerous to interrupt to insert any metering.

So, ultimately it means in real life, that you measure any 2 that's convenient and accurate to measure, and you can derive the third and calculate any expression of it you need.