0
$\begingroup$

![ have two points both of which I find correct. But I also find that both can't be correct at the same time. Because they contradict each other.

  1. The V in VI and the V in V²/R 'are the same thing' (in all cases) : because V²/R is derived simply repalcing the I( electricity through the resistor) with V/R where V is the voltage difference across the resistor.
  2. The V in VI and the V in V²/R 'are not the same thing' : If they were, the idea that they use to reduce the power loss in supply cables wouldn't have worked. But I have found on browsing that idea does work. So the Vs have to be different things in these formulas, like they are saying in the last paragraph in this book. One of my notions have to be wrong for the other to be correct. But I don't get which. Most probably it's the first one because I have found sources explaining the 'power loss prevention' the way they have explained on this page in the picture. But the 1st point can't be wrong either, right? I am really confused(https://i.stack.imgur.com/QyCo6.jpg)
$\endgroup$
1
  • 1
    $\begingroup$ Whether V is the one that goes in the power formula depends entirely on what you defined V to refer to in your system. $\endgroup$
    – The Photon
    Nov 13 '21 at 4:34
4
$\begingroup$

The $V$ in $VI$ and the $V$ in $V^2/R$ are indeed the same thing.

Your confusion comes because there is more than one $V$ and $I$ and $R$ in a typical circuit. In your supply cables example there is a voltage across the cable itself, a voltage across the load, and the source voltage. None of them is equal to the others.

If you want to apply $V^2/R$ for the cables then you must know the voltage across the cable, not the source voltage.

$\endgroup$
3
  • $\begingroup$ It's my first time asking a question here with a picture and I should have checked well enough if the link to my picture was working. I edited it just now and now it works. $\endgroup$
    – 123t
    Nov 13 '21 at 11:37
  • $\begingroup$ I am really sorry that you had to answer this question without the picture. My question sounds quite irrelevant to what I am actually looking for until you see it. So, your answer doesn't really address my confusion. Sorry to have wasted your time $\endgroup$
    – 123t
    Nov 13 '21 at 11:38
  • 1
    $\begingroup$ @123t my answer applies directly to the picture. The V of interest for finding the power lost in the cable is the voltage across the cable itself. With that V the formula works just fine. $\endgroup$
    – Dale
    Nov 13 '21 at 19:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.