Question about P=V*I, P=R*I^2 and V=R*I (electricity)

[zooflute]

This is my first post, and hopefully it's in the right section (my apologies if not!)...I have posted this on Answers, but there weren't any helpful responses there, so I though this might be a better crowd to ask.

Ok, we were taught in our science (electricity) class that in order to reduce power loss (P) on high voltage power lines, they increase the voltage (V) and reduce the current (I), which makes P go down by the square of I. So with this formula:

1) If V goes up by 10, then I goes down by 10, since P=VI
So if I goes down by 10, then P goes down by 100, since P=R*I^2

But this doesn't make sense when you put it into Ohm's law:

2) If V goes up 10 and R is constant, then I goes up 10. Using V=I*R
So if I goes up by 10 then P goes up by 100, since P=R*I^2

Also, in 1), why are the two "P"s different? (in P=V*R, P should be constant, but in the P=R*I^2 part, P goes up 100) Are they different types of P?

Any insight on this we be really appreciated!

 

[gneill]

First, you should distinguish between the power being delivered to the load
and power being 'wasted' in the power distribution line.

The power delivered to the load, call it Pd, is V*I at the receiving end.
The power lost in transit, your P, is the I^2*R loss on the line.

The idea is to reduce the I^2*R loss on the transmission line for a given constant
power Pd being delivered to the load. The I^2*R loss on the line is accompanied
by an end-to-end voltage loss on the line: DV = I*R. Note that DV is not V.

A careful analysis will reveal that it is a tad more complicated, because the voltage
delivered at the receiving end is actually V - DV, so the power delivered to the load
will really be (V - DV)*I. We want DV to be small compared to V.

 

[stevenb]

It's nice to see a good question and a good answer come from two first time posters.

 

[Stonebridge]

Good explanation from gneill
This diagram may help to see what's happening.
The figures are just chosen for illustration.
Cable has 10 ohm resistance, supply has 1000W power.
It could be supplied at, for example, 1000V x 1A (top) or 500V x 2A (bottom)

 

[zooflute]

Wow thank you both so much! Your answers put together helped me understand perfectly.
Good stuff ^^