Solving for Power Loss in a Transmission Line: (P-P')=(P^2)R/(V^2)

[Fisicks]

Homework Statement

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A power plant can produce electricity at fixed power P but the operator is free to choose the voltage. The electricity is carried with a current I through a transmission line ( resistance R) from the plant to the user where the user receives power P'. Show the reduction in power (P-P') is equal to (P^2)R/(V^2)

Homework Equations

P=V^2/R

The Attempt at a Solution

I seem to be missing some key idea because when I mess around with (P-P')=(P^2)R/(V^2)
I get that P' is zero which is obviously wrong. Is it not true that R/V^2 = 1/P?

 

[gneill]

I seem to be missing some key idea because when I mess around with (P-P')=(P^2)R/(V^2)
I get that P' is zero which is obviously wrong. Is it not true that R/V^2 = 1/P?

That is true if you're interested in the power being developed in the resistance R. In this case it's the power delivered to some (unspecified) load that we're interested in.

The power plant produces power P = I*V. It arrives at the far end with the same current, but at a reduced voltage due to the voltage drop across the transmission line resistance R. So the power available for delivery to the load is P' = I*V'. What's an expression for P' in terms of the current and resistance R?

 

[Fisicks]

I still don't quite understand but here it goes.

The only thing that has changed is V, so P'=I*V'=(I^2)*R
P-P'=I*V - (I^2)R=I(V-IR)

 

[gneill]

P' = I*V'. And V' is not V, since some voltage was dropped on the transmission line. How much was dropped? (Hint: you have the current I which remains the same, and the resistance R of the transmission line).

 

[Fisicks]

Ok i understand now. The voltage drop is IR, so P-P'=IV=I(IR)=(P/V)(P/V)R