Power line transmission question

[maccha]

Okay I'm just very confused by a statement in my textbook and could use some clarification. It says that power is transmitted at high voltages in order to minimize heat energy loss. It then says "it must be noted that this is because the higher the voltage is, the lower the current". Well doesn't this contradict Ohm's law, that I= V/ R? From that equation wouldn't it seem that the higher the voltage, the higher the current? I'm confused.

 

[tquad]

This relationship has to be considered for a fixed level of power (P), since that is what the electric company seeks to provide the customer. P = EI, hence the same P in watts can be obtained by:
(1 amp x 10 volts) = (5 amps x 2 volts) = 10 watts
A current of 5 amps requires a larger diameter wire than does 1 amp for a fixed amount of resistance and heat loss.

When viewed within the form of Ohm's Law E = IR when R is constant then yes I increases with E. However, when the voltage is fixed, the required increased transmitted current through the transmission line when demand increases, then there will be more work required with each of the customers applications and R increases as well:
(1.1 amp / 0.01 Ohms = 110 volts) versus
(5.5 amps / 0.05 Ohms = 110 volts)
So the electric company gates the flow of amperage to maintain a fixed voltage.

The confusion comes from not considering the overall context of what needs to be determined. This is not always explicitly stated when questions are brought up, but assumes an understanding of the broader context of the variables involved. When the obvious doesn't fit with what is stated or proposed, then some other parameters are being overlooked.

 

[Hoss]

Ok i have a question regarding the same thing. Consider this:

The variables:
P_s - the power supplied by the generator
V₁ - the voltage in the primary coil of the transformer
V₂ - the voltage in the secondary coil of the transformer
I₁ - the current in the primary coil
I₂ - the current in the secondary coil
R_L - the resistance of the transmission lines
R_c - the resistance of the consumer stuff
V_L - the voltage drop across the transmission lines
V_c - the voltage drop across the consumer stuff


This is what i did,

P_s = V₁*I₁ = V₂*I₂

V₂ = V_L + V_c
P_s / I₂ = I₂*R_L + I₂*R_c
P_s = (I₂^2)*R_L + (I₂^2)*R_c


So what will happen to the power supplied if the turns of the secondary coil are increased? Does it remain constant or does it decrease because I₂ decreases? And how does increasing the voltage reduce power losses relative to the total power supplied? Don't the power loss in the lines and the power dissipated at the consumers both decrease? I think I'm doing something wrong here but i can't figure out what. I've always had problems with this power transmission thing. I keep getting weird things when i play around with substitution with the equations.

 

[mgb_phys]

From that equation wouldn't it seem that the higher the voltage, the higher the current? I'm confused.

It confuses a lot of people!
In V=IR , V is the voltage difference ACROSS the resistor.

That's why we use high voltages, remember power = I*V so we can trade voltage against current to send the same power.
If the transmission line has a big R, then if we use a large I then all their is a large voltage difference across the resistor and so no volts left for the consumer.
If instead we use large V and small I then there is only a small voltage drop and lots of volts (and power) left at the consumer end