# Homework Help: Power line transmission question

1. Jan 6, 2009

### 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.

2. Jan 6, 2009

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.

3. Jan 24, 2009

### Hoss

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

The variables:
Ps - 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
RL - the resistance of the transmission lines
Rc - the resistance of the consumer stuff
VL - the voltage drop across the transmission lines
Vc - the voltage drop across the consumer stuff

This is what i did,

Ps = V1*I1 = V2*I2

V2 = VL + Vc
Ps / I2 = I2*RL + I2*Rc
Ps = (I2^2)*RL + (I2^2)*Rc

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.

4. Jan 24, 2009

### mgb_phys

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