Power Transmission: How High Voltage/Low Current Works

[MotoPayton]

From what I have read electrical power is transferred at high voltage and low current in order to reduce energy loss in the form of i^2R. High voltage at low current is the same power as low voltage at high current.

My question is how do they do it. If they apply a higher voltage and the resistance in the transmission wire is not changing how is the current made smaller.

From V=IR applying a higher voltage should increase the current and the power.

I'm confused.

 

[rock.freak667]

Usually they would step down the voltages/currents using a transformer.

 

[russ_watters]

Ultimately, it is the load that pulls however much it needs.

 

[sophiecentaur]

From what I have read electrical power is transferred at high voltages in order to reduce energy loss in the form of i^2R. High voltage at low current is the same power as low voltage at high current.

My question is how do they do it. If they apply a higher voltage and the resistance in the transmission wire is not changing how is the current made smaller.

From V=IR applying a higher voltage should increase the current and the power.

I'm confused.

They choose a very high voltage to transmit the power - so the I (in the VI=P) can be low. Most transmission uses AC so, at the consumer end, a transformer will reduce the volts to a manageable value for the short distance from transformer to consumer.

 

[MotoPayton]

So before the electrical power is transferred over a large distance a transformer will be used to to step up the voltage. This in turn will decease the current since the power going into the transformer is the same going out assuming no energy loss.

So now my question leads to... How can a transformer step up the voltage to a high amount yet only allow a small current. Because form the external circuit point of view all it sees is a high voltage and should obtain a higher current consistent with V=IR.

I guess the answer lies in the back EMF of the transformer?

 

[sophiecentaur]

Yes - the current that the load is taking (from the secondary of the transformer) produces a back emf and the high voltage supply 'sees' a much higher resistance than is actually connected to the secondary. Stepping down the volts by a factor of ten has the effect of increasing the apparent resistance by a factor of 100 - thus keeping VI the same. It's called Impedance Transformation.

 

[sophiecentaur]

Also see "en.wikipedia.org/wiki/Solar_sail" .

 

[taterz]

The proper equation to analyzer this is Ohm's law for power loss in a resistor:

P = i*i*r

 

[sophiecentaur]

The proper equation to analyzer this is Ohm's law for power loss in a resistor:

P = i*i*r

To analyse what? We already know about IsquaredR losses in the supply cable. The issue is how to reduce the I in the cable without reducing the amount of Power transmitted.

 

[taterz]

Power loss in the cable is given by P(loss)=i*i*r. The power delivered to the load P(load) = v*i where v is the voltage across the load. So for any specific P(load) there is less loss transmitting to the load if "v" is made larger and "i" is made smaller.