Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Energy Losses in Currents
Name:  Tony P.
Status: student
Age: 18
Location: N/A
Country: N/A
Date: 2000-2001


Question:
It was said that in a Transformer, in power station the voltages are stepped up so that little current flow through and therefore less heat loss..from the equation..P=VI..so when the current is reduced the voltsge is increased...however from what we know..voltage is the amount of energy converted by 1 coulomb of charge passing through a conductor...therefore, P=VI althogh the currrent is smaller but the V is greater compensating it so as tho get same output power..since same power the amount of energy used in 1 second is still the same....so what difference does it makes by steping up votage in power sation just to decrease current when when the amount of energy lost per second is still the same?


Replies:
You are absolutely right in that upping the voltage in a transformer just means that the same power is transmitted since the current is reduced proportionately and P = IV is not changed.

Ther advantage of upping the voltage comes when the power must be transmitted over a long distance. The wires in the transmission lines have resistance (assuming they are not superconducting) and the energy lost when transmitting power over a transmission line whose resistance is R is given by P(lost) = IIR (I squared R).

So if the voltage is increased by a factor of 10, the current needed to transmit the same power is reduced by a factor of 10, and the energy lost to the resistance of the wires is reduced a factor of 100. This is an important result of great practical value and explains why long distance transmission lines all use very high voltages -- 1 million volts is not uncommon.

Note that if the wires were superconducting so R = 0, NO energy would be lost regardless of the current and the expense of increasing and decreasing the voltage would be unnecessary.

Let me know if this is not clear or if you would like further information.

Best, Dick Plano...


There are two places that the energy can go .... The first is to the appliance that you want to operate. This is the desired use of the electricity. The second is to the power lines that carry the electricity. The energy that goes into the power lines is wasted in heating the power lines and the air around them.

It is true that changing the voltage does not change the power (or energy per unit time) delivered by the source. But it does change the ratio of the power delivered to the appliance vs the power lines. With a higher voltage, less power is lost to heating the power lines and thus more power is delivered to the appliance where it can perform useful work..

Greg Bradburn


The energy lost in resistive heating is not the same. It is I^2/R, and it decreases when I decreases.

Tim Mooney


In an world without friction or resistance, it makes no difference. In the real world, however, there is electrical resistance. As electrons move through the wire, they collide and cause the wire to heat up, resulting in conversion of some electrical power to heat power.

So, we need another equation, the one for the power loss due to resistance:

Pl = I*I*R (Power loss = I squared times R).

From this equation, it is immediately apparent that the greater the current, the more power will be lost due to resistance.

In essence, it is saying that resistors change some of the power into heat, so that some of the power will be lost as the electricity is changed.

As you have noted, P = VI is the equation for the electrical power, and current and voltage are interchangeable. So, given an amount of power that you want to transfer from one place to another, it becomes apparent that it the amount of power does not require any specific voltage or current. In essence, you can pick what you want.

The amount of power lost, however, does depend on the amount of current. The higher the current, the more power will be lost. So, in order to minimize power loss, the amount of current is minimized.

Since there is a large amount of power to transmit, if I is small, then V must be large.

Eric Tolman
Computer Scientist


Hi, Tony !!!

Good question !! But it is necessary to consider that a conductor has internal resistance that opposes to the flow of electrons. As a consequence of such internal resistance, you lose energy in the form of heat.

As you said, but in other words : difference of voltage between two points 1 and 2 is the difference of work done on an electrical charge to bring it from infinity to the point 1 and bring it to the point 2.

In the real world, you have to add a little more energy to overcome this loss. If there were no loss of energy as heat, than everything would be OK !!! But...there is. And the lesser the current, the lesser such a loss. And besides, you know that the power lost in this case is : P = R.I^2 (square of current). So, when you transport energy through wires, the higher the voltage the lesser the current, and lesser lost of energy as heat. Just take a look again in point where you say : "..since same power the amount of energy used in 1 second is still the same.... " The power consumed should be greater in 1 second in the real case.

Best regards

Alcir Grohmann
Beschaffung SAM



Click here to return to the Physics Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs

NEWTON AND ASK A SCIENTIST
Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory