

Energy Losses in Currents
Name: Tony P.
Status: student
Age: 18
Location: N/A
Country: N/A
Date: 20002001
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
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Update: June 2012

