### Re: Why is high voltage more efficient at transferring electricity?

Date: Mon Apr 19 17:21:15 1999
Area of science: Physics
ID: 924429150.Ph
Message:
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Ben,

Thanks for the question.  The formula that gets quoted to you is probably I^2 R
= P(line).  The context for this equation is a part of an electrical circuit,
like a transmission line, which has a current denoted I flowing through it.  The
wire in the line has a resistance R.  What the equation is saying is that the
power lost to the resistance of the wire is I multiplied by itself (squared) and
then multiplied by R.

This equation has some implications.  Obviously, if the resistance is big, then
you will lose a large amount of power to the line; it heats up.  The I^2 part
means that having a large quantity of current flow through the line is really
bad for power loss, doubling the current makes the power loss 4 times as big.
To minimize power loss you want the current to be as low as practical.

Now on to your question.  The power that is delivered by a transmission line is
described by the equation VI = P(delivered).  Here V is the voltage of the line
and I is the current flowing throuhgh the line and into whatever is using the
power delivered.  For a fixed amount of power that is needed, V and I can take
on many pairs of values, just as long as the their product VI is the power
needed.

Remember that to keep the power loss in the line low it is best to make the
current I small.  When we do this it means that the voltage V has to be large so
that the delivered power is the needed amount.

We don't have to use high voltages to transmit electrical power.  We could go
with low voltages and high currents.  However, this would mean that the power
loss in the lines would be large unless the resistance was small.  This would
then mean that we would have to use very thick and expensive wires to carry
electricity, so that the lines wouldn't heat up and melt.

transmission.

Regards,

Everett Rubel

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