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

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.

I hope that this answers your question about high voltage electrical power 
transmission.

Regards,

Everett Rubel