# Power dissipation in High Voltage Cables

I was doing the following physics problem in physics class:

You have two dimensionally identical pieces of metal, one made from aluminium the other made from iron. It is given to us that aluminium has a lower resistively then iron. Which metal glows first when they are connected in parallel to a battery. What about if they are connected in series?

If they are connected in parallel the voltage across them is the same however the current going thru the aluminum by ohms law is higher. Therefore the aluminum glows first.

When they are connected in series the current thru them is the same therefore they should glow at the same time. But then I remembered that power dissipated across a resistor is $P=VI$ and the voltage drop across the iron is greater - so the iron glows first.

Then I tried to look for the source in my mind for the first reasoning - hat heat losses depend on the current and not the voltage. And this is the point where the high voltage transmission lines come in.

I was taught that we transmit electricity at a high voltage (and then transform it down for home usage) to allow for a lower current and therefore decreases power dissipated. But now that I think about it increasing voltage to decrease current wouldn't work as what we same in a lower current we lose by a higher voltage according to $P=VI$

What is going on here, can someone please explain?

But recall that power dissipated

$P= VI$

is also , from Ohm's law, expressible as

$P = I^2 R$

So the dependency of power dissipated is linear in voltage, but quadratic in current, given the same resistance.

Also remember that the voltage supplied from the power station, and the voltage drop across the transmission line - which is what is important in power loss- , are not the same voltage. The former is considerably larger than the latter.

To see why, consider supplying a fixed amount of power at the end of a transmission line with a supply voltage $V_s$ and supply current $I$.

You would use the first equation, $P= V_s I$, to compute that power.

But the voltage drop across the transmission wire is $V_{drop} = IR$, which is less than the supply voltage. They are different quantities. The power dissipation is only quadratic in $V_{drop}$, not $V_s$.

Of course, if we are talking about transmission lines, the above is a vast oversimplification, since transmission lines carry AC power. For a short enough transmission line , it can be modeled as a resistance and inductance in series. For longer transmission lines, capacitive effects come into play. But still the qualitative picture of current dominating the loss over voltage

There is a report from Purdue University at this link that covers transmission line power loss in considerably more detail than there is room for here.

• But we can also express it $P=V^2/R$ also making it quadratic in voltage! In the end the current also depends on the voltage. Nov 2, 2014 at 2:33
• The supply voltage and the voltage drop across the transmission line are not the same thing- there are other sources of resistance in the power circuit. Just as if you connected two resistors in a series circuit with a 12-volt battery. Nov 2, 2014 at 2:41