# Power dissipated in the wires

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1. Jun 8, 2015

### skepticwulf

In a general homework question I often encounter terms like "a wire of resistance R is stretched, what factor the power dissipated changes?"
What exactly is this power dissipated ? Is it the one "wasted away" by heating up the wire due to the wire's resistance? or is it the power "used up" or transferred by the current conducted in the wire at certain voltage?

For instance in the example above if the length of the wire is doubled uniformly , the power dissipated is then decreased. So, when we use a longer wire did we decrease the power dissipation?

Do we have two power values for the same situation? P as I x V-power transferred- and P' as V^2/R-power dissipated- ?

2. Jun 8, 2015

### Simon Bridge

Yes to all of the above.
A load (i.e. a real wire) converts some of the electrical energy to a different form. This "uses up" the energy.

So if the power dissipated decreases, the wire is using less power.
Note: if the length of a wire were doubled uniformly, then the resistance is also doubled.
For the same voltage drop, the current is halved so power is halved.

The two power rules you list are the same rule writ two different ways.

Last edited: Jun 8, 2015
3. Jun 8, 2015

Thank you.

4. Jun 8, 2015

### CWatters

Yes. This is partly a result of the circuit that your wire is in. For example if the ends of the wire are connected to a voltage source the power dissipated in the wire is given by

P = V2/R

So if you make the wire longer increasing R then the power dissipated decrease.

However in many other circuits that's not what will happen. Consider a power transmission line to a city. The city needs a certain amount of power to arrive at the city at a constant voltage. The voltage drop in the wire is compensated for by increasing the voltage at the power station end.

Pcity = Icity * Vcity
or
Icity = Pcity/Vcity

The power dissipated in the wire is

Pwire = Icity2 * Rwire

Now if the resistance of the wire is increased the power dissipated in the wire is increased.

This is not inconsistent with your answer, its the result of the circuit being different.