Do longer wires consume more power?
Power can be expressed as \$P = IV\$. We also know that Ohms law says \$V = IR\$. Putting these two together, we get the expression:

$$P = \frac{V^2}{R}$$

In your case you have a fixed voltage source - so \$V=12\mathrm{V}\$. Put that in to the equation we get:

$$P = \frac{12^2}{R} = \frac{144}{R} = \frac{144}{R_{wire}+R_{load}}$$

As your wire gets longer, the resistance of it increases. As a result, the total power dissipated in the entire circuit is going to decrease. The power dissipated in the wire itself may go up, or it may go down.



How the current is affected will depend on what the circuit is. For example if you have a simple resistive load, then from ohms law we can find out how the current will be affected:

$$I = \frac{V}{R} = \frac{12}{R_{wire} + R_{load}}$$

If you increase the wire resistance, the total current will decrease. Depending on the ratio between wire resistance and load resistance, the power dissipated in the wire itself may go up, or it may go down.



However, if your load was, say, a constant current load, then things get a little more tricky. So long as there is enough voltage across the load to maintain the constant current, then we can say for sure that the power dissipated in the wire will increase. Going back to the first two equations, we can also derive that power is:

$$P = I^2R$$

So if the current stays the same, but the resistance of the wire is increased, then the power dissipated in the wire is going to be larger.