# Power loss and Joule's law of heating

I really can't understand the power loss law. If we have a wire carrying a $15\:\mathrm{V}$ and $1\:\mathrm{ A}$ going into an inverter, giving $150\:\mathrm{V}$ and $0.1\:\mathrm{A}$, if we want to calculate the power loss as heat in wires, we use $P = I^2 R$ or simple $P=VI$, does that mean that all power is lost?

With respect to power loss concept, when we say that the power is dissipated (or lost as you call it) it means that power was dissipated (or spent) as something else which might be useful (as an example power dissipated in a perfect lamp where all power is converted into radiation) or not useful (example is portion of power lost in heating the motor of fan, nobody want that). So I hope it is clear now. The term lost is typically used to describe the power dissipated non-usefully.

In your example about inverter, the voltage and current you gave suggest that the power lost is zero. If you wanted to count the power lost in wires you have to include the wires resistance in your computation, which means the power (voltage and current ratings) you get after inversion is less than the power before inversion.

In your inverter example, the useful power is the inverted voltage and current, the non-useful power is the one dissipated in resistors as heat. The total power should be:

$$P_{in} = V_{in} I_{in} = V_{out}I_{out} + I_{out}^2R_{wires} = P_{out} + I_{out}^2 R_{wires}$$

I hope that made it clear

• it's perfect thank you, but just one last thing , the wires leading to the inverter from battery, they wont heat up? – user28324 Nov 28 '13 at 23:12
• i mean what about I^2(in) R wires – user28324 Nov 28 '13 at 23:27
• Of course they will heat up, I mentioned that in my last paragraph just before equation. The power lost as heat is represented by the last term in the right hand side of my equation, does that answer your question?? @user28324 – Gotaquestion Nov 29 '13 at 0:01