# Power in electrical circuits

Why does $P=i^2R$ apply only to transfer of electrical potential energy to thermal energy in a device with resistance while $P=iV$ apply to electrical energy transfers of all kinds?

Why does P=i^2R apply only to transfer of electrical potential energy to thermal energy in a device with resistance

How would you expect a formula involving a term "R" to apply to something that doesn't have any characteristic called "R"?

while $P=iV$ apply to electrical energy transfers of all kinds?

This formula applies to all electrical devices (or more correctly, to any branch in a lumped circuit).

The alternate formula involving $R$ is derived from this one, based on Ohm's Law

$$V=iR.$$

If you substitute this into the power formula $P=iV$ you can get either

$$P=i(iR)$$

or

$$P = \left(\frac{V}{R}\right)V$$

Both of which simplify to well-known formulas for the power consumption of a linear resistor ($P=i^2R$ or $P=V^2/R$).

• If we connect the two terminals of a cell with an ideal conducting wire, to where is the the energy transferred to ( since according to $P=iV$ there is energy transfer) ? Jul 23, 2017 at 5:15
• @GauthamShankar, 1. There is no such thing as a perfect conducting wire, so some power will be converted to heat in the wire. 2. There is no such thing as a perfect voltage source, so some power will be converted to heat in the cell itself. Practically speaking, the majority of the heat will typically be generated in the cell. Jul 23, 2017 at 5:55
• If you had a perfect voltage source, and want to think about what happens if you get close to an ideal wire, use the form $P=V^2/R$ and take the limit as $R\to 0$. Jul 23, 2017 at 5:57

Remember the definition of voltage as energy change upon moving a charge. So you know the energy changes by QV for every charge that moves across. That gives a rate IV. It didn't have to be an Ohms law resistor in this derivation.