# Heat dissipated by resistor formula question

The heat dissipated by a resistor is given by the formula $$I^2 R$$

Any circuit portion has an equivalent resistance. Does that mean that the heat dissipated by any circuit portion, doesn't matter what it is, can be calculated using that formula?

If not, please explain why and how does it correlate to the equivalent resistance.

If yes, then the following doesn't make sense:

If an electric motor running with 230V AC draws 100 W: the current is: $$I = P/V = 100/230=0.43 A$$ the resistance is: $$R = V/I = 230 / 0.43 = 529Ohms$$ Therefore, the heat dissipated by it is the same as its wattage: $$H = I^2*R = 0.43^2*529 = 100 W$$ If it dissipates all it consumes as heat, then all the torque it produces would be "free energy", which, obviously, doesn't make sense.

Clearly I'm missing something here. Because either answer doesn't make sense to me and one of them must be correct because they are the opposite of each other.

• The resistance is not $V/I$, as most of the voltage is due to the back-EMF and not the resistance of the wire. – mike stone May 30 at 21:07
• An electric motor (or a generator) is not "just a resistor". To understand how it works you need to consider how the magnetic fields interact between the rotating and static parts of the motor. You can't even analyze the circuit using the methods you use in a first course on "AC circuit analysis" and including the inductance of the coils in the motor, because that ignores the key fact that the rotor and stator are moving relative to each other. – alephzero May 30 at 21:59

## 1 Answer

The heat dissipated by a resistor is given by the formula $$I^2R$$

No it's not. $$I^2R$$ represents the power delivered to a resistance by a DC source. DC refers to current that is constant in direction. If all of the power delivered to the resistor is stored as internal energy and transferred out of the system by heat, then the energy transferred to the surroundings is $$Q = I^2R\Delta t$$ where $$\Delta t$$ is the time interval for which the power has been delivered.

$$H = I^2R = 100W$$

Heat has units of joules, not watts, leading to the conclusion that the formula is incorrect. Moreover, the current delivered is AC, so it's direction varies periodically and this equation cannot be directly applied to obtain the power delivered either. The motor can also not be modelled as a simple resistor, hence the mistakes.

• There is no requirement for a steady state when using this formula. The instantaneous power and current will be given by this formula just fine. There is a subtlety if you want to use the averaged quantities, which you normally do when dealing with AC circuits, as the square of the average is not equal to the average of the square. To get the power you need to use the average squared power on the right hand side of the formula and so $I$ must be the root mean square power. – By Symmetry May 31 at 8:55
• @BySymmetry Thanks for pointing the errors out, I made appropriate corrections. – Cross May 31 at 9:14