# Does resistor power rating mean 'real' power?

Given a mixed capacitive-resistive circuit, I have a $$\36 \Omega\$$ resistor through which a current of $$\9.22 \angle 86° \text{mA}\$$ passes.

My AC theory is rusty, but wouldn't I do:

$$\P = I^2 R = 3.06 \angle 172° \text{mW}\$$

$$\P_{real} = P \cos 172° = -3.03 \text{mW}\$$

This is a negative value, which is probably wrong. Which figure should I be comparing against a resistor specsheet's power rating? (I realize nearly all resistors will accommodate 3mW; this is a question to straighten out my theory.)

# Edit

I'd totally forgotten about the conjugate rule. With the adjusted calculation, I get

$$\(9.22 \text{mA})^2 \cdot 36 \Omega = 3.06 \text{mW (peak)}\$$

$$\(6.52 \text{mA})^2 \cdot 36 \Omega = 1.53 \text{mW (RMS)}\$$

• Resistors don't create reactive power, so what else would it mean? – immibis Sep 24 '18 at 5:05

## 1 Answer

I have a 36 ohm resistor through which a current of 9.22mA @ 86º passes.

If the frequencies involved are not too high, the voltage associated with this current will also have a phase angle of 86º.

As a result, the power of the resistor will be 100% real power. There will be no reactive power. In order to have reactive power, you must have reactive elements: inductors or capacitors.

P = I^2 R = 3.06 mW @ 172º

The formula $$\P=I^2R\$$ isn't meant for phasor cases. For phasors, you should use

$$P=I^\star IR$$

where $$\I^\star\$$ is the complex conjugate of $$\I\$$.

In general, the power consumed by a component with phasor voltage $$\V\$$ and current $$\I\$$ is

$$S=VI^\star$$

giving complex power $$\S\$$, with $$\S=P+jQ\$$ where $$\P\$$ is real power and $$\Q\$$ is reactive power.

With this correction you will get a real result, as expected.

Which figure should I be comparing against a resistor specsheet's power rating?

Since the power used by a resistor is entirely real power, you need to calculate the real power and compare it to the rating.

If the frequency applied is very low, you may need to compare the peak power rather than the average power, since it could be possible for the resistor to burn out at the peak of the waveform.

• Awesome! I've fixed my calculation in the question. Hopefully what I've done for RMS conversion also makes sense. – Reinderien Sep 24 '18 at 4:23