# Resistor power rating in an AC circuit

Looking at this question and answer:

Source: Practical Electronics for Inventors, Page 91, Example 5

Shouldn't we use the peak voltage instead of the RMS voltage to calculate the minimum power rating? Over time, when using the 400 ohm resistor, the avg power would be 1/8 watt, but it will be oscillating between 0 and 1/4, exceeding the power rating, so shouldn't we use an 800 ohm resistor so that the power oscillates between 0 and 1/8 instead?

Is the power rating dependent on the maximum power through the resistor or the avg power? Does the oscillation give the resistor a chance to "cool down"?

• Mohamed Hatem - Welcome. (a) Thanks for including the required reference for the copied material. FYI you put it in the usually-invisible HTML Alt text part of the image syntax. It needs to be visible, so I copied it to the normal text that you see. (b) Please see the main site rules in the tour & help center as they differ from typical forums. Commented Apr 17 at 2:11
• Is the thermal time constant of the resistor shorter than the line frequency cycle? Commented Apr 17 at 12:41

In principle, your consideration is correct, that during the peaks the resistor dissipates twice the power rating, thus exceeding its specs.

But the reason for this power rating is heat dissipation. The frequency is given as 1000 Hz, so the power dissipation varies between 0 and 0.25 W and back to 0 within 0.5 ms. This is completely averaged out by the heat capacity of the resistor, so its temperature will be effectively constant, and correspond to the average ("RMS") power.

If we were talking about e.g. 0.01 Hz, then one power cycle would last 50 seconds, and then the resistor would probably have enough time to overheat during the power peaks.

You are correct, there are many ways of calculating what is correct and what isn't correct.

It depends on your resistor, and the data sheet of the resistor will tell in which way the power handling capability is defined.

For example you can apply DC and the power limit is as specified.

You can also apply a pulse of 4x the rated power voltage for quarter of the time, so the average is the same, but your power still momentarily was 4x the rated power handling.

Depending on how slow or fast the pulse was, the resistor might handle it without damage or they will damage.

Some resistors are especially designed and rated for handling high pulsed power peaks as long as they are within limits.

So what you saw on the website is just calculating the RMS average power dissipation. It may or may not be suitable way to calculate it in all cases, depending on the resistor. To be fair, the frequency is 1000 Hz which makes the peak to be very short, in comparison to mains frequency of 50 or 60 Hz, so you might get away in this case by just calculating the RMS power, but same resistor might damage when used at 10 Hz.

How you need to calculate it depends on the resistor. It will be safer if you simply use the peak power as the limit.

But, these calculations also do not consider that the power handling capability of a resistor depends on it's temperature too, so even if the rated power handling is listed at 25 °C, it does not apply when resistor has heated itself to 100 °C.

The example seems to assume the temperature does not need to be considered, even though in real life it does.

The RMS voltage value of a waveform (sine, triangle, some random voltage) is the voltage that will cause the same power dissipation in a resistor as a DC voltage that is equal to the RMS voltage.

Thus, the 20 Vpp sine wave will have an RMS voltage of $$\ {Vp \; \over \sqrt{2}} = {{20 \over 2} \over\sqrt{2}} = 7.07 \; V\$$.

• Yes the DC and RMS values are equal. But for a sine wave, the instantaneous power dissipation will exceed the rated value for 25% of the cycle and peaks at 2x the rated value. Commented Apr 17 at 10:04

Well, notice that the average power in a circuit is given by:

$$\overline{\text{P}}=\lim_{\text{n}\space\to\space\infty}\frac{\displaystyle1}{\displaystyle\text{n}}\int\limits_0^\text{n}\frac{\displaystyle\left(\hat{\text{u}}_\text{i}\sin\left(\omega t+\varphi\right)\right)^2}{\displaystyle\text{R}}\space\text{d}t=\frac{\displaystyle\hat{\text{u}}_\text{i}}{\displaystyle2\text{R}}\tag1$$

Now, I'll let you solve:

$$\overline{\text{P}}=\frac{1}{8}=\frac{\displaystyle\frac{20}{2}}{\displaystyle2\text{R}}\space\Longleftrightarrow\space\text{R}=\dots\space\Omega\tag2$$