# Why does RMS power produce the same heat as the equivalent DC power?

1. Jan 27, 2017

### tjosan

Hello.

I was browsing online to answer the question to why RMS voltage is used as opposed to the average voltage (and by average I mean over half a period, roughly 0.64V_peak), and the explanation I found was that the RMS voltage would produce the same heat as equivalent DC through a resistor. But how do you prove this analytically?

2. Jan 27, 2017

### cnh1995

Solve
Heat H= ∫[V(t)]2dt/R between 0 to t, where V(t) is the voltage as a function of time. For example, you can take V(t)=Vmsin(ωt).

3. Jan 28, 2017

### tjosan

Thank you. Using RMS felt so abstract, but now when I understand why it doesn't :)

Edit: shouldn't there be a square root also?

Last edited: Jan 28, 2017
4. Jan 28, 2017

### cnh1995

No.
H=V2t/R.

5. Jan 28, 2017

### LvW

For my opinion, you must not ask WHY both voltages "produce the same heat". It`s just the other way round:
We want that in both cases the same power (heat) is produced - and this equality leads to the definition of the rms value.

6. Jan 28, 2017

### Averagesupernova

If a person truly understands what RMS is instead of just multiplying peak volts by .707 I think it makes perfect sense.

7. Jan 28, 2017

### Staff: Mentor

The 0.707 is valid only for sinusoidal signals. I think it is interesting that RMS is so general that it applies to non-sinusoidal signals, non-repetitive signals, and even to non-integrable signals such as white noise. So even if you can't find an analytic expression, you could take samples, treat each sample as a DC value constant for a short time step, sum up the $II^2 R$ or $V^2 /R$ values, and find that the sum is the same as the RMS.

So RMS is a mathematical "trick" but one which is valid and very useful.

8. Jan 28, 2017

### tjosan

Okay I attempted to solve the integral. If I want to integrate over the one period the limits are $t=0$ and $t=2pi/\omega$.

$\int_0^{2\pi/\omega} H dt=\frac{1}{R}\int_0^{2\pi/\omega} V_p^2 sin(\omega t)^2 dt = \frac{V_p^2}{R}\int_0^{2\pi/\omega} sin(\omega t)^2 dt$
RHS integration:
$\int_0^{2\pi/\omega} H dt=H \frac{2\pi}{\omega}$
LHS integration:
$\frac{V_p^2}{R}\int_0^{2\pi/\omega} sin(\omega t)^2 dt = \frac{\pi V_p^2}{R\omega}$
Which yields:
$H \frac{2\pi}{\omega}=\frac{\pi V_p^2}{R\omega} \Rightarrow H =\frac{V_p^2}{2R}$

So what did I do wrong? There should be a square root in there.

Edit: I just realized that the voltage should be squared in the formula, so this is correct.

9. Jan 28, 2017

### sophiecentaur

But what if the waveform is not symmetrical about 0V or even not repeated? "Half a cycle" demands certain particular conditions. You need to integrate 'something' over a long period of time. That something has to give a positive answer over the whole integration time.
The power delivered is the time integral of VI i.e. the total of each infinitesimal packet of power over the time of the integration. It's as simple as that and, if the load is resistive, I = V/R so you can re arrange that to give the time integral of I2R or V2/R
That's true for any waveform. If the load is not purely resistive then the Energy delivered is still ∫VIdt .
The 0.707 figure is, unfortunately, the first thing that they tell you and that is the confusing bit. It only applies to a sinusoid and sinusoids are actually hard to find in real power engineering. Near-square waves are pretty common, actually and then the factor is 1.

10. Feb 1, 2017

### Joseph M. Zias

Be careful not to use the term RMS power - there is no such quantity. There is average power that as seen can be calculated for sinusoids using the traditional RMS voltage and RMS current. For other waveforms or pulses you will have to do the integral and that will lead to other form factors. You can find some good information on those techniques in old SCR (silicon controlled rectifier) manuals, e.g. GE, International Rectifier, Westinghouse. Power rectifier and transistor manuals would also have that information.

11. Feb 1, 2017

### sophiecentaur

You are absolutely right but it is not uncommon to find a switch position on equipment that has RMS in dB and that can be confusing. You have to be careful and understand what they mean. RMS is a term that's frequently used in HiFi Jargon about Power and it is a sort of attempt at yet another way of specifying an amp performance to make it seem bigger than it really is.

12. Feb 21, 2017

### olivermsun

So it seems there is such a term as RMS power (it even appears in CEA specs, according to Wikipedia!), but it comes from loose/bad terminology.

On the other hand, I've always thought RMS power ratings looked smaller than a "peak" power ratings, which were really inflated...

13. Feb 21, 2017

### Baluncore

I can represent a signal by a fourier series.
Is the RMS voltage of that signal equal to the DC component + √2 * Σ( independent sinusoidal fourier component amplitudes)?

14. Feb 22, 2017

### sophiecentaur

It has some use, as a term, as long as it comes with a health warning. An audio amplifier may be capable of producing high peak powers, briefly but limited due to thermal dissipation or power supply capacity. The same amplifier will also have a 'continuous' rating for an unvarying sinusoid. That measure is a bit pessimistic / conservative and certainly doesn't look good as a selling feature. If you plot the short term power over time, a normal audio programme will produce a variation in output power. To give a more honest idea of what the amplifier can achieve than quoting peak power, the term RMS Power was invented as an 'average' figure. But it's not necessary to use mean squares because the Power is always a positive quantity and it would probably be better to just quote the Mean Programme Power (possibly with some weighting factor over the frequency range). But HiFi is full of BS. Few people really want to listen at the levels that the HiFi salesmen would lead them to believe they need.