AC vs DC

[Pyro]

Hi all, I'm hoping somebody can help me to understand something regarding the level of energy for an equivalent circuit. Does an applied voltage of say 20VAC(peak to peak)contain less energy than 20VDC?

Would the RMS value of the AC be closer to the energy in a DC value?

Thanks in advance

[ammlione]

The RMS value is the equivalent DC value,
ie: if you apply 240VAC RMS to a resistor it will generate the same heat as if you applied 240VDC.

So, your problem now is, Is the RMS value bigger or smaller than the peak to peak voltage?
(google is your friend)
Cheers

[bob_m_54]

Have a read here

[lsemmens]

To determine the RMS (Root Mean Square) of an AC waveform all you need do is multiply the AC Peak by .7071 which will then allow you to determine the true power of the circuit.

[trash]

There are a couple of really easy ways to remember this. It all has to do with calculus.
Remember that subject in high school maths that you hated because your teacher couldn't make you understand it, it was boring, abstract and had no apparent uses.
Well this is where it will finally make sense and you'll realise it is actually very easy and useful.

OK... so Isemmens has mentioned the one everybody remembers ..... 0.707. I remember it because of the plane. Inverse is 1.414. (1/0.707=1.414 simples)
So a waveform with 100V peak is 70.7 volts RMS.

If you measure the AC from a power point, your multimeter will display about 240V. If however you measure it with a CRO (I don't recommend this) then the CRO will show you a sine wave with a peak of about 340V.

If you're studying for a Ham Radio licence, I can tell you that there is a trick RMS question in there.
100VAC_RMS is equivalent to 100V DC.

The question will ask something like ....
What is the equivalent DC voltage of 100V_RMS
A>141.4V
B>100V
C>70.7V
D>63.7V

You're natural response is to fall for the 1.414 or 0.707 answers.

So where does calculus come into this?
Well that is the really simple part. It is nothing but the area of the sine wave under the curve. The boring stuff ∫sin(x) dt.

This 'area under the curve' works for all waveforms. A really simple example is a square wave.
If we have a square wave with 1V peak and a frequency of 1Hz. Then for 500mS it is 1V and 500mS it is 0V.
(1V x 0.5s) + (0V x 0.5s) = 1V x 0.5s = 0.5Vrms

We can do this for a sawtooth wave with 1Vp and 1Hz frequency. /|/|/|/|
1V x 1s x 0.5 = 0.5Vrms (If we cut a square or rectangle down the diagonal, it is always half the area.)

We can also do a hyperbolic staircase example ... 0V 1V 2V 4V 8V with a frequency of 1Hz. (each step is 200mS)

(0V x 0.2s) + (1V x 0.2s) + (2V x 0.2s) +(4V x 0.2s) + (8V x 0.2s) = 0 + 0.2 + 0.4 + 0.8 + 1.6 = 3Vrms

Calculus does the same thing ... but does it the hard (easy) way. But it is just looking for the area under the waveform whatever shape it is.
∫Xv dt (t=0.2) [if X=2v] = 2²/2 = 2 x 0.2s = 0.4 etc


Oh... don't be fooled into thinking frequency has anything to do with it.
If we use a 1Hz square wave and a 2Hz square wave ...
1V x 0.5s = 0.5Vrms This is a shortcut ... the real formula is (V x t) x f where t is the partial period we are measuring f is the frequency in Hz
It is (1V x 0.5s) x 1 = 0.5Vrms

So
(1V x 0.25s) x 2 = 0.5Vrms

It is the shape of the wave that matters.
A pulse of 0.25s with a frequency of 1Hz.. (1V x 0.25) x 1 = 0.25Vrms
A pulse of 0.1s at 2Hz = (1V x 0.1) x 2 = 0.2Vrms

[trash]

Welcome to Quantum Physics. Where AC is DC !

As long as you have time, then DC is always AC. If you have time, then there was a point where the power was turned on and DC changed from 0V to something else. Hence it is always AC.

If time is quantised, then time comes in discrete chunks. Between these chunks, there is no time. If the voltage is changing (dv/dt aka AC) then it cannot change between quantised periods. It must do so in steps.
If the eletromagnetic force is quantised (and it is) then the voltage too can only rise and fall in steps.

Therefore, AC is nothing but DC when you look at it close enough.

[loopyloo]

I'm confused ... is this what you guys are talkin' about ?

[kolithawick]

thanks trash it remembered me my good old school days.

[Pyro]

But where does the alternating current part for into that quantum model trash?

[trash]

It's all about whether something is smooth or stepped.

Think of it like this. Vdt = Vt2 - Vt1 This little formula describes a voltage changing with time Vdt = an AC voltage.
Vt1 and Vt2 are DC voltages, they aren't changing. (dt=0, time is fixed)

The quantised world is made up of very small DC steps. It's not smooth at the finest scales.
The closer you look at something in the quantum world, the more it looks like something it is not.

[Pyro]

Slightly off topic Slash, but in the quantum world, if you looked at an object in motion, for a very, very brief moment in time, would it appear stationary?

[Skepticist]

This could put me in a coma almost as quickly as my early introductions into calculus
But once you have a real-world application for it, well that's very different.

RMS - the square root of the mean of squares of instantaneous values distributed over the waveform at equal time intervals (sounds dreadful but very useful for complex waveforms) - twas a nightmare in the days of slide rules and pencils but now a CPU can do it in uSeconds
For a sinewave all you need to do is divide the peak value by the square root of 2

[lsemmens]

Which is my point, Skepticist, .7071 is (roughly) equal to sqrt(2)